Category: Class 11

These MP Board Class 11th Biology Notes for Chapter 7 Structural Organisation in Animals help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 7 Structural Organisation in Animals

→ The term morphology refers to the study of external structure. Earthworm is an annelid that has segmented body and looks like a snake. It has about 100 to 120 segments with metameric segmentation.

→ The mouth is situated in the first segment. A single median female genital pore opens on the ventral side of the 14th segment. A pair of male genital pore is present on the ventral side of the 18th segment. A prominent band encircles 14th, 15th and 16th segments. This is known as clitellum.

→ Cockroach has the characteristic jointed legs and is nocturnal in its habit. The body is segmented externally and divisible into a number of segments under head, thorax and abdomen.

→ The head is somewhat pear shaped which ariculated with the thorax by flexible neck. It bears compound eyes, antennae, mouth parts. Thorax is three segmented bearing jointed appendages. There are ten segments in abdomen.

→ Alimentary canal is well developed and is divisible into foregut, midgut and hind gut, Malpighian tubules are present at the junction of fore and mid gut and help in excretion. Respiration occurs by trachea. The blood vascular system is of open type. Fertilization is internal.

→ Frog is an amphibian that lives in water or on land near water. Its skin is soft, smooth and moist. The male frog bears copulatory pads and well developed vocal sac on the ventral surface of body cavity.

→ Body is divisible into head and trunk. It can respire through skin in water and through lungs on land. Circulatory system is closed with single circulation. The male reproductive organ is a pair of testes. The female reproductive organ is a pair of ovaries.

→ Rat is a mammal and its body is divided into head, neck, trunk and tail. Integument is made up of epidermis, dermis and their derivatives. Closed and double circulatory system is present.

→ Heart is four chambered. Six to eight young ones are produced in a litter.

→ Glucose is the main nutrient in the blood.

→ The total volume of E.C.F. (extracellular fluid) in adult human being is about 15 litres i.e., about 45% of the total body water.

→ Spongy bone occurs in the deeper central parts of bones. It has no concentric organization like haversian system. It consists of a network of many fine irregular bony plates called trabeculae.

→ In whale thick layer of adipose tissue is called blubber.

→ Ligament is a strong band of elastin connecting the two bones at a joint and holds them in position preventing dislocation.

→ Tendon is a collagenous connective tissue which connects a muscle to a bone or cartilage.

→ The average lifespan of R.B.Cs. in human beings is 120 days.

→ Increase in the number of R.B.Cs. in our body is called as polycythemia.

→ The membrane which covers the muscle fibres is called sarcolemma.

→ Largest muscle of human body is buttock muscles (Gluteus maximus).

→ Smallest muscle of human body is stapedius. .

→ Riger mortis : The rigidity and non-elasticity of muscles after death. It is first seen in jaw muscles.

→ Myology : Study of muscles.

→ The process of the formation of blood corpuscles is known as haemopoiesis.

→ Decomposition of blood cells is called as haemolysis.

→ Collagen and elastin is the fibrous protein found in connective tissue.

→ The amount of haemoglobin is 15 gm per 100 ml of blood.

→ Abnormal decrease in the number of R.B.Cs. in the blood is called as anaemia.

→ Excessive stretching of ligaments is called sprain.

→ Haversian systems are absent in spongy bones of mammals.

→ Myoglobin protein is called as muscle haemoglobin.

→ Platelets occur in mammals only.

→ Muscular tissues are formed from the mesoderm layer of embryo.

→ Striated muscles contain acetylcholine which stimulates their contraction.

These MP Board Class 11th Biology Notes for Chapter 6 Anatomy of Flowering Plants help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 6 Anatomy of Flowering Plants

(A) Plant Tissue And Anatomy

→ A group of cells having a similar structure and function is called a tissue.

→ In plants two types of tissues are present: (i) Meristematic tissue and (ii) Permanent tissue.

→ The tissues that are capable of cell division are called meristematic tissue.

→ There are three groups of meristematic tissues : (i) Apical meristem, (ii) Intercalary meristem, (iii) Lateral meristem.

→ Apical meristem is found in the tips of shoot and roots, intercalary meristem occurs below the nodes and lateral meristem occurs on the lateral portion of the plant body.

→ Tissues which have no capacity to divide are called permanent tissues.

→ Callose is a polysaccharide which lines the pores of sieve plate.

→ The thickening of lignin and suberin on the inner tangential and radial walls of endodermis is called casparian strip.

→ The xylem having protoxylem towards the centre of the axis is called endarch.

→ The xylem having protoxylem towards periphery and metaxylem towards the centre is called
exarch.

→ The endodermal cells opposite the protoxylem groups that lack the thickening of lignin and suberin are called passage cells.

→ Tracheids are the elongated cells with tapering ends found in xylem of gymnosperms and angiosperms.

→ Vessel is a pipe like structure representing rows of cells, placed one above the other, in the xylem tissue.

→ Parenchyma and collenchyma are the living tissues of plant body.

→ The xylem of stem is endarch whereas it is exarch in roots.

→ Xylem is made up of tracheids, vessels, xylem fibres, and xylem parenchyma.

→ Xylem is the conductive tissue which plays a vital role in the conduction of water and minerals.

→ Phloem is made up of sieve tube, companion cell, phloem fibre and phloem parenchyma.

→ Phloem plays an important role in the conduction of organic food manufactured by leaves.

→ Xylem and phloem are the conductive tissues of plant which are collectively called as vascular bundle.

→ Simple tissues consist of single type of cells which are similar in origin, morphology and functions. Simple tissue may be (i) Parenchyma, (ii) Collenchyma, (iii) Sclerenchyma.

→ Parenchymatous cells act as storage tissue for food starch, fats, oil and proteins.

→ Collenchyma provides strength to the organs.

→ Sclerenchyma has thickened, usually lignified cell walls, which acts as supporting tissue.

→ Calyptrogen is a constituent of root apical meristem which occurs in monocots and is meant for producing the root cap.

→ The vascular bundle in which xylem and phloem are found in separate bundles present on different radii is called radial V.B. It is the characteristic of roots.

(B) Secondary Growth

→ Secondary growth is the increase in the girth of stems and roots.

→ Tissues formed from primary meristem are called as primary tissues, whereas tissues formed from secondary meristem are called as secondary tissues.

→ Increase in the girth of stems and roots due to secondary tissues is called as secondary growth.

→ Secondary growth is the characteristic feature of dicotyledonous stems and roots.

→ Except few cases, secondary growth is absent in monocotyledonous stems and roots.

→ Secondary growth takes place due to the activity of cambium and cork cambium.

→ Cambium present between xylem and phloem is called fascicular cambium.

→ Cambium present between two vascular bundles in ground tissue is called interfascicular cambium.

→ Fascicular and interfascicular cambium together constitute cambium ring.

→ Cambium ring cuts off secondary phloem on outer surface and secondary xylem on its inner surface.

→ During secondary growth, cells of the outermost layer of cortex become meristematic to produce cork cambium.

→ During secondary growth due to increase in the girth of stems and roots, the epidermis ruptures and cells become dead to produce bark.

→ Cork cambium cuts off cork cells on outer surface and secondary’ cortex on inner surface.

→ Cork cells and secondary cortex together constitute cork which is used for making plug of bottles.

→ Secondary xylem forms the maximum part of wood.

→ Lenticels are the pores on the bark of stem, composed of loosely arranged parenchymatous cells having intercellular spaces for gaseous exchange.

→ Secondary wood formed during a single year or one growth period is called annual ring.

→ The age of the plant can be calculated by the counting of annual ring.

→ The bladder like out growths of parenchyma, through the pores in the lateral walls of xylem vessel cause plugging of them and are called as tyloses.

→ Heartwood is the dark coloured wood in the centre and is dead. It contains oils, gums, resins, tannins, etc.

→ Heartwood is more durable and resistant to micro-organisms and insects. It is good for furniture.

→ Sapwood is the peripheral light coloured living wood responsible for conduction of materials.

→ Sapwood is less durable and resistant to micro-organisms and insects. Good for fuel.

→ Heartwood is a porous wood as it has vessels.

→ Early wood is formed during favourable seasons of spring and summer when cambium is active and produce large vessels or tracheids.

→ Late wood is produced during adverse conditions of climate (autumn and winter) in which tracheids and vessels are smaller.

→ Wood is the name of secondary xylem of the stems and main roots but the term wood can also be used for all types of xylem.

→ During leaf fall, cork cambium is formed in the leafbase which results in the formation of an abscission layer.

→ The cells formed on inner surface of the phellogen are called phelloderm or secondary cortex.

→ Dracena is a monocot plant, whose stem exhibit abnormal secondary growth.

These MP Board Class 11th Biology Notes for Chapter 4 Animal Kingdom help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 4 Animal Kingdom

→ All eukaryotic, multicellular, chlorophyll less, heterotrophic, cell wall less and holozoic organ¬isms are placed under animal kingdom.

→ Animals possessing the capacity of sensitivity and irritability.

→ Animals are the most advanced organisms of the living kingdom.

→ Porifera is the simplest and chordata is the advanced phylum of the kingdom animalia.

→ Animal body construction is characterized by : (i) Body plan, (ii) Symmetry, (iii) Body cavity, (iv) Body segmentation and appendages, (v) Body support and protection.

→ Invertebrates and chordates are differentiated on the basis of absence and presence of noto¬chord, respectively.

→ Animals may be diploblastic or triploblastic.

→ Sponges are primitive sessile pore bearing, mostly marine animals with cell aggregate body plan.

→ Animals of phylum coelenterata are diploblastic, radially symmetrical, bearing stinging cells (nematocysts) exhibit blind sac body plan.

→ Herdmania (Urochordate) show retrogressive metamorphosis, i.e., its larva is more developed than adult.

→ Whale and dolphins are aquatic animals. Their forelimbs are modified as flippers. Hind limbs are absent. .

→ Blue whale is the largest animal of the world.

→ Tusks of elephants are modified incisors.

→ Bats are able to detect objects in the dark with the help of echo location.

→ Roundworm (phylum nemathelminthes) are parasite in animal and plant or live in soil. These are triploblastic, pseudocoelomate and with a tube within tube body plan.

→ The nervous system of annelids comprises preoral ganglia jointed by circumoesophageal commissures to a double ventral ganglionated nerve cord.

→ Phylum arthropoda is the largest phylum of the animal kingdom. Their body is metamerically segmented.

→ The body of arthropods is differentiated into head, thorax and abdomen.

→ Prototheria the egg laying mammals like duck-billed platypus and echidna.

→ Eutheria are the true placental mammals.

→ The pouched mammals (marsupials) give birth to weak young one. These are kept in a marsu- pium present on the abdomen of the female.

→ The primates includes the prosimians such as lemurs, tarsiers and lorises and simians such as old world and new world monkeys.

→ Amphiblastula is the flagellated larva found in the porifera. At metamorphosis the flagellated cells move to the interior and become choanocytes.

→ Bipinnaria larva is the larval form found in asteroid echinoderms.

→ Choanocyte is the flagellated collar cell found as a lining of the internal cavities of the porifera.

→ Diploblastic : Animals having two layers of cells in the body wall.

→ Haemocoel is a cavity filled with blood which replace perivisceral coelom, specially in arthro-pods and molluscs.

→ Pelagic animals living in the surface waters of the sea.

→ Pericardium is a cavity in which heart is located.

→ Triploblastic animals having three layers of cells in the body wall.

→ Tube-feet is a tentacle like outpushings from the water vascular system which may be used for locomotion in the echinodermata.

→ Viviparous animals giving birth to living young.

→ About 75% animals of the animal kingdom belonging to phylum arthropoda.

→ Man {Homo sapiens) is the highly advanced organism of the animal kingdom.

→ Platyhelminthes are bilaterally symmetrical, dorsiventrally flattened acoelomate worm like animals.

→ The Mollusca are soft bodied, non-metameric, triploblastic coelomate animals consisting of anterior head, a ventral muscular foot and dorsal mass surrounded by a thin fleshy envelope.
The mantle generally sheltered in an external calcareous shell of their own secretion.

These MP Board Class 11th Biology Notes for Chapter 3 Plant Kingdom help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 3 Plant Kingdom

→ Annuals : Plants which live for one season and complete their life cycle in a single favourable season.

→ Algae : They are large assemblage of autotrophic plants which are predominantly aquatic.

→ Land plants are divided into two divisions, i.e., Bryophytes and tracheophytes.

→ Bryophytes are the amphibians of plant kingdom. In them, the gametophyte is the dominant phase and sporophyte is short-lived.

→ Tracheophytes include pteridophytes, gymnosperms and angiosperms.

→ Tracheophytes are characterized by the presence of vascular tissue.

→ Gymnosperms are woody plants with naked seed but without flowers.

→ Angiosperms are the flowering plants and their seeds are covered by the ovary wall.

→ The flower is a special modified reproductive shoot and is composed of four whorls namely calyx, corolla, androecium and gynoecium.

→ Bryophytes serve as a connecting link of aquatic algae and terrestrial pteridophytes.

→ Thallophyta have thallus like body and constitute algae and fungi.

→ Bryophytes may be thalose or foliose and found in moist and shady places.

→ In bryophytes, zygote develops into sporogonium which is further distinguished into food, seta and capsule.

→ On the basis of number of cotyledons, angiosperms are divided into two following major groups : (i) Monocotyledons : Bear only one cotyledon and (ii) Dicotyledons : Bear two cotyledons.

→ Heterospory : The production of spores of two different sizes and two different developmental patterns is called as heterospory. e.g., Selaginella.

→ Formation of fruit without fertilization is called parthenocarpy and the fruits are called parthenocarpic fruits.

→ Bicollateral vascular bundles : When both phloem and cambium occur twice once on outer side of xylem and then again on its inner side. The sequence is outer phloem, outer cambium, xylem, inner cambium and inner phloem.

→ Concentric vascular bundles : Vascular bundles in which xylem is covered by phloem or vice versa are called as concentric vascular bundles.

→ Radial vascular bundle : Vascular bundles in which xylem and phloem occur in separate patches on alternate radii as in roots are called as radial vascular bundle.

→ Polyembryony : Presence of more than two embryo in a seed is called as a polyembryony.

→ All eukaryotic, photosynthetic (autotrophs) multicellular organisms are included in kingdom plantae.

→ Plant kingdom is classified into three groups : (i) Algae, (ii) Bryophyta and (iii) Tracheophyta. Algae is divided into three subdivisions : (i) Rhodophyta, (ii) Phaeophyta and (iii) Chlorophyta.

→ Bryophyta and tracheophyta together are called as embryophyta.

→ Reproductive organs of embryophyta are multicellular.

→ Bryophytes are amphibian.

→ Higher vascular plants are evolved from bryophytes.

→ Flowering plants are called as angiosperms.

→ Flower of rafflesia is the largest flower.

→ Bryophytes serves as link between aquatic algae and terrestrial pteridophytes.

→ Main plant of bryophytes are gametophyte.

→ Main plant of pteridophytes are sporophyte.

→ Gymnosperms are naked seeded plants.

→ The flower is a special modified reproductive shoot and is composed of four whorls namely, calyx, corolla, androecium and gynoecium.

→ The ovary develops into fruit and the ovule develops into seed.

These MP Board Class 11th Biology Notes for Chapter 2 Biological Classification help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 2 Biological Classification

→ The bacteria constitute the most ancient, smallest, simplest organisms which are classified on the basis of structure and activities.

→ Bacteria often reproduce by binary fission.

→ Many eubacteria are disease producers in man, animals and plants. Some bacteria produce antibiotics.

→ Cyanobacteria are ancient photosynthetic prokaryotes. They liberate oxygen during photosynthesis.

→ Bacteria are photoautotrophic, chemoautotrophic or heterotrophic.

→ Archaebacteria are living fossils.

→ Plasmids are the extranuclear circular DNA found in bacteria.

→ In Moneran, 70S type ribosomes are found.

→ The cell wall of bacteria contain polysaccharide, lipids, proteins and mucopeptides.

→ Photosynthetic bacteria does not ‘split water’ to obtain reducing power. Instead of water they obtain H2 from H2S or other sulphur containing compounds.

→ Use of more concentration of salt or sugar in food preservation kills the bacteria by plasmolysis thus prevent the food from decaying.

→ Joker of plant kingdom : Mycoplasma is also known as Joker of plant kingdom.

→ Trichome : A filament without mucilage sheath is called trichome.

→ ‘Germ theory of disease’ is proposed by Louis Pasteur.

→ Louis Pasteur (1822-1895) is the ‘founder of bacteriology’ and regarded as ‘father of microbiology.

→ Endospore : A resting spore produced by few bacteria under unfavourable conditions.

→ Antibiotics : A chemical substance excreted by micro-organisms which inhibit the development and growth of other organisms.

→ Bacterial cell wall: It is made up of peptidoglycan.

→ Ehrich (1854-1915) is regarded as ‘father of chemotherapy’.

→ Fungi are heterotrophic thallophytes because of the absence of chlorophyll.

→ The body of fungi is made up of thin filaments. A single filament of fungi is called as hyphae.

→ The group of filaments forming a network is called as mycelium.

→ The cell wall of fungi is made up of chitin.

→ Glycogen is the stored food material of fungi.

→ Kingdom fungi is classified into five classes :

1. Class 1. Mastigomycotina or Eumycotina.
2. Class 2. Zygomycotina.
3. Class 3. Ascomycotina.
4. Class 4. Basidiomycotina.
5. Class 5. Deuteromycotina.

→ Asexual reproduction in fungi mostly takes place by spores or conidia, whereas the sexual reproduction may be isogamous or heterogamous. In lower forms, it takes place by conjuga¬tion.

→ The higher fungi are included under ascomycetes and basidiomycetes.

→ The most distinctive feature of ascomycetes is the formation of sac like ascus in which sexually produced ascospores develop. In some members, asci are borne in a fruiting body called the ascocarp.

→ An association of algae and fungi is called as Lichen.

→ Mushrooms are the fungi used as food.

→ Mycology is the branch of biology deals with study of fungi.

These MP Board Class 11th Biology Notes for Chapter 1 The Living World help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 1 The Living World

→ Hippocrates and Aristotle tried for the first time to classify the organisms systematically.

→ Theophrastus is known as ‘father of botany’.

→ Carolus Linnaeus is called as ‘father of taxonomy’.

→ Viruses are the border line of living and non-living things.

→ John Ray was first to describe the natural system of classification.

→ Euglena can live double lives. In presence of light and absence of organic matter it become green photosynthetic like plants, but in presence of organic matter or in the absence of light energy they absorb or ingest food like the protozoans. Due to this ability they were earlier included in the plant kingdom by botanists and in the animal kingdom by zoologists.

→ Protists known as ‘slime moulds’ are devoid of a cell wall in the vegetative phase and ingest particular matter. But in the reproductive phase they develop a cell wall as in some fungi.

→ The evolutionary history of any group of organisms is called phylogeny.

→ Viruses are the connecting link of living and non-living things.

→ At present more than 1-7 million species of organisms have been identified and named.

→ The basic unit of classification is species.

→ Viruses are made up of protein and nucleic acid, RNA or DNA.

→ Arranging living organisms in different groups based on their morphology is called as classification.

→ Smallest group of a classification is called as Species and Largest category is called as Kingdom.

→ Scientific name of living organisms consists of two words. First word is called as Generic name and second word is called as Specific name.

→ Scientific name of man is Homo sapiens.

→ Linnaeus was the first to establish and define hierarchy of taxonomic categories.

→ Full form of ICBN is “International Code of Botanical Nomenclature.”

→ Full form of ICZN is “International Code of Zoological Nomenclature.”

→ Taxon : The Taxa are the groups of organisms, generally group of species.

→ Species : Species is a population of interbreeding individuals.

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 13 Hydrocarbons which are most likely to be asked in the exam.

MP Board Class 11th Chemistry Important Questions Chapter 13 Hydrocarbons

Hydrocarbons Class 11 Important Questions Very Short Answer Type

Question 1.
Frans-alkene is formed as a result of reduction of alkyne in liquid ammonia by sodium. Will butene formed by such type reduction of 2-butyne show geometrical isomerism?
Answer:

Trans-2-butene formed by the reduction of 2-butyne shows geometrical isomerism.

Question 2.
In spite of the -I effect, halogen in haloarene compounds is o-(ortho) and p-(para) directive. Give reason.
Answer:
Halogen is a highly reactive group. Due to strong -I effect electron density decreases on benzene ring. But due to resonance, compared to o- and p-place, electron density is more at m-place due to which halogens are o- and ρ -directive.

Question 3.
Alkenes are more reactive than alkanes. Why?
Answer:
In alkanes there is only a-bond between C – C but in alkenes there C = C has one σ and one π-bond. Due to lateral overlapping, the π-bond is weaker than the σ-bond. Hence, alkenes are more reactive than alkanes. The bond energy of π-bond (25 kJ mol^-1) is lower than the bond energy of a -bond (347 kJ mol^-1). Due to this difference of bond energy alkenes are more reactive as compared to alkanes.

Question 4.
What is meant by asymmetric carbon atom?
Answer:
Asymmetric carbon atom: If all the four valencies of carbon atom in a regular tetrahedron compound are linked to four different groups, then the carbon atom of tetrahedron is referred as asymmetric carbon atom or chiral carbon.

Question 5.
Draw Newman’s projection figures of Staggered and Eclipsed conformations of Butane.
Answer:
structural formula of Butane :

Question 6.
What is meant by Chirality?
Answer:
Molecules which are non-superimposable on their mirror images are called chiral molecules and this property is known as chirality. Chiral molecules are optically active. Chirality in the molecule is due to asymmetrical carbon atom. Chiral molecules are optically active, but molecules with asymmetric carbon may be optically active or may not.

Question 7.
What are Alkanes? What type of bond is present in them?
Answer:
Alkanes are saturated hydrocarbons. They are also known as paraffins. They are less reactive and more stable. Their general formula is C_nH2_n+2.
In alkanes, single σ-bond is present between carbon-carbon and carbon-hydrogen bond. Each carbon in alkanes is sp³ hybridized.
Example : Methane CH₄, Ethane C₂H₆.

Question 8.
What are Alkenes? In which hybrid state is carbon present in them?
Answer:
Alkenes are unsaturated hydrocarbons. They are also known as olefins. They are more reactive than alkanes and less stable than them. They easily show addition reactions. Their general formula is C_nH2_n. Carbon-carbon double bond is present in alkenes and carbon is in sp2 hybridized state.
Example: Ethylene H₂C =CH₂, Propylene CH₃ -CH =CH₂.

Question 9.
What are Alkynes? What type of bond is present in them?
Answer:
Alkynes are unsaturated hydrocarbons. They are highly reactive and are very unstable. Their general formula is C_nH2_n-2.
Triple bond is present between carbon-carbon, of which one is σ and two are π bonds and both carbons are in sp hybridised state.
Example: Acetylene HC ≡ CH, Propyne CH3 – C ≡ CH.

Question 10.
Write the isomers of pentane.
Answer:
Pentane (C₅H₁₂) forms three isomers.

Question 11.
What is cis-trans isomerism ? Explain with example.
Answer:
cis-trans isomerism is also known as geometrical isomerism which is shown by such double bond compounds in which the double-bonded carbon atoms are linked to two different atoms or groups. When similar groups or hydrogen atoms are on the same side of double bond, the compound is cis isomer and if situated on opposite sides, then the compound
is trans isomer.
Example: CH₃CH =CH – CH₃ :

Question 12.
Why is benzene extraordinarily stable though it contains three double bonds?
Answer:
Due to Resonance and displacement of electrons, benzene molecule is extraordinarily stable.

In hybrid state, the dotted ring represents the 6 electrons displaced between 6 carbon atoms of benzene ring. Thus, due to delocalized 6 electrons benzene is stable.

Question 13.
How will you obtain nitrobenzene from acetylene?
Answer:

Question 14.
Electrophilic and Nucleophilic reactions occur between electron-deficient and electron-rich species. Thus, their tendency is to attack electron rich and electron-deficient nucleus. Classify the following species into electrophiles and nucleophiles :
(i) H3CO^-,
(ii)
(iii) Cl,
(iv) Cl₂C:,
(v) (CH₃)3C⁺,
(vi) Br^–,
(vii) CH₃OH,
(viii) R-NH-R.
Answer:
Electrophiles are electron deficient species. These are neutral or positively charged.
(iii) Cl, (iv) Cl₂C: and
(v) (CH₃)3C⁺ are electrophiles.
Nucleophiles are electron rich species. These are neutral or negatively charged species.
(i) CH3O^–,
(ii)
(vi) Br^–,
(vii) img and
(viii) RNHR are nucleophiles.

Question 15.
Cyclopropane is more reactive as compared to cyclohexane.
Answer:
The bond angle in cyclopropane C-C-C is 60° due to which the molecule is in more strain (less stable) and hence more reactive, whereas in cyclohexane, the bond angle is very close to 109°28′ due to which the molecule is in less strain and hence its reactivity decreases.

Question 16.
What is meant by functional isomerism? Explain with example.
Answer:
Functional isomerism: When two compounds have same molecular formula but have different functional groups are called functional isomers. For example, dimethyl ether and ethyl alcohol. Ether contains -0- group while alcohol contains -OH group. Molecular formula of both is C₂H₆O.

Question 17.
What is Resolution?
Answer:
Process of separation of racemic mixture into d and / forms or optically active forms, is called Resolution. Resolution can be done by biochemical or chemical methods.

Question 18.
Which type of isomerism is found in arenas? Explain with example.
Answer:
In Arenes, position isomerism is found.
Example: Linking of one or more methyl group to arenes gives rise to position isomerism. Xylene (C₈H₁₀) is a dimethyl disubstituted product of benzene.
(i) When two substituent groups are present 1,2 i.e., adjacent position, then it is called ortho (or o) isomers.
(ii) When the position is on 1 and 3 (alternate) then it is called meta (m).
(iii) When the position is on 1 and 4 (diagonally opposite to one another) then it is called para (p).

Question 19.
Write a reaction to show acidic nature of alkynes.
Answer:
Alkynes whose molecules possess a triple bond at the terminal position act as weak acid. Such alkynes release hydrogen when treated with reactive metals like Na, Ca and their derivatives are obtained which are known as acetylides.

Question 20.
What is Baeyer’s reagent? How is it used to test unsaturation?
Answer:
Alkaline KMnO₄ is known as Baeyer’s reagent. On reacting unsaturated hydrocarbons with Baeyer’s reagent, alkaline KMnO₄ decolourises. Therefore, it is used to test unsaturation.

Question 21.
Write action of Baeyer’s reagent on aikene. Give equation also.
Answer:
Alkenes react with Baeyer’s reagent (1 % basic KMnO₄) and form ethylene glycol. The colour of KMnO₄ disappears. This is also called as Baeyer test for unsaturation.

Question 22.
What do you understand by Peroxide effect? Explain with example.
Answer:
Peroxide effect: Addition of HBr to propene in presence of organic peroxides takes place against the Markownikoff’s rule and this effect of peroxide which changes the direction of reaction and proceeds against Markownikoff’s rule is called peroxide effect.

Di-tertiary butyl peroxide (CH₃)₃CO – OC(CH₃)₃ and benzoyl peroxide (C₆H₅COO)₂ are organic peroxides. It should be remembered that peroxide effect on alkenes takes place only during addition of HBr. Addition of HCl, Hl, H₂SO₄ and water are not affected by presence or absence of peroxide.

Question 23.
How can you differentiate between Ethane, Ethylene and Acetylene?
Answer:
Differences between Ethane, Ethylene and Acetylene

Question 24.
What is Berthelot synthesis?
Answer:
On passing a stream of hydrogen gas through an electric arc between two carbon electrodes, Acetylene is obtained.

Question 25.
What is Westron or Westrosol ?
Answer:
1. Compound formed by the addition of halogen on acetylene in the presence of CCl₄ is known as Westron.

2. When Westron is heated with alcoholic KOH, then due to dehydrohalogenation Westrosol is obtained.

Question 26.
What is Prototrophy?
Answer:
By the hydrolysis of Propyne, enol and keto forms are obtained. It represents tautomerism. This type of isomerism is shown by such compounds which contain at least one hydrogen which can transfer from one position to another. This is known as prototrophy.

Question 27.
What is conformational isomerism?
Answer:
Various spatial arrangements produced due to rotation of carbon-carbon double bond are called conformations and molecular orientation of conformers is known as conformational isomerism.

Question 28.
Explain polymerisation reaction with example.
Answer:
Polymerisation reactions: In polymerisation, many simple molecules of a substance combine together to form a huge molecule. The simple molecule is called monomer and the huge molecule as polymer or macromolecule. Rubber, nylon, bakelite, terylene and P.V.C. are examples of high polymers.
Polymerisation of alkenes takes place in presence of Lewis acid BF₃ or AlCl₃ or organic and inorganic peroxides. Organic peroxides used are benzoyl and acetyl peroxide.

Question 29.
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitution reactions with difficulty?
Answer:
Due to the presence of delocalized 6 π-electrons, benzene behaves like an electron-rich source. Thus, it attracts electrophilic reagents (electron-deficient species) but repel nucleophilic reagents (electron-rich species). This is the reason that benzene undergoes electrophilic substitution reactions easily and nucleophilic substitution reactions with difficulty.

Question 30.
Alkenes exhibit Electrophilic addition reactions whereas arenes exhibit Electrophilic substitution reactions. Give reason.
Answer:
Alkenes are rich source of loosely bonded π-electrons due to which they exhibit electrophilic addition reactions. Huge energy change takes place in electrophilic addition reactions due to which they are more effective in energetic form rather than electrophilic substitution reactions.

In Arenes, during electrophilic addition reactions aromatic nature of benzene ring is destroyed whereas in electrophilic substitution reactions it remains constant. Electrophilic substitution reactions of arenes are energetically more effective than electrophilic addition reactions.

Question 31.
Write the necessary conditions for geometrical isomerism.
Answer:
The following conditions are required for geometrical isomerism:
1. At least one carbon-carbon double bond should be present.
2. The double-bonded carbon atoms should contain different atoms or groups linked to them.

Question 32.
Write name and structural formula of cyclic compounds which exhibit geometrical isomerism.
Answer:
Some cyclic compounds, due to restricted rotation of carbon-carbon single bond exhibit geometrical isomerism.
Example:

Question 33.
What is Saytzeff’s rule? Explain with example.
Answer:
The rule states that “During dehydrohalogenation of an alkyl halide, that alkene is formed in which the double-bonded carbon atom is more alkylated i. e., hydrogen is eliminated from that carbon atom which has lesser number of hydrogen.”

Question 34.
Write a note on octane number.
Answer:
The resistance of a sample of petrol to check knocking is measured in terms of octane number. For this purpose two substances viz iso-octane and n-heptane have selected and arbitrarily rated, n-heptane has no resistance to check knocking and has been rated as octane No. zero. Iso-octane has 100% resistance to check knocking and is rated as 100.

Octane number of a petrol is defined as percentage of iso-octane by volume in a mixture of iso-octane and n-heptane which gives the same knocking as oil under investiga¬tion operated in a standard engine and under standard conditions.

Question 35.
Compare paraffins, olefins and acetylene on the basis of the following points :
1. IUPAC name, 2. General formula, 3. Reactivity, 4. Reaction with Br2 water.
Answer:

Hydrocarbons Class 11 Important Questions Short Answer Type 

Question 1.
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also, give reason for this behaviour.
Answer:
Hybridisation of carbon in the given compound is as follows :

Acidic character increases with the increase in 5-character of orbital. Thus, decreasing order of acidic behaviour of benzene, n-hexane and ethyne is as follows :
Ethyne > Benzene > n-hexane.

Question 2.
What effect does branching of alkane chain has on its boiling point?
Answer:
As a result of increase in branching in the chain, surface area decreases. Thus, due to increase in branching in the chain van der Waals’ force of attraction decreases due to which boiling point also decreases.

Question 3.
What are the necessary conditions for a system to be aromatic?
Answer:
The necessary conditions for a system to be aromatic are as follows :

• Molecule should be planar.
• Molecule should be cyclic with alternate single and double bonds, there should be complete delocalisation of n electrons in the ring.
• There should be (4n + 2) n electrons in the ring of the molecule where n- 0,1,2,3,……………………….. (Huckel’s rule)
  If the molecule does not satisfy anyone or more conditions, then it is non-aromatic.

Question 4.
What is Metamerism? Explain with example.
Answer:
Metamerism: Isomerism which exists among same class of compounds is known as metamerism. Alkyl groups attached to functional group are different.
Example :
(a) C₄H₁₀O :

(b) C₅H₁₀O:

Question 5.
What is Tautomerism? Explain with example.
Answer:
Tautomerism: It is a special type of functional isomerism in which same compound represents both of isomers which remain in equilibrium. It is due to migration of an H atom from one polyvalent atom to other. It involves change of bonds also.

Dyad system: It involves oscillation of hydrogen atom between only two polyva¬lent atoms e.g., Hydrocyanic acid:

Triad system: It involves the oscillation of hydrogen atom between three polyva¬lent atoms:

This is also known as Keto-Enol Tautomerism.

Question 6.
What is Newman’s Projection formula?
Answer:
If a single bond is present between two carbon atoms, then both the carbon atoms are represented by a circle. Hydrogen atoms linked to them are represented from the centre. If both the carbon atoms are one behind the other, then only front carbon atom is seen and its three hydrogens are joined from the centre and the hydrogens of the carbon atom behind are linked to the circumference.

Question 7.
State Markownikoff’s rule. Explain with example.
Answer:
Markownikoff’s rule: Markownikoffs rule is for addition of polar molecules to unsymmetrical alkenes.
According to this rule, in case of unsymmetrical olefines, the negative part of adding molecule goes to that carbon atom to which less number of hydrogen atoms are attached.
These reactions take place in presence of ether but in absence of organic peroxides. For example, when HBr reacts with propene, the main product is 2-bromopropane (isopropyl bromide). Small amount of 1 -bromopropane i.e., n-propyl bromide is formed.

Question 8.
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher boiling point and why?
Answer:
Hex-2-ene is CH₃ — CH₂ — CH₂ — CH = CH — CH₃ . Its cis and transform are as follows:

cis isomer is more polar than transform, therefore dipole moment of cis form is more than that of trans form. Thus, boiling point of cis form is higher than transform because dipole-dipole interaction is more, transform of hex-2-ene is non-polar.

Question 9.
An alkene, ‘A’ on ozonolysis gives a mixture of ethanal and pentane-3-one. Write structure and IUPAC name of ‘A’.
Answer:
The structures of the products formed by ozonolysis are written in such a way that their oxygen atoms point towards each other. After removing both the oxygen, both ends are joined by double bond. When structure of Alkane ‘A’ is obtained.

Question 10.
Explain Racemic mixture with example.
Answer:
Racemic mixture: When equal quantity of dextrorotatory and laevorotatory isomers are present in a sample, it will not show optical activity because both enantiomers rotate the plane-polarized light to the same extent in opposite direction and hence net rotation is zero. Such an optically inactive mixture is represented by dl or ±.
Example: Racemic lactic acid: It is an equimolar mixture of d and l forms. It has 50% d form and 50% l form. This form is optically inactive and it can be separated into d and / forms. It is represented by dl or ±.

Question 11.
Give substitution reaction of alkanes with example.
Answer:
The reaction in which any atom present in the compound (e.g., H or other atoms) is replaced by other atom or radical in such a way that there is no change in external structure is called substitution reaction. Examples of substitution reaction are halogenation, nitration, sulphonation, etc. e.g.,
Halogenation reaction: Mixture of mono, di, tri, tetra haloalkane is formed.

Question 12.
An alkene ‘A’ contains three C – C, eight C-H σ bonds and one C-C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44u. Write IUPAC name of ‘A’.
Answer:
Aldehyde with molar mass 44u is ethanal, CH₃CHO. Thus, form its two moles

Thus, in But-2-ene, there are three C – C, eight C – H σ-bonds and one C – C π-bond are present.

Question 13.
Propanal and pentane-3-one are the ozonolysis products of an alkene. What is the structural formula of the alkene?
Answer:

Question 14.
Suggest the name of a Lewis acid other than anhydrous AICl₃ which can be used during ethylation of benzene.
Answer:

In the ethylation of benzene, ethyl group gets linked with benzene ring. This is done by Friedel-Crafts ethylation reaction of benzene. Instead of anhydrous AlCl₃, anhydrous FeCl₃ or SnCl₄ can be used as catalyst (Lewis acid).

Question 15.
What is B.H.C. ? State its uses.
Answer:
On treating benzene with Cl₂ in the presence of sunlight B.H.C. is formed.

It is available in eight three-dimensional forms but only four isomeric forms α,β,γ and δ can be obtained in pure state. The various isomeric forms are obtained due to different arrangements of hydrogen and chlorine. γ isomeric form is more stable and a powerful insecticide. It is also known by the name of Lindane or Gammaxene or 666. It is used as an insecticide.

Question 16.
How will you convert the following? Give only chemical equations.
(i) Methane to Ethane,
(ii) Ethane to Methane
(iii) Acetylene to Benzene.
Answer:
(i) Methane to Ethane :

(ii) Ethane to Methane


(iii) Acetylene to Benzene.

Question 17.
Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking an example.
Answer:
For the preparation of alkane with odd number of carbon atoms, two different alkyl halides should be taken and two different alkyl halides will react in three ways as a result of which mixture of alkanes are obtained.
Example:

This is the reason that Wurtz reaction is not preferred for the preparation of Alkanes containing odd number of carbon atoms.

Question 18.
Out of benzene, m-dinitrobenzene and toluene which will undergo nitra-tion most easily and why ?
Answer:
Nitration is Electrophilic substitution reaction. Reactivity of benzene nucleus towards E+ decreases in the presence of electron attracting group whereas it increases in the presence of electron releasing group.
Electron attracting groups (-I) like (-NO₂) deactivate electrophilic substitution reaction.
Electron releasing groups (+I) like (-CH₃) activate benzene nucleus towards electrophilic reactions.
Thus, rate of nitration is as follows:

Thus, Toluene will undergo nitration easily.

Question 19.
How will you obtain (Give only chemical equations):
(i) B.H.C. from benzene
(ii) Acetophenone from benzene
(iii) P.V.C. from chloroethene
(iv) Teflon from tetrafluoroethene.
Answer:
(i) B.H.C. from benzene :

(ii) Acetophenone from benzene :

(iii) P.V.C. from chloroethene

(iv) Teflon from tetrafluoroethene :

Question 20.
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile E⁺ :
(a) Chlorobenzene, 2,4-dinitrochiorobenzene,p-nitrochlorobenzene.
(b) Toluene, p – H₃C – C₆H₄ – NO₂, p – O₂N – C₆H – NO₂.
Answer:
In the presence of electron releasing groups (or active group) in benzene nucleus, electron density increases. Thus, electrophilic reagents can easily attack on benzene nucleus.
Example : (-CH₃)
But electron attracting groups (like : NO₂) decrease the electron density on benzene nucleus. Thus, electrophilic reagents attack on benzene nucleus with difficulty.
Thus, decreasing order of relative reactivity of the above compounds towards electrophiles (E+) is as follows :
(a) Toluene > p -nitrochlorobenzene > 2,4-dinitrochlorobenzene.
(b) Toluene > p – H₃C – C₆H₄ – NO₂ > p – O₂N – C₆H – NO₂.

Question 21.
Write structures of all the alkenes which on hydrogenation give 2-methyl- butane.
Answer:
Structure of 2-methyl butane is as follows :

In this structure, double bond is introduced in various places and tetravalency of each carbon is satisfied by hydrogen atoms.

Question 22.
Explain geometrical isomerism with an example. Or, Geometrical isomerism is found in which type of compounds? Explain with example.
Answer:
Geometrical isomerism: This type of isomerism is exhibited by those compounds in which two carbon atoms are joined by a double bond. Both carbon atoms are linked by two different groups or atoms and their general formulae (a)(b)C=C(b)(a) can be represented by two forms. When identical groups are present on the same side then it is called cis type and when identical groups are present on the opposite sides then it is referred as trans type of isomerism.

Question 23.
What is Huckel’s Law?
Answer:
Huckel stated that “All planar cyclic compounds possess aromatic character, whose cycle contain (4n + 2) n electrons.”
Here n is a whole number whose value is 0,1,2,………………….
Thus, cyclic compounds which contain 6(n = 1), 10 (n = 2), 14 (n = 3) electrons possess aromatic character.
Aromatic compounds are Of two types :
1. Benzenoids: Compounds which contain one or more benzene rings are called benzenoids.

2. Non-benzenoids: Compounds which do not contain benzene ring.

Question 24.
Express Friedel-Crafts reaction with equation.
Answer:
Friedel-Crafts reaction: When benzene is treated with alkyl halide in presence of anhydrous AlCl₃, then alkyl benzene is formed.
Alkylation :

Acetylation: When benzene is treated with acetyl chloride in presence of anhydrous AlCl₃, acetophenone is formed. In this, H-atom of benzene is displaced by acyl group.

Question 25.
For the nomenclature of geometrical isomerism what are E and Z symbol?
Answer:
When double-bonded carbon atom has all four groups different, then geometrical isomers are represented by E and Z symbols. It is based on the following rules:
Rule I. All atoms or groups attached to doublebonded carbon atom is the same, the atom with higher atomic number gets highest priority and atom with lower atomic number gets less priority. If the first atom is same then second atom or third atom is considered.
Rule II. If the two higher priority groups are on opposite side of the double bond, the isomer is E and if the two higher priority groups are on the same side of the double bond, the isomer is called Z.
Example :

Question 26.
What are Dienes? How many types of dienes are there? Explain with example.
Answer:
Dienes are unsaturated hydrocarbons which contain double bond betweenm carbon-carbon atoms and based on the position of double bond, they are of three types:
1. Isolated dienes: Dienes in which one or more single bonds are present between two double bonds are isolated dienes.
CH₂=CH – CH₂ – CH=CH₂

2. Conjugated dienes: Dienes in which two double bonds are conjugated or in alter nate position are conjugated dienes.

3. Cumulated dienes: Dienes in which two double bonds are in close position.
CH₂=C= CH-CH₃
CH₃-CH= C = CH₂.

Question 27.
Write Diels-Alder reaction with example.
Answer:
When conjugated dienes are heated with an alkene or substituted alkene, six membered cyclic compound is formed, therefore this reaction is known as cyclic reaction, in it 4n electrons join to the system. This type of reaction is known as Diels-Alder reaction.

Question 28.
What is Lindlar’s catalyst? Write its uses.
Answer:
Mixture of Palladium supported on Barium sulphate or Calcium sulphate is called Lindlar’s catalyst. Here sulphur or quinoline acts as catalytic poison and reduce alkyne only up to alkene state.
Uses: Reduction of alkyne by hydrogen to alkene is done in presence of Lindlar’s catalyst. This catalyst does not allow alkene further to reduce to alkane.

Question 29.
Write four facts in favour of Kekule’s structure for benzene.
Answer:
Kekule’s structure for benzene: Kekule proposed a six-membered ring structure for benzene. In this ring, all the six carbon atoms occupy comers of a hexagon and hydrogen is linked with each carbon atom.
In order to meet the fourth valency of carbon, Kekule suggested presence of alternate single and double bonds.

Facts in favour of Kekule’s structure :
(i) Benzene reacts with three molecules of hydrogen to form cyclohexane which proves the presence of three double bonds.

(ii) Benzene gives addition reaction with 3 molecules of chlorine to form benzene hexachloride, hence presence of double bonds is confirmed.

(iii) With ozone, it forms a triozonide which on hydrolysis gives three molecules of glyoxal.

(iv) Three molecules of acetylene polymerize in a red hot tube to form benzene.

Question 30.
Give reasons :
1. Boiling point of branched compounds is less than straight chain compounds. Why?
2. Melting point of compounds with odd number carbon atoms is less than the compounds with even number carbon atoms.
Answer:
1. Surface area of Linear chain organic compounds is more and intermolecular attractive force is high. Whereas in case of branched chain carbon atoms, the atoms come closer and intermolecular attractive force between them is less, therefore boiling point of branched compounds is less than straight chain compounds.

2. In alkanes with odd number of carbon atoms, the end carbon atoms lie on the same side whereas in alkanes with even number of carbon atoms, the end carbon atoms lie in opposite sides. Packing of alkanes with even number of carbon atoms is more dense than that of odd number carbon atoms.
Therefore, melting point of odd number alkanes is less than that of even number alkanes.

Question 31.
Write chemical equation for the combustion reaction of the following hydrocarbons :
(i) Butane,
(ii) Pentene,
(iii) Hexyne,
(iv) Toluene.
Answer:
All hydrocarbons on complete combustion give CO₂ and H₂O.

Question 32.
Write notes on :
(i) Sabatier and Senderens reaction,
(ii) Wurtz reaction,
(iii) Duma reaction.
Answer:
(i) Sabatier and Senderens reactions: In this method alkanes are formed by hydrogenation of unsaturated hydrocarbons in presence of Ni as catalyst.

 

(ii) Wurtz reaction: When an alkyl halide (bromide or iodide) is heated with sodium in presence of dry ether, an alkane is obtained. The number of carbon atoms in alkane is double than that of alkyl halide. This is called Wurtz reaction.

This method is suitable for preparing alkanes containing even number of carbon atoms like CH₃-CH₃, C₂H₅-C₂H₅,etc.

(iii) Duma reaction: Sodium salt of carboxylic acid reacts with soda lime when due to decarboxylation, alkanes are obtained.

Question 33.
Write Kolbe’s electrolysis with equation.
Answer:
By the electrolysis of concentrated aqueous solution of sodium or potassium salt of monocarboxylic acid, alkanes are obtained at anode.

Electrolysis occurs as follows:
2CH₃COONa ⇌ 2CH₃COO^–+2Na⁺

At Anode: Acetate ion loose electron and become neutral and then dissociate to form alkane.
2CH₃COO^– 2e^– → 2CH₃COO → C₂H₆ + 2CO₂

At Cathode: Hydrogen ion gains electron to give hydrogen gas.
2Na⁺ + 2HOH + 2e^– → 2NaOH + H₂.

Question 34.
Give only equations for the following reactions :
(1) Calcium carbide with water.
(2) Reaction of bromine water on ethylene.
(3) On heating ethylene with alkaline KMnO4.
(4) On heating benzene with cone. HN03 and cone. H2SO4.
(5) On heating benzene with CH3Cl in presence of anhydrous AlCl3.
Answer:
1. Reaction of calcium carbide with water :
CaC₂ + 2H OH → CH ≡ CH + Ca(OH)₂

2. Reaction of ethylene with bromine water :

3. On heating ethylene with alk. KMnO4:

4. On heating benzene with conc. HNO₃ and conc. H2SO₄:

5. On heating benzene with CH₃Cl in presence of anhydrous AlCl₃:

Question 35.
How are the following obtained :
1. Acetaldehyde from Acetylene,
2. Mustard gas from Ethylene,
3. Ethane from Grignard reagent,
4. Cuprous acetylide from acetylene,
5. Methane from Aluminium carbide.
Answer:
1. Acetaldehyde from Acetylene :

2. Mustard gas from Ethylene:

3. Ethane from Grignard reagent:

4. Cuprous Acetylide from Acetylene :
CH≡ CH + Cu₂Cl₂ + 2NH₄OH → Cu – C ≡ C – Cu + 2NH₄Cl + 2H₂O

5. Methane from Aluminium carbide :
Al₄C₃ + 12H₂O → 3CH₄+4Al(OH)₃.

Question 36.
A primary alkyl halide when subjected to Wurtz reaction forms CgHlg as the only product. By the monobromination of this alkane, an isomer of tertiary bromide is obtained. Identify the alkane and tertiary bromide.
Answer:
Since alkane CgHlg on monobromination forms an isomer of a tertiary bromide, thus tertiary hydrogen should be present in alkane. It is only possible when primary alkyl halide (Which participate in Wurtz reaction) contain a tertiary hydrogen.

Question 37.
Write the intermediate hydrocarbon radical formed by the monochlorination of 2-methyl propane. Which of these is more stable? Give reason also.
Answer:
2-methylpropane gives the following two types of radicals which are as follows :

Radical (I) is more stable, because it is 3° free radical and stabilizes nine hyper can- jugated structures (because it contains 9- α hydrogen).

Radical (II) is less stable because it is 1° free radical and it stabilizes only one hyperconjugated structure (because it contains only 1α -hydrogen).

Question 38.
Explain Dehydrohalogenation and Dehalogenation with example.
Answer:
Dehydrohalogenation: Elimination of a hydrogen halide molecule from an organic compound is known as dehydrohalogenation. During this reaction, hydrogen is eliminated from β-position. Therefore, it is known as β-elimination.
Example: On heating normal propyl chloride with alcoholic KOH, alkene is obtained.
CH₃– CH₂ – CH₂ – Cl + KOH → CH₃— CH = CH₂ + KCl + H₂O.

Dehalogenation: On heating vicinal dihalide with Zn powder in presence of methyl alcohol alkene is obtained. During this reaction, only halogen is eliminated, therefore it is known as dehalogenation reaction.

Question 39.
What happens when :
(1) On heating sodium acetate with soda lime.
(2) On heating 2-propanol with alumina at 300°C temperature.
(3) On passing acetylene through ammoniacal AgN03 solution.
(4) On heating ethyl iodide with sodium.
(5) By reacting hypochlorous acid on Ethylene.
Answer:
1. On heating sodium acetate with soda lime :

2. On heating 2-propanol with alumina at 300°C temperature :

3. On passing acetylene through ammoniacal AgNO₃ solution :
CH ≡ CH + 2 AgNO₃ + 2NH₄OH → Ag- C = C – Ag + 2NH₄NO₃ + 2H₂O

4. On heating ethyl iodide with sodium :

5. On reacting hypochlorous acid with ethylene :

Question 40.
Write notes on :
(1) Hydroboration
(2) Epoxy reaction.
Answer:
1. Hydroboration: When alkene is treated with diborane, double bond of alkene undergoes addition reaction and trialkyl borane is formed which on hydrolysis gives alcohol. This is known as Hydroboration reaction.

2. Epoxy reaction: Alkene oxide is known as Epoxide. Formation of epoxide by the oxidation of alkene is known as Epoxy reaction.
Example: Lower alkenes combine with oxygen at a temperature of200-400°C in the presence of silver catalyst to form epoxide.
Silver catalyst

Question 41.
In the alkane H3C — CH2 — C(CH3)2 — CH2— CH(CH3)2. Identify 1°, 2°, 3° carbon atoms and give the number of H atoms bonded to each one of these.
Answer:

Question 42.
In the reaction of HI, HBr and HCl with propene same intermediate carbocation is formed. Bond energies of HCl, HBr and HI are mainly 430-5 kJmol^-1 363^-1 kJmol-1 and 296-8 kJmor¹. What will be the order of reactivity of halogen acids?
Answer:
Addition of alkene with halogen acids is an Electrophilic addition reaction.

Since first step is a slow step, thus it is a rate-determining step. Rate of this step, depends upon the availability of proton, which again depend on the dissociation energy of H-X molecules. Lower the bond dissociation energy of H – X molecule, higher is the reactivity of halogen acid. Thus, decreasing order of reactivity of halogen acid is as follows : HI(296-8kJ) > HBr(363-7kJ) > HCl(430-5kJ).

Question 43.
What is Cracking? Give an example.
Answer:
Cracking: When fuel oil or lubricating oil is heated at a high temperature of 770 K in absence of air C—C and C-H bonds break to form lower paraffins or olefins. The method is called cracking or pyrolysis and has great importance in petroleum refining. In short conversion of less volatile higher hydrocarbon into more volatile lower hydrocarbons is called Cracking.

Cracking is of two types :
(a) Thermal cracking: When diesel, oil etc. are heated at high temperature, petrol is obtained. By cracking of kerosene oil, gas is prepared.

(b) Catalytic cracking: In presence of catalyst (Al₂O₃, SiO₂ etc.) cracking is achieved at low temperature. By catalytic cracking of lubricating oil, petrol is obtained.
Uses:
1. Formation of gas from kerosene oil.
2. Formation of petroleum gas from petrol.

Question 44.
A hydrocarbon ‘A’ (Vapour density = 14) decolourises Bayer’s reagent and reacts follows :

Write the name and chemical formula of A, B, C and D.
Answer:
Hydrocarbon ‘A’ with vapour density 14 has molecular mass 28, therefore it is an alkene containing 2 ‘C’ and 4 ‘H’ i.e., CH₂ = CH₂.

Hydrocarbons Class 11 Important Questions Long Answer Type

Question 1.
Explain the laboratory method of preparation of ethylene under the follow¬ing heads:
(i) Chemical equation,
(ii) Labelled diagram,
(iii) Procedure,
(iv) Two main pre¬cautions.
Answer:
Laboratory preparation of alkene (ethylene): Ethylene is obtained in laboratory by heating ethyl alcohol with excess of cone. H₂SO₄ at 170°C.
(i) Chemical equation:

(ii) Labelled diagram :

(iii) Method: 50 cc of ethyl alcohol and 100 cc of conc.H₂SO₄ is taken in a flask, then 8 gm anhydrous Al₂ (SO₄)₃ and 50 gm of sand is added. The mixture of Al₂ (SO₄)₃ and sand checks up formation of foam and facilitates the reaction to take place at 140° C. Now the flask is fitted with a thermometer, an exit tube and a dropping funnel. Flask is placed on a sand bath and fixed with a stand.

The other end of exit tube dips in wash bottle containing NaOH. Another tube from wash bottle leads to a behive shelf placed in a trough of water. A water jar is inverted over behive shelf. Flask is heated at a temperature of 150° C and simultaneously mixture of alcohol and cone. H₂SO₄ is added dropwise into the flask. Along with ethylene, CO₂ (by oxidation of alcohol) and SO₂(by reduction of H₂SO₄) are also present as impurities in flask. These impurities get adsorbed in NaOH solution and pure ethylene is collected in gas jars by downward displacement of water. Ethylene prepared by this method, is pure.

(iv) Precautions :
1. H2S04 should be more in quantity otherwise ethylene gets converted to ether.
2. On mixing small amount of Al₂(SO₄)₃ in the mixture of ethyl alcohol and cone. H₂SO₄the reaction occurs at 140°C.

Question 2.
Explain the laboratory preparation of acetylene under the following heads :
(i) Chemical equation,
(ii) Labelled diagram,
(iii) Procedure,
(iv) Two main pre¬cautions.
Answer:
Laboratory method of preparation of acetylene: The first member of alkyne homologous series, i.e., ethyne or acetylene is prepared by dropping water on calcium carbide. Acetylene obtained by this method contains impurities of PH₃ and NH₃. These are removed by passing the gas through CuSO₄ solution.
(i) Chemical equation :
CaC₂ +2H₂O → Ca(OH)₂ +CH≡ CH
Calcium carbide

(ii) Labelled diagram :

(iii) Method: Calcium carbide is placed on sand taken in a flask. The flask is fitted with a dropping funnel and a delivery tube. The air in the flask is displaced by oil gas. Water is added dropwise from the dropping funnel. The reaction is vigorous and produces much heat. Acetylene is formed. It is passed through acidified CuSO₄ solution to remove impuri¬ties and collected by downward displacement of water.

(iv) Precautions : (i) Before starting the experiment, air in the flask is replaced by oil gas because acetylene forms explosive mixture with air.
(ii) Water is added dropwise.
(iii) Sand is kept at bottom in the flask and then CaC₂ kept over it.

Question 3.
Describe laboratory method of preparation of methane with labelled diagram.
Answer:
Alkanes are obtained by heating sodium salts of fatty acids with soda lime at 630K. Mixture of caustic soda and quick lime is known as soda lime. Only NaOH takes part in the reaction and CaO keeps NaOH dry and decreases its intensity by which reaction does not occur on glass.

Mixture of sodium acetate and soda-lime is taken in a hard glass tube fitted with a delivery tube with the help of a cork. Other end of the delivery tube is immersed in a gas jar filled with water. Glass tube containing the mixture is heated when methane gas is collected over water in the gas jar.

Question 4.
An unsaturated hydrocarbon ‘A’ joins two molecules of H2 and gets reduced. It then undergo ozonolysis and gives Butane-1,4-dial, Ethanal and Propanone. Write the structure and IUPAC name of compound ‘A’. Explain the reactions.
Answer:
Hydrocarbon ‘A’ can add two molecules of H₂. Thus, ‘A’ should be alkane, diene or Alkyne. After ozonolysis reduced ‘A’ gives three ‘species’ of which one is dialdehyde, this means the molecule breaks at both the ends. Thus, A should contain two double bonds. The three species are as follows :

Question 5.
Explain Acidic nature of Acetylene with Example.
Answer:
Acidic nature of acetylene can be explained on the basis of orbital structure. In C₂H₂ carbon atom is sp hybrid state, sp hybrid carbon is smaller and hence its electronegativity is greater than hydrogen. So it pulls electron pair closer to it.

Due to this polarisation of C-H bonds occurs and partial positive charge develops on hydrogen. Therefore, hydrogen is replaced in the form of proton, easily and thus acetylene exhibits acidic nature.

Example: 1. Reaction with AgNO₃: On passing acetylene gas through ammoniacal silver nitrate solution, white precipitate of silver acetylide is formed.

2. Reaction with ammoniacal cuprous chloride: On passing acetylene gas through ammoniacal cuprous chloride solution, red precipitate of cuprous acetylide is formed.

Question 6.
What is confirmation? Describe conformation found in ethane and cyclohexane.
Answer:
Conformation: C- C bonds present in alkanes are sigma (σ) bonds which are formed by axial overlap of two hybridised carbon atoms.
Distribution of electrons in the bond due to planar nature of molecular axis can present the movement of carbon atoms on all the four sides of the bond. Keeping one carbon atom stationary, the second carbon can be rotated on all the four sides of bond axis in space.

Due to this form both the carbon atoms have various states of regular tetrahedrally linked atoms or group, various such forms in space are called conformation.
On account of rotation between the carbon-carbon single bond various spatial ar¬rangements are possible which are called as conformation.

Confirmation of C₂H₆: Two main types of conformation of ethane are :
(i) Eclipsed conformation
(ii) Staggered conformation.
The two conformation of any molecule cannot be superimposed but can be converted into one another.
If the position of one carbon atom of ethane is fixed in space and the other carbon atom is rotated in all the four sides of single bond then various conformations of ethane are possible. Out of these, the conformation which has the lowest energy is called staggered and the one having greatest energy as the eclipsed conformation.

(i) Eclipsed conformation : In eclipsed conformation all the C-H bonds of one methyl group are covered up by C-H bonds of second methyl group.

(ii) Staggered conformation: In staggered conformations the H atoms of both the methyl groups occupy position at maximum distance out of eclipsed and staggered form. It is the staggered conformation which is more stable by 12.6 kJ mol^-1. Due to less difference in energies of staggered and eclipsed forms they change one into other readily and cannot be separated.

Question 7.
In the presence of peroxide, addition of HBr in propene follows Anti- Markownikoff’s rule, but HC1 and HI does not exhibit peroxide effect.
Answer:
Addition of HCl and HI in propene does not exhibit peroxide effect. This is because H-Cl bond (430.5 kJ mol^-1) being stronger than H-Br bond (363.7 kJ mol-1) does not cleave by free radical. On the other side H-I bond (296.8 kJ mol^-1) is weak and iodine instead of joining to free radical double bond combines to form iodine molecule.

This can also be discussed with the help of thermodynamic calculations. Free radicals of hydrogen halide in addition reactions undergo enthalpy change in the followings steps :

It is clear from the above data, that both the steps of addition of HBr are exothermic which is favourable condition for the reaction. But in HCl and H one step is endothermic which is not a favourable condition for any chemical reaction from thermodynamic point of view.

Question 8.
Explain conformational isomerism in cyclohexane.
Answer:
Conformation of cyclohexane: In addition to open chain alkanes the closed chain alkanes i.e., cycloparaffins also exhibit conformational isomerism. The bond angles in cyclopropane (C₃H₆) and cyclobutane (C₄H₈) respectively are 60° and 90° therefore both the rings are under strain and are more reactive. Further they are planar molecules.

The bond angle in cyclopentane is 108° which is quite close to tetrahedral bond angle but cyclohexane ring is free from strain and its each C-C-C bond angle is 109° 28′ hence it is stable. It can be shown by two conformation in space, which are called as chair conformation and boat conformation.

Chair conformation: If is the most stable conformation of cyclohexane in which all bond angles are equal to tetrahedron bond angles and all the adjacent carbon atoms of C-H bonds are in staggered position. This conformation is free from all sorts of strain and has less energy.

Boat conformation: This is the second conformation of cyclohexane and is less stable than the chair conformation. All the carbon (C-C) bonds are present tetrahedrally in space but the H atoms present on adjacent carbon atoms are in eclipsed state. In boat confor¬mation both the H atoms shown as H„ repel each other. There is no repulsion between the H atoms of chair conformation. There is difference of 44 kJ mol^-1 energy between the two conformations.

Question 9.
Write staggered and eclipsed forms of Newman and Sawhorse. Which form will be more stable and why?
Answer:
Staggered form of Ethane is about 12-5 kJ mol^-1 of energy more stable than the eclipsed form. This is because in staggered form the hydrogen atoms of the two carbon atoms are maximum distance apart whereas in eclipsed form they are very close (sometimes overlapped atoms). Thus, in staggered form has minimum repulsive force, minimum energy and maximum stability.

Question 10.
Explain Resonance structure of Benzene.
Answer:
Resonance theory: Bond length of each C – C bond in benzene is 1.39 Å which is intermediate between single bond (1.54Å) and double bond (1.34Å).
Thus each carbon atom in benzene is supposed to be linked by partial double bond. Thus it is supposed that benzene is a resonance hybrid of following structures :

Formula (I) and (II) are Kekule structure compared to other structure their contribution towards resonance is more. These structures participate 80% to resonance. Thus benzene is supposed to be the resonance hybrid of mainly these two Kekule structures.

Energy of Dewar’s structure is high due to presence of para bond. Therefore, it contributes only 20% to resonance whereas Kekule structure is more stable and contributes 80% to resonance.

Question 11.
Explain molecular orbital structure (Model) of Benzene.
Answer:
Orbital Model of Benzene: In benzene, each carbon is in state of sp² hybridization which means there are three sp2 orbitals on each carbon atom which form an angle of 120° and the fourth unhybrid pz orbital remains perpendicular to the plane of ring.

Two sp² orbitals of each carbon undergo axial overlap with sp² orbital of neighbouring carbon atoms to form sigma bonds. The third sp² orbital overlaps with s-orbital of hydrogen to form sigma bond. As a result of this overlapping, a regular hexagon is formed which has an internal angle of 120°. It has six C – C sigma bonds and six C – H sigma bonds.

Now, there remains one p_z-orbital in each carbon atom. These p_z-orbitals are parallel to each other and perpendicular to the plane of ring.

Question 12.
Explain the free radical mechanism of halogenation.
Answer:
Halogenation reaction in alkanes occur by free radical mechanism.
1. Chain initiation: The reaction is initiated by homolysis of chlorine molecule in presence of heat and chlorine free radical is obtained.

2. Chain propagation step: Chlorine free radical displaces hydrogen and alkyl free radical is obtained.


This methyl free radical reacts with chlorine molecule and forms alkyl halide and a chlorine free radical.
This free chlorine radical again reacts with alkane to give methyl free radical. In this way, the reaction is repeated.

3. Chain termination step: At the end both the free radicals react mutually and end the chain

Free radical of halogen attacks alkyl halide and form a new free radical.

This free radical reacts with halogen to form dihaloalkane. This reaction takes place slowly till all the hydrogen atoms get displaced.

Question 13.
Explain three examples of electrophilic substitution reactions in aromatic hydrocarbon with their mechanism.
Answer:
All electrophilic reactions complete in the following steps:
(i) Electrophilic reagent is obtained by the fission of reagent.

(ii) This electrophilic reagent attacks benzene nucleus and positive charge is produced on the ring which achieve stability by resonance.
Example: 1. Halogenation: At normal temperature and in presence of a halogen carrier benzene reacts with halogen to form halobenzene.

(iii) A proton is removed from positively charged carbonium ion and product is formed.

Example: 1. Halogenatíon: At normal temperature and in presence of a halogen carrier benzene reacts with halogen to form halobenzene.

Mechanisam:
(i) Cl₂+FeCl₃ → Cl⁺+FeCl₄


2. Nitration: Nitrobenzene is produced by the reaction of benzene with cone. HNO₃ and cone H₂SO₄.
Mechanism :
HNO₃ +2H₂SO₄ →NO₂⁺+ H₃O⁺ + 2HSO₄


3. Sulphonation: Benzene sulphonic acid is produced by heating benzene with cone H₂SO₄.
Mechanism:
(i) 2H₂SO₄ → SO₃ + HSO₄^– +H₃O⁺

Question 14.
Write IUPAC name of the following compounds :
(a) CH3CH = C(CH3)2
(b) CH2== CH — C = C —CH3



Answer:

Question 15.
Give Electronic specification of directive effect of groups.
Answer:
If a substituent group is already present in benzene ring then as a result of substitution reaction, the position of the incoming group is decided on the basis of the substituent present. This is known as directive effect of that group.
Directive effect of group on the basis of electronic concept:
1. ortho and para directing group : These are such groups which being already present on the benzene ring direct the incoming group to ortho and para place.
Example : – OH, – OCH₃, – CH₃, – NH₂.
These groups due to their electron releasing tendency increase the electron density at the ortho and para place.

Due to the increase in electron density, reactivity of benzene nucleus also increases and electrophilic reagents due to high electron density enter into these places.

2. meta directing effect: These are such groups which being already present on the benzene ring direct the incoming group to meta place.
Example : – N0₂, – COOH, – SO₃H, – CN.
These groups when present in benzene ring due to mesomeric effect pull the electrons as a result of which electron density decreases mainly at ortho and para place whereas it remains unaffected at meta position i.e. at met a place electron density is high. Therefore electrophilic reagent enters at meta place.

Question 16.
Explain the halogenation in benzene.
Answer:
Example: 1. Halogenatíon: At normal temperature and in presence of a halogen carrier benzene reacts with halogen to form halobenzene.

Mechanisam:
(i) Cl₂+FeCl₃ → Cl⁺+FeCl₄

2. Nitration: Nitrobenzene is produced by the reaction of benzene with cone. HNO₃ and cone H₂SO₄.
Mechanism :
HNO₃ +2H₂SO₄ →NO₂⁺+ H₃O⁺ + 2HSO₄

3. Sulphonation: Benzene sulphonic acid is produced by heating benzene with cone H₂SO₄.
Mechanism:
(i) 2H₂SO₄ → SO₃ + HSO₄^– +H₃O⁺

Question 17.
How will you convert benzene into the following :
(i) p-nitro bromobenzene
(ii) wi-nitrochlorobenzene
(iii) p-nitrotoluene
(iv) Acetophenone.
Answer:
(i) Bromo group is o/p directing whereas nitro group is m-directing. Thus, to obtain p-nitrobromobenzene bromination and nitration are performed.

(ii) To obtain m-nitro chlorobenzene, chlorination is performed after nitration. Chloro group is o/p directing and nitro group is m-directing.

(iii) To obtain nitro toluene, nitration is performed after Friedel-Crafts reaction.

(iv) To obtain acetophenone, Friedel-Crafts Acylation (-COR) is performed.

Question 18.
Write structural formula and IUPAC name of all possible position isomers of the following compounds in which number of double and triple bonds are repre¬sented :
(a) C₄H₈ (one double bond)
(b) C₅H₈ (one triple bond).
Answer:
(a) Isomers of C₄H₈ which possess double bond are as follows :

(b) Isomers of C₄H₈ which possess triple bond are as follows :

Question 19.
How will you convert the following compounds to benzene :
(i) Ethyne,
(ii) Ethene,
(iii) Hexane.
Answer:
(i) Benzene from Ethyne :

(ii) Benzene from Ethene:

(iii) Benzene from Hexane :

Question 20.
Which of the system is not aromatic? Give reason.
(i)

Answer:

Reason: Due to presence of sp3 hybridized carbon atom, molecule is not planar. It contains 6π-electrons but its π-electron of meta, does not surround all the carbon atoms. Thus, it is not aromatic.

Reason: Due to the presence of sp³-hybridised carbon atom, the molecule is not planar. It has 4π-electrons. Thus, molecule is not aromatic because its (4n + 2)π-electrons do not possess planar cyclic electron cloud. ”

Reason : Cyclo-octatetraene is non-planar with 87r-electrons. Thus, it is not aromatic.

Question 21.
Relative reactivity of 1°, 2°, 3° halogens towards chlorination is 1: 3.8: 5. Determine the percent of all monochlorinated products of 2-methyibutane.
Answer:

Relative amount of monochlorinated products=Amount of hydrogen x Relative reactivity 1° (For monochlorinated product) = 9 ×1=9 2° (For monochlorinated product) = 2×3.8 = 7.6 3° (For monochlorinated product) =1×5 = 5
Total number of monochlonnated compounds = 9+ 7.6 +5=21.6
% amount of 1° monochlonnated product = \(\frac{9 \times 100}{21 \cdot 6} \) = 41.6%
% amount of 20 monochlonnated product =\(\frac{7.6 \times 100}{21 \cdot 6} \) = 35.18%
% amount of 3° monochLorinated product =\(\frac{5 \times 100}{21 \cdot 6}\) = 23.15%.

Question 22.
Which product will be obtained as a result of the following reaction and why :

Answer:

When Friedel-Crafts reaction takes place with higher alkyl halide like n-propyl carbocation (1 ° carbocation) rearranges to more stable isopropyl carbocation (2° carbocation) as a result of which main product obtained is isopropyl benzene.

Question 23.
Arrange the following compounds in the decreasing order of their relative reactivity towards electrophilic reagent:

Answer:
-OCH₃ (Methoxy group) is electron releasing group. It increases the electron density over benzene nucleus due to Resonance effect (+R effect). Due to this anisole is relatively more reactive towards electrophilic reagents than benzene.

In case of aryl halides, halogens .due to their -I effect are extremely unreactive by which total electron density in benzene nucleus decreases. Due to this substitution decreases.
-NO₂ group is electron attractive group. This is due to strong -I effect decreases the electron density in benzene nucleus. Due to this nitrobenzene becomes less reactive. Thus, order of reactivity of these compounds towards electrophilic reagents is as follows :

Hydrocarbons Class 11 Important Questions Objective Type

1. Choose the correct answer:

1. Bond angle H-C-H in methane is:
(a) 100.5°
(b) 109.0
(c) 109°28’
(d) 180°.
Answer:
(c) 109°28’

Question 2.
C≡C contains:
(a) 3 σ bonds
(b) One σ and three π-bonds.
(c) 3 n-bonds
(d) One σ and 2π-bonds.
Answer:
(d) One σ and 2π-bonds.

Question 3.
Ethene has:
(a) Sσ and one z-bond
(b) 6σ-bonds
(c) 4σ and 2π-bonds
(d) Sσ-bonds.
Answer:
(a) Sσ and one z-bond

Question 4.
Which is most reactive:
(a) C₂H₂
(b) CH₄
(c) C₂H₄
(d) C₂H₆.
Answer:
(a) C₂H₂

Question 5.
Represents geometrical isomerism:
(a) But-2-ene
(b) But-2-yne
(c) Butan-2-ol
(d) Butanol.
Answer:
(a) But-2-ene

Question 6.
Molecular formula of all the following compounds is C4H8. One which represents geometrical isomerism:
(a)

(b)

(c)

(d)
Answer:

(a)

Question 7.
The hybridized orbital of carbon in benzene is :
(a) sp³
(b) sp²
(c) sp
(d) dsp³.
Answer:
(b) sp¹

Question 8.
C – C single bond carbon atoms is HC ≡ C – CH = CH2 have the hybridization : ’
(a) sp³– sp³
(b) sp²-sp³
(c) sp³ – sp
(d) sp- sp².
Answer:
(d) sp- sp².

Question 9.
The number of sigma bonds in 1-butene is :
(a) 8
(b) 10
(c) 11
(d) 12.
Answer:
(c) 11

Question 10.
The double bond present between two carbon atoms in ethylene is :
(a) Two perpendicular sigma bonds
(b) One sigma and one pi bond
(c) Two perpendicular pi bonds
(d) Two pi bonds at an angle of 60°.
Answer:
(b) One sigma and one pi bond

Question 11.
The reagent that differentiates ethylene from acetylene is :
(a) Aq. alkaline KMnO₄
(b) Cl₂ dissolved in CCl₄
(c) Ammoniacal Cu2Cl₂
(d) Cone. H2SO₄.
Answer:
(c) Ammoniacal Cu2Cl₂

Question 12.
In an engine knocking is produced when the fuel:
(a) Bums slowly
(b) Bums fast
(c) Contains water
(d) Is mixed with machine oil.
Answer:
(b) Bums fast

Question 13.
The compound that is mixed in fuel to reduce knocking is :
(a) PbBr₂
(b) ZnBr₂
(c) PbO
(d) TEL (Tetraethyl lead).
Answer:
(d) TEL (Tetraethyl lead).

Question 14.
The process which can be used to prepare methane is :
(a) Wurtz reaction
(b) Kolbe reaction
(c) Reduction of alkyl halide
(d) Hydrogenation of alkene.
Answer:
(c) Reduction of alkyl halide

Question 15.
Acetylene reacts with HCl in presence of HgCl₂ and forms :
(a) Methyl chloride
(b) Acetaldehyde
(c) Vinyl chloride
(d) Formaldehyde.
Answer:
(b) Acetaldehyde

Question 16.
Propyne reacts with aqueous H₂SO₄ in presence of HgSO₄. The main product is:
(a) Propanal
(b) Propyl hydrogen sulphate
(c) Acetone
(d) Propanol.
Answer:
(c) Acetone

Question 17.
The reagent to differentiate propene from propyne is :
(a) Bromine
(b) Alkaline KMnO₄
(c) Ammoniacal AgNO₃
(d) Ozone.
Answer:
(b) Alkaline KMnO₄

Question 18.
The reaction is called:
(a) Wurtz reaction
(b) Kolbe’s reaction
(c) Sabatier-Senderens reaction
(d) Carbylamine reaction.
Answer:
(c) Sabatier-Senderens reaction

Question 19.
Ammoniacal silver nitrate solution reacts with C₂H₂ to form :
(a) Silver mirror
(b) Silver oxide
(c) Silver formate
(d) Silver acetylide.
Answer:
(d) Silver acetylide.

Question 20.
An unknown compound A has a molecular formula C4H6. When A is treated with an excess of Br₂ a new substance B with formula C₄H₆Br₂ is formed. A forms white precipitate ammoniacal AgNO₃ solution. A may be :
(a) But-1 -yne
(b) But-2-yne
(c) But-1 -ene
(d) But-2-ene.
Answer:
(b) But-2-yne

Question 21.
Group which increase activity :
(a) o,p-directing group
(b) m-directing group
(c) NO₂
(d) None of these.
Answer:
(a) o,p -directing group

Question 22.
Which of the following alkane is not obtained by Wurtz reaction :
(a) CH₄
(b) C₂H₆
(C) C₃H₈
(d) C₄H₁₀.
Answer:
(a) CH₄

Question 23.
Which of the following is non-aromatic :
(a) Benzene
(b) Cyclo-octadiene
(c) Tropolium cation
(d) Cyclopentadienal cation.
Answer:
(d) Cyclopentadienal cation.

Question 24.
Which of the following is non-aromatic :
(a)
(b)
(c)
(d)
Answer:
(c)

Question 25.
Of the following whose dipole moment is zero:
(a) cis-2-butene
(b) trans- 1 -butene
(c) 1 -butene
(d) 2-methyl-1 -butene.
Answer:
(b) trans-1 -butene

Question 26.
Strongest acid is:
(a) HC ≡ CH
(b) C₆H₆
(c) C₂H₆
(d) CH₃OH.
Answer:
(d) CH₃OH.

Question 27.
Butene -1 is converted to butane:
(a) Pd/ H₂
(b) Zn / HCl
(c) Sn / HCl
(d) Zn – Hg.
Answer:
(a) Pd/ H₂

Question 28.
Number of carbon-carbon triple bond in CaC₂ is:
(a) One σ, one π
(b) One σ, two π
(c) One σ,\(1 \frac{1}{2} \) π
(d) One σ bond.
Answer:
(b) One σ, two π

Question 29.
Reagent which distinguishes 1-butyne and 2-butyne is:
(a) Br₂ in CCl₄
(b) Dil.H₂SO₄ / HgSO₄
(c) H₂ Lindlar’s reagent
(d) Ammoniated CaCl₂.
Answer:
(d) Ammoniated CaCl₂.

Question 30.
Is produced by the reaction of n-propyl bromide and alcoholic KOH:
(a) 1 -Butene
(b) 1 -Butanol
(c) 2-Butene
(d) 2-Butanol.
Answer:
(c) 2-Butene

2. Fill in the blanks:

1. Kerosene oil is a mixture of ……………….. .
Answer:
Alkane

2. Carbon-carbon bond length is minimum in ……………….. .
Answer:
Ethyne

3. ……………….. reagent is used to distinguish between propene and propyne.
Answer:
Baeyer’s (Alkaline KMnO₄)

4. Teflon is a polymer of ……………….. .
Answer:
Tetrafluoroethylene

5. Dehydration of ethanol forms ……………….. .
Answer:
Ethene

6. ……………….. is formed by addition of H₂ on Benzene.
Answer:
Cyclohexane

7. Geometrical isomerism is found in ……………….. .
Answer:
Alkene

8. Most stable configuration is in ……………….. form.
Answer:
Staggered

9. trans isomer is ……………….. stable than cis isomer.
Answer:
More

10. ……………….. is obtained by the reaction of aluminium carbide with water.
Answer:
CH₄

11.
Answer:
C₂H₄.

3. Match the following:

I.

‘A’	‘B’
1.	(a) Anti-Markownikoff’s rule
2.	(b) Baeyer’s reagent
3.Mixture of KMnO4 + KOH	(C) Kolbe reaction
4.	(d) Dehydration
5.	(e) Decarboxylation.

Answer:
1. (e) Decarboxylation
2. (d) Dehydration
3. (b) Baeyer’s reagent
4. (C) Kolbe reaction
5. (a) Anti-Markownikoff’s rule.

II.

‘A’	‘B’
1. Method of preparation of Alkane, Alkene and Alkyne	(a) Electrophilic substitution reaction
2. Ethane is not formed with water	(b) Kolbe reaction
3. On reaction of Aluminium carbide	(c) sp2 hybridization
4. Nitration in benzene	(d) By Wurtz reaction
5. In benzene molecule each carbon atom is	(e) Methane.

Answer:
1. (b) Kolbe reaction
2. (d) By Wurtz reaction
3. (e) Methane
4. (a) Electrophilic substitution reaction
5. (c) sp² hybridization.

III.

‘A’	‘B’
1. π bond	(a) Electrophilic substitution
2. Nitration in Benzene	(b) Unsaturated hydrocarbon
3. Amine	(c) Liquefied petroleum gas
4. Alkene	(d) cis-trans isomers
5. LPG	(e) Sidewise overlapping
6. CNG	(f) O, p-group
7. Geometrical isomerism	(g) Compressed natural gas.

Answer:
1. (e) Sidewise overlapping
2.(a) Electrophilic substitution
3. (f) O, p-group
4. (b) Unsaturated hydrocarbon
5. (c) Liquefied petroleum gas
6. (g) Compressed natural gas,
7.  (d) cis-trans isomers.

4. Answer in one word/sentence:

1. Full name of TEL is.
Answer:
Tetraethyl lead

2. Which product is formed when ethylene dibromide is heated with Zn powder?
Answer:
Ethylene

3. Name the smelling agent in LPG.
Answer:
Ethyl mercaptan

4. Ethane is formed by the electrolysis of aqueous solution of potassium acetate.
Answer:
Kolbe reaction

5. Name of the reaction CH₂ =   is
Answer:
Sabatier and Senderens reduction reaction

6. write the name of the products formed in the reaction.
Answer:
(A) CH ≡ CH,
(B) CH₃ – CHO,
(C) CH₃CH₂OH,

7. Name the reaction by which Alkane is prepared by the reaction of alkyl halide and sodium.
Answer:
Wurtz-Fittig reaction.

5. State true or False: 

1. Pi ( π) bond is stronger than sigma (σ) bond.
Answer:
False

2. Alkyl group in benzene is O-,ρ- directing.
Answer:
True

3. Benzene is very reactive due to the presence of double bond.
Answer:
False

4. Due to presence of directing group benzene nucleus becomes more reactive.
Answer:
False.

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry which are most likely to be asked in the exam.

MP Board Class 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry Class 11 Important Questions Very Short Answer Type

Question 1.
How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Solution:
Molar mass of Na₂CO₃ = (2 × 23) + 12+ (3 × 16)
= 106 g mol^-1
Mass of 0.50 mol Na₂CO₃ = Number of moles × Molar mass
= 0.50 × 106 = 53g Na₂CO₃
Thus, 0.50 M Na₂CO₃ means, 53 gm of Na₂CO₃ is present in 1 litre solution.

Question 2.
What will be the mass of one ¹²C atom in g?
Solution:
Mass of 1 atom of ¹²C = \(\frac{\text { Atomic mass of } \mathrm{C}}{\text { Avogadro number }}\)
\(=\frac{12}{6 \cdot 022 \times 10^{23}}\)
= 1.9927 × 10^-23g .

Question 3.
What is the law of constant proportion?
Answer:
The law states that, “A chemical compound is always found to be made up of the same elements combined together in the same proportion by weight.”
Example : Water from different sources is collected and decomposed. After decomposition in each sample of water, it is found that the weight of hydrogen and oxygen is according to the ratio 2 : 16 or 1 : 8.

Question 4.
N₂ and H₂ combine to form NH₃. Under same condition of temperature and pressure ratio in the volume is 1 : 3 : 2. Write the name of the law which proves it.
Or,
Write Gay Lussac’s Law of gaseous volume.
Answer:
Gay Lussac’s law of combining gaseous volume :
When gases react together, they do so in volumes which bear a simple ratio to one another and to the volumes of the gaseous products provided all measurements were made under similar condition of tem¬perature and pressure.
For example,

The simple ratio in their volume is 1 : 3 : 2.

Question 5.
Write the definition of the following :
(i) Gram atomic mass,
(ii) Gram molecular mass.
Answer:
(i) Gram atomic mass.:
Gram atomic mass of an element may be defined as the mass, which is numerically equal to its atomic mass, expressed in atomic mass units (a.m.u.).

(ii) Gram molecular mass :
The gram molecular mass of an element is the mass of the substance in gram, which is numerically equal to the molecular mass of substance in atomic mass unit. It is also called one gram molecule of the substance.

Question 6.
What is empirical formula?
Answer:
Empirical formula is the simple ratio of the atoms of various elements present in one molecule of the compound.
Significance : It represents the ratio between the different atoms of the compound.
Example : Molecular formula of glucose is C₆H₁₂O₆ ratio between its elements is 1:2:1. Therefore its empirical formula is CH₂O.
Molecular formula = (Empirical formula)_n
Where, n is an integer i.e.,n= 1,2, 3 etc.

Question 7.
Define Molality of a solution.
Answer:
Molality of a solution is the number of moles of solute present in 1000 gm of solvent. It is represented by m.

Question 8.
What is mole?
Answer:
The number of elementary particles present in 12 grams of C-12 is called mole. It is equal to 6.023 x 10²³ particles. This is known as Avogadro number.
6.023 x 10²³ atoms, molecules or ions are called one mole atom, one mole molecule or one mole ions respectively.

Question 9.
What is Limiting Reagent?
Answer:
The reactant which is present in lesser amount gets consumed first. Such a reactant is called limiting reagent because it limits the product formed.

Question 10.
Write the following in scientific notation :
(i) 0.0048
(ii) 234000
(iii) 8008
(iv) 500.0
(v) 6.0012.
Answer:
(i) 4.8 × 10^-3
(ii) 2.34 × 10⁵
(iii) 8.008 × 10³
(iv) 5.0 × 10²
(v) 6.0012 × 10°

Question 11.
Calculate the molecular mass of the following :
(i) H₂O
(ii) CO₂
(iii) CH₄.
Answer:
(i) Molecular mass of H₂O = 2 × Atomic mass of Hydrogen + 1 × Atomic mass of Oxygen
= 2 × 1 + 1 × 16 = 18

(ii) Molecular mass of CO₂ = 1 × Atomic mass of Carbon + 2 × Atomic mass of Oxygen
= 1 × 12 + 2 × 16
= 44

(iii) Molecular mass of CH₄ = 1 × Atomic mass of Carbon + 4 × Atomic mass of Hydrogen
= 1 × 12 + 4 × 1
= 16.

Question 12.
What do you understand by Significant figures?
Answer:
Number of significant figures in a certain value are all the certain digits plus one uncertain digits, For example, number of significant figures in 4.028 are 4.

Question 13.
Tell the number of significant figures in the following :
(i) 0.0025,
(ii) 208,
(iii) 5005,
(iv) 126.000,
(v) 500.00,
(vi) 2.0034.
Answer:
(i) 2,
(ii) 3,
(iii) 4,
(iv) 3,
(v) 4,
(vi) 5.

Question 14.
How are significant figures rounded up?
Answer:
For rounding up a number to three significant figures, if fourth number is smaller than 5 then it is left but if the fourth number is 5 or more than 5 then 1 is added to the third significant figure.

Question 15.
Round up the following into three significant figures :
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808.
Answer:
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810.

Question 16.
What are the S.I. Units of mass, Length, Time, Temperature and amount of substance?
Answer:
Unit of mass is kilogram (kg), length is metre (m), time-second (s), temperature is kelvin (K) and amount of substance is mole (mol).

Question 17.
Convert the following into basic units :
(i) 28.7 pm
(ii) 15.15 μs
(iii) 25365 mg.
Solution:
(i) Basic unit of length is metre (m)
2.87 pm × \(\frac{10^{-12} \mathrm{~m}}{1 \mathrm{pm}}\) = 2.87 × 10^-11m

(ii) Basic unit of Time is second (s)
15.15,_μs × \(\frac{10^{6} \mathrm{~s}}{1 \mu \mathrm{s}}\)= 1.515 × 10s

(iii) Basic unit of mass is kilogram (kg)
25365 mg × \(\frac{1 \mathrm{~g}}{1000 \mathrm{mg}} \times \frac{\mathrm{lkg}}{1000 \mathrm{~g}}\) = 2.5365 × 10^-2 kg.

Question 18.
If 10 volume of dihydrogen reacts with 5 volume of dioxygen gas then how many volume of water vapour will be obtained?
Solution:

According to Gay Lussac’s law of gaseous volume
Volume ratio 2 : 1 : 2
According to the question 10 : 5 : 10
Thus, 10 Volume H₂ will react with 5 volume O₂ to form 10 volume water vapour.

Question 19.
What is the law of conservation of matter? What is the present position of this theory?
Answer:
Law of conservation of matter :
In 1744, Lavoisier gave the law of conservation of matter based on the experiments.
This law states that when a chemical reaction takes place, the total mass of the system remains the same before and after the reaction. This law is also referred to as the law of indestructibility of matters, which implies that matter can neither be created nor destroyed by any physical or chemical means.
This law can be proved by experiments.

In the light of modern investigations, law of conservation of matter :
From modern investigations, it is proved that matter can be changed to energy. In some reactions, some energy is produced which causes loss in mass.
According to Einstein’s famous equation of energy-mass relation,
E = mc²
Where, E = energy (in erg), m = mass of the substance (in gm), c = velocity of light (in cm/sec).
In reactions, which we perform in laboratories, very less quantity of energy is absorbed or produced, so the change in mass is very less and is negligible. Such types of changes can be seen clearly in nuclear reaction (atomic bomb). For ordinary reactions, the law is quite valid.

Question 20.
What is the Law of Reciprocal proportion? Explain with example.
Answer:
Law of reciprocal proportion or Law of Equivalent proportion: This law was put forward by Richter in 1792. It states that:
When two different elements unite with the same quantity of a third element the proportions m which they do so will be the same as, or some simple multiple of the proportions in which they unite with each other.
Or
Weights of two elements (say A and B) which separately combine with a fixed weight of a third element (C) are either the same or simple multiples of the weights (of A and B) in which they combine with each other.

This law can be explained with the help of the following examples:
Consider three elements carbon, oxygen and hydrogen. Element carbon and oxygen combines with a definite mass of hydrogen in the ratio given below :
In methane (CH₄), H : C = 4 : 12
and in water (H₂O),H : O = 2 : 16 or 4 : 32
According to this law if carbon and oxygen combines with each other then their ratio should be 12 : 32 or some simple multiple of it.
We know that in carbon dioxide, CO₂, ratio of C : O = 12 : 32 which is according to this law. Hence, it illustrates law of reciprocal proportion.

Question 21.
In which of the following number of atoms is maximum :
(i) 1gAu_(s)
(ii) 1g Na_(s)
(iii) 1g Li_(s)
(iv) 1g Cl_2(g).
Answer:
Number of moles of substance = \(\frac{\text { Mass of substance }(\mathrm{g})}{\text { Molar mass }}\)
Number of Atoms = Number of moles × Avogadro number
(i) 1gAu_(s) = \(=\frac{1}{197}\) Au mole atoms =\(=\frac{1}{197}\) × 6.022 × 10²³ Au atoms
= 3.057 × 10²¹ atoms

(ii) 1g Na_(s) = \(\frac{6 \cdot 022 \times 10^{23}}{23}\) = 2.618 × 10²² atoms

(iii) 1g Li_(s) = \(\frac{6 \cdot 023 \times 10^{23}}{7}\) = 8.604 × 10²² atoms

(iv) 1g Cl_2(g) = \(\frac{1}{71}\)Cl₂ mole molecule = \(\frac{1}{71}\) × 6.022 × 10²³ Cl₂ molecule
∴ (1 mole of chlorine contains 2 atoms)

= \(\frac{2}{71}\) × 6.022 × 10²³ Cl atoms
= 1.697 × 10²² atoms
Thus, maximum number of atoms are in 1g Li_(s)

Question 22.
How is molecular formula of a substance related to its Empirical formula?
Answer:
Molecular formula is represents the actual number of atoms of various elements present in one mole of a compound.
It is either similar to the empirical formula or a simple multiple of it.
Molecular formula = n (Empirical formula)
n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}\)
Example: The empirical formula of a compound is CH₂O and its molecular mass is 180.
n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}\)
\(=\frac{180}{12+12+16}=\frac{180}{30}\) = 6
Thus, Molecular formula = n x Empirical formula
= 6(CH₂O ) = C₆H₁₂O₆.

Question 23.
(i) What is law of multiple proportion? Explain with an example.
(ii) Why is it necessary to balance chemical equations?
Answer:
(i) Law of multiple proportion :
When two elements combine to form two or more compounds, different weights of a particular element which combines with definite amount of other element, bear a simple ratio to one another.
Example : Carbon monoxide (CO) and carbon dioxide (CO₂) are two compounds of elements carbon and oxygen. The weight of oxygen which combines with a definite weight of carbon (i.e., 12) are 16 and 32 respectively. Thus, the simple ratio of weight of oxygen is 1 : 2, which supports law of multiple proportion.

(ii) It is necessary to balance a chemical equation in order to satisfy the law of conservation of mass according to which the total mass of reactants is equal to the total mass of products.

Question 24.
What is the law of constant composition? How has this law been affected by the discovery of isotopes?
Answer:
This law was stated by French chemist Louis Proust (1799). It states that ‘A pure chemical compound, regardless of its source, is always made up of the same elements combined together in the fixed proportion by weight.’
Example: Regardless of its source, water is always composed of hydrogen and oxygen combined together in the ratio of 1 : 8 by weight.

Present position of the law :
Discovery of isotopes has proved that an element can have different atoms with different atomic masses but identical properties. Thus there cannot be fixed composition of any compound.
Example : The composition of water is given below :

These figures are contradictory to the law, however the law is valid as the distribution of isotopes in an element is uniform in the nature.

Question 25.
Use the data given in the following table for calculating the molar mass of Argon available in nature :

Isotopes	Isotopic Molar mass	Abundance
36 Ar	35.96775g mol-1	0.337%
36 Ar	37.96272g mol<sup	1 – 0.063%
36 Ar	39.9624 gmol-1	99.600%

Solution:
Average molar mass of Ar = Σf_i × A_i
= (0.00337 × 35.96755) + (0.00063 × 37.96272) + (0.99600 × 39.9624)
= 0.121 + 0.024 + 39.803
= 39.948 g mol^-1.

Question 26.
How many significant figures siinuid he in the following calculations.
(i) \(\frac{0 \cdot 02850 \times 298 \cdot 15 \times 0 \cdot 112}{0 \cdot 5785}\)
(ii) 5 × 5.364
(iii) 0.0125+0.7864+0.0215.

Solution:
(i) In multiplication and di ision in minimum exact number 0.112 is 3 significan
figures. Thus, three (3) significant figures should he in tLe tesult.
\(\frac{0 \cdot 02856 \times 298 \cdot 15 \times 0 \cdot 112}{0 \cdot 5785}\) = \(\frac{0 \cdot 953698}{0 \cdot 5785}\)
= 1.64857
Three significant figures = 1.65

(ii) In multiplication exact number is 5 and in the second number exact number is four. Thus, there should be 4 significant figures in the result.
5 × 5.364 = 26.82 (4 significant figures)

(iii) In addition and subtraction, in the result should not be more than the number beyond the decimal after addition (or subtraction). Thus, there should be 4 significant figures in the result.
0.0125 + 0.7864 + 0.0215 = 0.8204 Four significant figures.

Question 27.
Explain modern concept of Dalton’s atomic theory.
Answer:
Modem concept of Dalton’s atomic theory are

• According to Dalton, atom is indivisible, but now it is established that atom is divisible. ,
• It is not necessary that weights of atoms of the same element may be same, their weights may be different Amins of the same element havina different weights are called isotopes.
• It is not necessary that atomic massof different elements may be different, e.g., argon and calcium have same atomic mass i,. e., 40 a.m.u. .
• Atoms always do not combine in small numbers and simple ratio. For example, the ratio of C, H and O in sugar (C₁₂H₂₂O₁₁) molecule is 12 : 22 : 11 which is not a simple ratio.
• Atom is the smallest unit which can participate in chemical reaction.
• According to Dalton’s idea, atom is indivisible and it cannot be created: Nowadays breaking of atoms and formation of new atoms is possible by means of nuclear transmutation.

Question 28.
Write the difference between Atom and Molecule.
Answer:
Differences between Atom and Molecule

Atom Molecule	Molecule
1. Smallest particle of an element, which generally cannot exist in free state.	Smallest particle of an element compound which can exist in free state.
2. Participates in chemical reactions.	A part of molecule participates in chemical reactions.
3. It cannot be divided in chemical reactions.	It can be divided in chemical reactions.
4. All atoms of any element are same.	It is formed by combining similar or dissimilar atoms.

Question 29.
Differentiate between Molarity and Molality.
Answer:
Differences between Molarity and Molality

Molarity	Molality
1. Molarity is related to volume of solution	Molality is related to mass of solvent.
2. Molarity depends upon temperature because volume changes with the change in temperature.	Molality does not depend on temperature because mass is not affected by temperature.
3. It is denoted by ‘M’	It is denoted by ‘m’

Some Basic Concepts of Chemistry Class 11 Important Questions Long Answer Type

Question 1.
What is chemical equation? Write the ways to balance it with example.
Answer:
Method of representing a chemical reaction with the help of their chemical formulae and symbols, is called chemical equation.

Balancing of a Chemical Equation :
In general, two methods are used for balancing chemical equations :
1. Hit and trial method :
(i) First of all atoms of that element which covers smallest space are balanced on both sides of the equation.
(ii) Those equations in which oxygen, nitrogen, hydrogen, etc. gases are formed, first of all they are written in atomic state. Such an equation obtained is called atomic equation. Atomic equation is multiplied by 2 to get molecular equation.
Example 1.
Potassium chlorate on heating gives potassium chloride and oxygen.
(a) KClO₃ → KCl + O (skeleton equation)

(b) In this equation, right side oxygen atom is multiplied by 3 to get same number of
oxygen atoms on both sides.
KClO₃ → KCl + 3O

(c) But in this equation, oxygen is in atomic form. To convert it into molecular form,
the equation is multiplied by 2.
Thus, 2KClO₃ → 2KCl + 3O₂ (balanced equation)

Example 2.
Copper reacts with sulphuric acid to give copper sulphate, sulphur dioxide
and water.
(a) Cu + H₂SO₄ → CuS₄O + SO₂ + H₂O (skeleton equation) .

(b) To balance sulphur atoms on both sides, H₂SO₄ is multiplied by 2.
Cu + 2H₂SO₄ → CuSO₄ + SO₂ + H₂O

(c) Now to balance hydrogen atoms, H₂O is multiplied by 2.
Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O (balanced equation)

2. Partial equation method:
It is generally difficult to balance the complex equations by Hit and Trial method. It is observed that complex equations gets completed in two or more steps. These equations are balanced. If necessary, then partial equation is multiplied
by a whole number. Then, all the partial equations are added in such a way that the intermediate product which is not obtained in the last step gets cancelled. This way, the last equation obtained is correct and balanced.

Example : On heating Cu with cone. H₂SO₄, copper sulphur, sulphur dioxide and water is formed. This equation can be represented by the following partial equations.

Intermediate products 2H and O being present on both the sides of the equations are cancelled.

Question 2.
What is the law of multiple proportion? Explain its situation in relation to modern research.
Answer:
Law of multiple proportion :
This was stated by Dalton in 1803, and was experimentally proved by Berzelius and Stass. It states that, “When two elements combine with each other to form two or more compounds the masses of one of the elements which combine with a fixed mass of the other bear a simple whole number ratio to each other”. For example, Nitrogen forms 5 oxides.

Various masses of oxygen 8,16,24,32,40 which combine with a definite mass 14 of nitrogen is in the form of a simple ratio.

Law of multiple proportion in relation to Modem Research:

• Law of multiple proportion can be applicable in that situation when a mass of isotope of an element combines with various masses of a definite isotope of a second element.
• This law is not true for organic compounds.

Question 3.
Explain mole and Avogadro’s number. Write the definition of Atomic mass and molecular mass on the basis of mole concept.
Answer:
Mole concept:
Atom and molecule are extremely minute particles. In a very small amount of a substance also number of atoms and molecules is very large. For example One milligram (0.001 gram) of carbon contains total 5.019 × 10¹⁹ atoms. Generally in chemical studies large amount of substances are used in which number of atoms or molecules is extremely high. It is impossible to count these molecules but its knowledge is essential. Just as 12 pen is a dozen pen, 1000 metre is a kilometre, similarly group of 6.023 × 10²³ particles is called one mole. These particles can be molecules atoms or ion.

Thus mole is a unit which represents 6.023 × 10²³ particles (molecule, atoms or ion)
Number of molecules in one mole molecule = 6.023 × 10²³
Number of atoms in one mole atom = 6.023 × 10²³
Number of ions in one mole ion = 6.023 × 10²³
It also represents molar mass :
One gram mole oxygen i.e., 32 gram oxygen contain total 6.023 × 10²³ molecules. One gram mole sodium i.e., 23 gram sodium contain 6.023 × 10²³ atoms.
Formula mass of sodium chloride is 58.5. Thus 58.5 gram sodium chloride contain one mole sodium ion and one mole chloride ion.
“At same temperature and same pressure equal volume of gases contains equal number of atoms.” According to this law, at N.T.P. one gram of all gases contain 6.023 × 10²³ atoms. This number is known as Avogadro’s number and it is denoted by N.

Question 4.
What is Stoichiometry? How are the problems based on chemical equations solved by it?
Answer:
Stoichiometry is derived from the Greek word ‘stoicheion ’ which means element and ‘ metron ’ meaning measure. The word stoichiometry means measurement of amount of element or compound.

In a balanced chemical equation, a quantitative relation exists between the molecules, masses, moles and volume of the reactants and products participating a chemical equation. Problems related to these calculations are of three types :

(a) Involving Mass-Mass relationship :
In such type of problems mass of a product or reactant is given and that of the other is calculated.

(b) Involving Mass-Volume relationship:
In such type of problems mass/volume of a reactant or product is given and that of the other is calculated.

(c) Involving Volume-Volume relationship :
In such type of problems volume of
one reactant or product is given and that of the other is calculated.
To solve the above type of questions (Problems) the following procedure is adopted:

• Balanced chemical equation is written in its molecular form.
• Select the symbols and formulae of those species (atoms or molecules) whose mass/volume are either given or are known by calculation.
• Atomic mass/molecular mass/mole/molar volume of molecules and atoms included in calculation are written.
• By common mathematical calculations amount of the selected substance is calculated.
  Think about the following balanced equation :

Number of molecules (mole) is known as stoichiometric coefficient.

Question 5.
What is a chemical equation? What its limitations? How can it be made more informative?
Answer:
Chemical equation :
Method of representing a chemical reaction with the help of chemical formula and symbol of species involved in the reaction called chemical equation. For example, zinc reacts with dilute hydrochloric acid give zinc chloride and hydrogen can
be represented as
Zn + 2HCl → ZnCl₂ + H2↑

Limitations Of a chemical equation :
Following informations cannot be obtained from a chemical equation:

• Physical state of reactants and products.
• Concentration of reactants and products cannot be furnished.
• Does not give information about the condition of the reaction,
• No knowledge is obtained about the velocity of the reaction.
• No knowledge about the absorption or evolution of heat of the reaction is obtained.
• Chemical equation does not give any information whether a gas is evolved or a precipitate is formed.
• Chemical equation does not furnish any information about what precautions are necessary for reaction.
• Chemical equation does not furnish any idea about the method of the chemical reaction.
• No information regarding the nature of reaction, i.e., reversible or irreversible.
• It does not provide information about phosphorescence or explosion.
• How much time it will take in completion, this type of information is not obtained.

A chemical equation can be made more clear and informative and useful by balancing
the chemical equations and using the suitable signs. The following are the advantages of it:
(a) Number of reactants and products participating in the reaction.
(b) If atomic mass of the elements are known, then molecular mass of reactants and products can be calculated.
(c) If reactant and products are gaseous, then their volume can be calculated.
(d) It proves the law of conservation of matter.
(e) Valency and Equivalent mass of the elements can be calculated.

Some Basic Concepts of Chemistry Class 11 Important Numerical Questions

Question 1.
Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄).
Solution:
Mass percent of Element \(=\frac{\text { Mass of the element in the compound }}{\text { Molar mass of the compound }}\) × 100
Molar mass of Na₂SO₄ = (2 × 22.99) + 32.06 + (4 × 16.00)
= 142.04g
Mass percent of Sodium = \(\frac{45 \cdot 98 \times 100}{142 \cdot 04}\)
= 32.37%
Mass percent of Sulphur = \(\frac{32 \cdot 06 \times 100}{142 \cdot 04}\)
= 22.57%
Mass percent of Oxygen \(=\frac{64 \times 100}{142 \cdot 04}\)
= 45.06%.

Question 2.
Determine the Empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Solution:

Question 3.
Calculate the amount of carbon dioxide that could be produced when :
(i) 1 Mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
Solution:

(i) According to the above equation on burning 1 mole of carbon completely 44g of CO₂ is produced.
(ii) According to the above equation, for burning 1 mole of carbon 32g of oxygen is required. But only 16gm of oxygen is available. Thus, oxygen is the limiting reagent.
Thus, CO₂ obtained from 16g of oxygen = \(\frac{44 \times 16}{32}\)
= 22 gm.

Question 4.
Calculate the mass of sodium acetate (CH₃COONa) required to make 500 ml of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^-1.
solution:
Molarity M = \(\frac{w \times 1000}{M \times \text { Volume of solution (in ml) }}\)
Where w = Mass of solute
M = Molar mass of solute
Given
Molarity of solution = 0.375 M
Molar mass of solute = m = 82.0245g mol^-1
Volume of solution = 500 ml
Mass of solute w = ?
Mass of solute w = \(\frac{m \times M \times V(\text { in } m l)}{1000}\)
\(\frac{m \times M \times V(\text { in } m l)}{1000}\)
=15.379g
= 15.38g.

Question 5.
How much copper can be obtained from 100g of copper sulphate (CuSO₄)?
Solution:
Molar mass of CuSO₄ = 63.54 + 32.06 + (4 × 16)
= 159.6g mol^-1
159.6g of CuSO₄ contains Cu = 63.54g
∴ 1 g of CuSO₄ will contain Cu = \(\frac{63 \cdot 54}{159 \cdot 6}\) g
∴ 100g of CuSO₄ will contain Cu = \(\frac{63 \cdot 54}{159 \cdot 6}\) × 100 = 39.81g

Question 6.
Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively.
Solution:
First of all Empirical formula is determined,

The values given are same.
Empirical formula of iron oxide = Fe₂O₃
Calculation of Empirical formula mass = (2 × 55.85) + (3 × 16)
= 159.7gmol^-1
n = \([latex]\begin{array}{c}
\text { Molar mass } \\
\hline \text { Empirical formula mass }
\end{array}\)[/latex]
= \(\frac{159 \cdot 7}{159 \cdot 7}\) = 1
∴ Molecular formula = (Empirical formula),,
= (Fe₂O₃ )₁
= Fe₂O₃.

Question 7.
Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41gm^-1 and the mass percent of nitric acid in it being 69%.
Solution:
Molarity = \(\frac{w \times 1000}{\mathrm{M} \times \mathrm{V}_{(\mathrm{ml})}}\)
Given, d = 1.41 gml^-1 mass percent of HNO₃ = 69%
69% HNO₃ means, In 100g solution of HNO₃, 69g of HNO₃ is present.
Thus, mass of HNO₃ (solute) w = 69g
Molar mass of HNO₃ = 1.0079 + 14.0067 + (3 × 16.00)
= 63.0146 gmol^-1
Density d \(=\frac{m}{V}\)
or
V = \(\frac{m}{d}=\frac{100 \mathrm{~g}}{1 \cdot 41 \mathrm{gml}^{-1}}\)
Molarity = \(\frac{69 \times 1000 \times 1 \cdot 41}{63 \cdot 0146 \times 100}\) =15.439 M

Question 8.
If the speed of light is 3.0 × 10⁸ ms^-1, calculate the distance travelled by light in 2.00ns.
Solution:
2.00ns = 2.00 × 10^-9s (1 ns=10^-9s)
Distance covered = Velocity × Time
=3.0 × 10⁸ms^-1 × 2.0 × 10^-9S
= 6.0 × 10^-1 = 0.6m.

Question 9.
Calculate the atomic mass (average) of chlorine using the following data :

	% Natural abundance	Molar mass
35Cl	75.77	34.9689
37CI	24.23	36.9659

Solution:
Average atomic mass \(\overline{\mathrm{A}}\), is equal to the product of relative abundance F_i and their corresponding molar mass A_i.
Average atomic mass \(\overline{\mathrm{A}}\) = Σ f_i A_i = F₁ × A₁ + F₂ × A₂ + ……………
Thus, average atomic mass of chlorine \(\overline{\mathrm{A}}\) = 0.7577 × 34.9689 + 0.2423 × 36.9659 g
= 26.4959 + 8.9568
= 35.4527u ≈ 35.5u.

Question 10.
What is the concentration of sugar (C₁₂H₂₂O₁₁) in moT1 if its 20g are dissolved in enough water to make a final volume up to 2L?
Solution:
Molar mass ofsugar(C₁₂H₂₂O₁₁)(M) = (12 x 12) + (22 x 1 .0079) + (11 x 16)
= 342 g mol^-1
Given, w = 20g and volume = 2L
Molarity (M) = \(\frac{w}{M \times \text { Volume of solution in litre }}\)
\(=\frac{20}{342 \times 2}\)
= 0.0292 mol L^-1

Question 11.
If the density of methanol is 0.793 kgL^-1, what is the volume needed for making 2.5 L of its 0.25 M solution?
Solution:
Given, d = 0.793 kgL^-1 = 0.793 × 10³ gL^-1
Final volume V₂ = 2.5 L
Final Molarity M₂ 0.25 M
Initial molarity of solution = M₁ = ?
Initial volume = V₁ = ?
Molar mass of methanol (CH₃OH) = (1 × 12) + (4 × 1.0079) + (1 × 16.00)
= 32.0416 ≈ 32gmol^-1
Molarity M₁ = \(\frac{0.793 \times 10^{3} g \mathrm{~L}^{-1}}{32 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 24.781 mol L^-1
∵ Thus, M₁V₁ ≡ M₂V₂
24.781 × V₁ = 0.25 × 2.5
V₁ = \(\frac{0 \cdot 25 \times 2 \cdot 5}{24 \cdot 781}\)= 0.02522L = 25.22 ml

Question 12.
Pressure is determined as force per unit area of the surface. The S.I. unit of pressure, pascal is shown below :
1 Pa = 1 Nm^-2
If mass of air at sea level is 1034 gcm^-2. Calculate the pressure in Pascal.
Solution:
Pressure is force or mass per unit area.
Pressure = \(\frac{\text { Force }}{\text { Area }}\)

= 101332.0 Nm^-2[IN = 1kg ms^-2]
= 1.01332 × 10⁵ Pa [1Nm^-2= 1 pa]

Question 13.
A sample of drinking water was found to be severely contaminated with chloroform CHC13, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.
Solution:
(i) 15 ppm means :
∵ 10⁶ gram solution contains 15 gm CHCl3
∴ 1 gram solution contains = \(\frac{15}{10^{6}}\)
∴ 100 gram solution contains = \(\frac{15 \times 100}{10^{6}}\) = 15 x 10^-14gm ≈ 1.5 x 10^-3%

(ii) Molar mass of CHCl₃ = 12 + 1 + (3 × 35.5)
= 119g mol^-1
1.5 × 10^-3% means 100g of sample contains
1.5 × 10^-3g CHCl₃
Molarity M = \(\frac{w \times 1000}{m \times \text { Volume of sample }}\)
For water density = \(\frac{1 \cdot 5 \times 10^{-3} \times 1000}{119 \times 100}\) = 0.000126 ,
= 1.26 × 10^-4M.

Question 14.
The following data are obtained when dinitrogen and dioxygen react
together to form different compounds :

Mass of dinitrogen	Mass of dioxygen
(i) 14g	16g
(ii) 14g	32g
(iii) 28g	32g
(iv) 28g	80g

(a) Which law of chemical combination is obeyed by the above experimental
data. Give its statement.
(b) Fill in the blanks in the following conversions :
(i) 1km = ………………. mm = …………….. pm
(ii) 1mg = ………………… kg = ……………….. ng
(iii) 1mL = ………………. L = ……………… dm³.
Answer:
(a) Law of multiple proportion :
According to this law,’ “If two elements combine together to form more than one compound then the masses of one of the element which combine with a fixed mass of the other element is in the form of simple multiple ratio”.
In the given example, in the four oxides if mass of nitrogen is fixed 28g then the ratio of oxygen which combine is 32, 64, 32, 80 which is a simple ratio is 2 : 4 : 2 : 5. Thus, the given data follows the law of multiple proportion.
1km = 1km × \(\frac{1000 \mathrm{~m}}{\mathrm{lkm}}\) × \(\frac{100 \mathrm{~cm}}{1 \mathrm{~m}}\) × \(\frac{10 \mathrm{~mm}}{\mathrm{lcm}}\) 10⁶mm
1km = 1km × \(\frac{1000 \mathrm{~m}}{\mathrm{lkm}}\) ×\(\frac{1 \mathrm{pm}}{10^{-12} \mathrm{~m}}\) = 10^-15
∴ 1km = 10⁶mm = 10¹⁵pm

(ii) 1mg = 1mg × \(\frac{\lg }{1000 \mathrm{mg}}\) × \(\frac{1 \mathrm{~kg}}{1000 \mathrm{~g}}\) = 10^-610kg
1mg = 1mg × \(\frac{\lg }{1000 \mathrm{mg}}\) × \(\frac{\ln g}{10^{-9} \mathrm{~g}}\) = 10⁶10ng
∴ 1mg = 10^-610kg = 10⁶10ng

(iii) 1mL = 1mL × \(\frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}\) = 10^-3L
1mL = 1cm³ = 1cm³ × \(\frac{1 \mathrm{dm} \times 1 \mathrm{dm} \times 1 \mathrm{dm}}{10 \mathrm{~cm} \times 10 \mathrm{~cm} \times 10 \mathrm{~cm}}\) = 10³dm³
∴ 1mL = 10^-3L = 10^-3dm³

Question 15.
In a reaction
A + B₂ → AB₂
Identify the limiting reagent, if any, in the following reaction mixtures :
(1) 300 atoms of A + 200 molecules of B
(ii) 2 mole A +3 mole B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mole A + 25 mole B
(v) 2.5 mole A +5 mole B.
Solution:
Limiting reagent first gets consumed (finishes) in a reaction.. Thus, to determine it
amount of A and 13 are compared.
A + B₂ → AB₂
(i) According to the above equation 1 atom of A reacts with I molecule of B. Thus,
200 atoms of Awil1 react with 200 molecules of B.
Thus, B is the limiting reagert and A will be in excess.

(ii) According to the above equation 1 mole of A reacts with I mole of B. Thus, 2 moles
of A will react with 2 moles of B.
In this condition, A will be the limiting reagent, and B will be in excess.

(iii) According to the above equation I atom of A reacts with 1 molecule of B.
Thus, 100 atoms of A will react with 100 molecules of A.
Thus, it is a stoichiometric mixture.
Thus, there is no limiting reagent neither A nor B.

(iv) In this condition 2.5 mole of A will react with 2.5 mole of B.
Thus, B is limiting reagent and A is in excess.
(v) In this condition A will be the limiting reagent and B will be in excess.

Question 16.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation :
N_2(g) + 3H_2(g) → 2NH_3(g)
(i) Calculate the mass of ammonia produced if 2.00 × 10³g dinitrogen reacts with
1.00 × 10³g of dihydrogen.
(ii) Will any of the two reactants remain unreacted.
(iii) If yes, which one and what would be its mass.
Solution:

∵ 28g of N₂ reacts with = 6gH₂
∴ 1g N₂ will reacts with = \(\frac{6}{28}\) gH₂
∴ 2000g N₂ will reacts with = \(\frac{2000 \times 6}{28}\) = 428.57gH₂
Thus, N₂ will be the limiting reagent.N₂ limits the amount of ammonia produced,
∴ 28g N₂ produces = 34g NH₃
∴ 1 gN₂ produces = \(\frac{34}{28}\)gNH₃
∴ 2000 g N₂ produces = \(\frac{34}{28}\) × 2000 = 2428.57g NH₃
= 2429g NH₃
(ii) H₂ will be left. It is in excess.
(iii) Remaining amount of H₂= 1000 – 428.57
= 571.43g.

Question 17.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38g carbon dioxide, 0.690g of water and no other products. A volume of 10.0L (measured at STP) of this welding gas is found to weigh 11.6g. Calculate
(i) Empirical formula,
(ii) Molar mass of the gas and
(iii) Molecular formula.
Solution:
(i) ∵ 44g CO₂ = 12g Carbon
∴ 3.38g CO₂ = \(\frac{12}{44}\) × 3.38g = 0.9218g Carbon
∵ 18g H₂O = 2g Hydrogen
∴ 0.690g H₂O = \(\frac{2}{18}\) × 0.690g = 0.0767g Hydrogen
Mass of compound = 0.9218 + 0.0767 = 0.9985g
Since, only carbon and hydrogen are present in the compound.
% Amount of C in the compound = \(\frac{0 \cdot 9218}{0 \cdot 9985}\) × 100 = 92.32%
% Amount of H in the compound = \(\frac{0 \cdot 0767}{0 \cdot 9985}\) × 100 = 7.68% .

Thus, Empirical formula = CH

(ii) Calculation of molecular mass of gas:
∵ At STP, mass of 10.L of gas = 11.6g
∴ At STP, mass of 22.4L of gas =\(\frac{11 \cdot 6 \times 22 \cdot 4}{10 \cdot 0}\) = 25.984g ≈ 26g mol^-1.

(iii) Calculation of Molecular formula:
Empirical formula mass (CH) = 12 + 1 = 13
n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}\) = \(\frac{26}{13}\) =2
Molecular formula = (Empirical formula)_n
=(CH)₂
=C₂H₂. [n = 2]

Question 18.
Calculate the molarity of a solution of ethanol in water in which mole fraction of ethanol is 0.040 (assume the density of water to be 1).
Solution:
Number of moles of solute (Ethanol) in 1 litre solution will be molarity.
1L ethanol solution (dilute solution) = 1L water.
Number of moles of H₂O in 1L water = \(\frac{1000 \mathrm{~g}}{18}\) = 55.55mol
In Binary solution, two components are present.
X₁+ X₂ = 1
X_H2O = 1 – X_C2H5OH
X_H2O = 1 – 0.040 = 0.96
X_H2O = \(\frac{\cdot n_{\mathrm{H}_{2} \mathrm{O}}}{n_{\mathrm{H}_{2} \mathrm{O}}+n_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}\)
0.96 = \(\frac{55 \cdot 55}{55 \cdot 55+n_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}\)
On cross multiplication,
(0.96 x 55.55 + 0.96n_C2H5OH = 55.55
53.328 + 0.96n_C2H5OH = 55.55
0.96n_C2H5OH = 55.55 – 53.328 = 2.222
n_C2H5OH = \(\frac{2 \cdot 222}{0 \cdot 96}\)
2.3145 mol.

Question 19.
Calcium carbonate reacts with aqueous HCl to give CaCl2 and C02 according to the reaction :
CaCO_3(s) +2HCL_(aq) → CaCl_2(aq) + CO_2(g) + H₂O_(l)
What mass of CaCO₃ is required to react completely with 25ml of 0.75 M HCl?
Solution:
(i) Mass of 25 ml of 0.75 M HCl
Molarity = \(\frac{w \times 1000}{m \times V(m l)}\)
0.75 = \(\frac{w \times 1000}{36 \cdot 5 \times 25}\)
Molecular mass of HCl = 1 + 35.5 36.5 g/mol
w = 0.75 × 36.5 \(\frac{25}{1000}\) = 0.6844g.

(ii) CaCO_3(s) + 2HCl_(aq) ) → CaCl_2(aq) + CO_2(g) + H₂O₍₁₎
100g 2 × 36.5 = 73g
∵ 73g HC1 completely reacts with = 100g CaCO₃
∴ 1 g HCl completely reacts with = \(\frac{100}{73}\) g CaCO₃
∴ 0.6844g HCl will completely reacts with = \(\frac{100 \times 0.6844}{73}\)
= 0.9375g CaCO₃.

Question 20.
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO) with aqueous hydrochloric acid according to the reaction:
4HCl_(aq) + MnO_(2s) → 2H₂O_(l) + MnCl_2(aq) + Cl_2(g)
How many grams of HCl reacts with 5.0g of manganese dioxide?
Solution:
MnO₂ + 4HCl → MnCl₂ +2H₂O+Cl₂
1 mole 4 mole
87g 36.5 × 4
= 146g
∵ 87g MnO₂ reacts with 146 g HCl
∴ 1 g MnO₂ will reacts with = \(\frac{146}{87}\) g HCl
∴ 5g MnO₂will react with = \(\frac{146 \times 5}{87}\) = 8.39g HCl.

Question 21.
Calculate the number of atoms in each of the following:
(i) 52 moles of Ar
(ii) 52u of He
(iii) 52g of He.
Solution:
(i) ∵ 1 mole Ar = 6.022 × 1o²³ atoms
∴ 52 moIes Ar = 52 × 6.022 × 10²³ atoms

(ii) ∵ 4u of He = I He atom
∴ 52u of He = \(\frac{52}{4}\) He atom = 13 He atoms.

(iii) 1 mole atom of He 4g = 6.022 × atoms²³
∴ 52g of He \(=\frac{52 \times 6 \cdot 022 \times 10^{23}}{4}\) = 78.286 × 10²³ atoms
= 7.82886 × 10²²⁴ atoms.

Some Basic Concepts of Chemistry Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Volume of 2.0 gram of hydrogen gas at N.T.P. is :
(a) 224 litre
(b) 22.4 litre
(c) 2.24 litre
(d) 112 litre.
Answer:
(b) 22.4 litre

Question 2.
Molarity of pure water is :
(a) 18
(b) 50
(c) 55.6
(d) 100.
Answer:
(c) 55.6

Question 3.
Number of atoms in 12 gm of ₆C¹² is :
(a) 6
(b) 12
(c) 6.02 × 10²³
(d) 12 × 6.02 × 10²³.
Answer:
(c) 6.02 × 10²³

Question 4.
Total mass of the product at the end of chemical reaction :
(a) Always increases
(b) Always decreases
(c) Does not change
(d) Always increases or decreases.
Answer:
(c) Does not change

Question 5.
Five oxides of nitrogen N₂O, NO, N₂O₃, N₂O₄ and N₂O₅ follow which law of chemical combination:
(a) Constant proportion law
(b) Reciprocal proportion law
(c) Gay Lussac’s law of gaseous volume
(d) Multiple proportion law.
Answer:
(d) Multiple proportion law.

Question 6.
Empirical formula of a compound is CH₂. Mass of one mole of a compound is 42 grams. Its molecular formula will be :
(a) CH₂
(b) C₃H₆
(C) C₂H₂
(d) C₃H₈.
Answer:
(b) C₃H₆

Question 7.
Who proposed law of definite proportion :
(a) Lavoisier 1774
(b) Dalton 1804
(c) Ritcher 1792
(d) Proust 1799.
Answer:
(c) Ritcher 1792

Question 8.
Percentage amount of oxygen in ZnSO₄.7H₂O will be :
(a) 22.65%
(b) 11.15%
(c) 22.30%
(d) 43.90%.
Answer:
(d) 43.90%.

Question 9.
One mola solution is that in which one mole solute is present in :
(a) 1000 gram solvent
(b) 1 litre solution
(c) 1 litre solvent
(d) 22.4 litre solution.
Answer:
(a) 1000 gram solvent

Question 10.
Mass of 22.4 litre SO₂ at N.T.P. will be :
(a) 44 gram
(b) 0.44 gram
(c) 64 gram
(d) 3 gram
Answer:
(c) 64 gram

Question 11.
What will be the percent mass of copper in cupric oxide :
(a) 22.2%
(b) 79.8%
(c) 63.5%
(d) 16%
Answer:
(b) 79.8%

Question 12.
Who is considered to be the father of modern chemistry :
(a) Priestly
(b) Lavoisier
(c) Robert Boyle
(d) Rutherford.
Answer:
(b) Lavoisier

Question 13.
Mass of 112 cm³ CH₄ at S.T.P. will be :
(a) 016 gram
(b) 0.8 gram
(c) 0.08 gram
(d) 1.6 gram
Answer:
(d) 1.6 gram

Question 14.
Sometimes it is observed that there is a decrease in mass in a chemical reaction. This is observed to be against which of the following laws :
(a) Constant proportion
(b) Multiple proportion
(c) Conservation of mass
(d) Reciprocal proportion.
Answer:
(c) Conservation of mass

Question 15.
Separation of two substances by fractional distillation method is based on the difference of their:
(a) Density
(b) Volatility
(c) Solubility
(d) Crystalline shape.
Answer:
(c) Solubility

Question 16.
Maximum number of molecules is :
(a) In 54 gram nitrogen peroxide
(b) In 28 gram carbon monoxide
(c) In 36 gram water
(d) In 46 gram carbon monoxide.
Answer:
(a) In 54 gram nitrogen peroxide

Question 17.
Molecular mass of O₂ and SO₂ are 32 and 64 respectively. If one litre O₂ at 13°C and 750 mm, pressure contain N molecules, then what will be the number of molecules in two litre SO₂ at the same temperature and pressure :
molecules in tw
(a) \(\frac{\mathrm{N}}{2}\)
(b) N
(c) 2N
(d) 4N.
Answer:
(c) 2N

Question 18.
Number of moles in 180 gram water is:
(a) 1 mole
(b) 10 mole
(c) 18 mole
(d) 100 mole.
Answer:
(b) 10 mole

Question 19.
If at normal temperature and pressure, two gases are placed in two containers of same volume, then in them :
(a) Number of molecules will be same
(b) Number of atoms will be same
(c) Their masses will be same
(d) Their densities will be same.
Answer:
(a) Number of molecules will be same

Question 20.
By the decomposition of 3.4 gram H₂O₂ at N.T.P., the following volume of O₂ will be obtained:
(a) 0.56 litre
(b) 1.12 litre
(c) 2.24 litre
(d) 3.36 litre.
Answer:
(b) 1.12 litre

Question 21.
By the reaction of 2 litres N₂ and 2 litres H₂ at N.T.P, volume of ammonia obtained will be:
(a) 0.665 litre
(b) 1.0 litre
(c) 1.33 litres
(d) 2.66 litre.
Answer:
(c) 1.33 litres

Question 22.
On burning 2 moles C₂H₅OH in excess of air, C0₂ obtained will be :
(a) 88 gram
(b) 176 gram
(c) 44 gram
(d) 22 gram.
Answer:
(b) 176 gram

Question 23.
12 gram Mg (atomic mass = 24) on reacting completely with acid produce H₂ whose volume at N.T.P. will be :
(a) 22.4 litre
(b) 11.2 litre
(c) 44.8 litre
(d) 6.4 litre.
Answer:
(c) 44.8 litre

Question 24.
12 gram At (atomic mass = 27) reacts with ………… gm oxygen completely :
(a) 8 gram
(b) 16 gram
(c) 32 gram
(d) 24 gram.
Answer:
(d) 24 gram.

Question 25.
2.76 gram silver carbonate (Atomic mass of Ag = 208) is heated vigorously. The
weight of the residue obtained will be:
(a) 2.48 gram
(b) 2.16 gram
(c) 2.32 gram
(d) 2.64 gram.
Answer:

Question 26.
Number of gram atoms of sulphur (atomic mass of S = 32) in 80 gram Sulphur is:
(a) 2.5
(b) 3.2
(c) 5
(d) 10.
Answer:
(a) 2.5

Question 27.
Ammonia gas is formed by the equation N₂ + 3H₂ → 2NH₃. Nitrogen gas required to prepare 10 litre ammonia will be:
(a) 5 litre
(b) 15 litre
(c) 75 litre
(d) 10 litre.
(a) 5 litre

Question 28.
Volume of 0.1 mole gas at N.TP. will be:
(a) 22.4 litre
(b) 2.24 litre
(c) 0.224 litre
(d) 22400 litre.
Answer:
(b) 2.24 litre

Question 29.
10 mole of water is:
(a) 10 gram
(b) 1oo gram
(c) 18 gram
(d) 180 gram.
Answer:
(d) 180 gram.

Question 30.
Information not obtained by chemical equation is :
Zn_(s) + 2HCl_(l) → ZnCl_2(l) + H_2(g)
(a) Physical state of reactants and products
(b) Concentration of reactants and products
(c) Heat change in the reaction
(d) Direction of reaction.
Answer:
(c) Heat change in the reaction

Question 31.
Atomic number olDeuterium is:
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(a) 1

Question 32.
Molarity of a solution formed by 7.1 gram Na₂SO₄ in 100 ml water is (Atomic mass of Na = 23,S = 32,O = 6):
(a) 2.0 M
(b) 1.0 M
(c) 0.5 M
(d) 0.05 M.
Answer:
(c) 0.5 M

Question 33.
A compound contains 50% C, 50% 0 and its molecular mass is about 290. Its
molecular formula is:
(a) CO
(b) C
(c) C₁₂ O₉
(d) C₃O₃.
Answer:
(c) C₁₂ O₉

Question 34.
A solution contains 20 moles of solute and total number of moles Is 80. Mole
fraction of the solute will be:
(a) 2.5
(b) 0.25
(c) 1
(d) 0.75.
Answer:
(b) 0.25

Question 35.
Number of molecules in 16 gram Methane is: ,
(a) 3.0 × 10²³
(b) 6.02 × 10²³
(c) \(\frac{16}{6 \cdot 02}\) × 10²³
(d) \(\frac{16}{3 \cdot 0}\) × 10²³
Answer:
(b) 6.02 × 10²³

Question 36.
Number of molecules (approx.) in 4.25 gm NH₃ is :
(a) 1 × 10²³
(b) 2 × 10²³
(c) 4 × 10²³
(d) 6 × 10²³
Answer:
(d) 6 × 10²³

Question 37.
Aqueous solution of 6.3 gm Oxalic acid (dihydrate) was diluted to 250 ml. To neutralize 10 ml of this solution completely, volume of 0.1 N NaOH required will be:
(a) 40 ml
(b) 20ml
(c) 10ml
(d) 4ml.
Answer:
(a) 40 ml

Question 38.
Weight of how many mole electrons will be 1 kilogram :
(a) 6.023 × 10²³
(b) \(\frac{1}{9 \cdot 108}\) × 10³¹
(c) \(\frac{6 \cdot 023}{9 \cdot 108}\) × 10⁵⁴
(d) \(\frac{1}{9 \cdot 108 \times 6 \cdot 023}\) × 10⁸
Answer:
(d) \(\frac{1}{9 \cdot 108 \times 6 \cdot 023}\) × 10⁸

Question 39.
Number of significant figures in 0.0500 is :
(a) One
(b) Three
(c) Two
(d) Four.
Answer:
(b) Three

Question 40.
Number of significant figures in a correctly written answer of sum of 29.4406, 3.2 and 2.25 will be :
(a) Three
(b) Four
(c) Two
(d) Five.
Answer:
(a) Three

Question 41.
Density of a solution prepared by dissolving 120 g. Urea (Molecular mass = 604)
in 1000 gm water is 1.15g/ml. Molarity of the solution is :
(a) 1.78 M
(b) 1.02 M
(c) 2.05 M
(d) 0.50 M.
Answer:
(c) 2.05 M

Question 42.
What will be the density (in gm L^-1) of 3.60 M sulphuric acid which according to mass is 29% ? (Given Molar mass of sulphuric acid = 98 gm mol^-1):
(a) 1.45
(b) 1.64
(c) 1.88
(d) 1.22.
Answer:
(d) 1.22.

Question 43.
Percentage of water of crystallization in a pure sample of green vitriol (FeS0₄.7H₂0) is:
(a) 45.00
(b) 54.23
(c) 45.32
(d) 44.22.
Answer:
(c) 45.32

Question 44.
Mass of electron is 9.108 × 10^-31 kg. Its significant figure is :
(a) One
(b) Two
(c) Three
(d) Four.
Answer:
(d) Four.

Question 45.
Which of the following number has maximum significant figures :
(a) 0.00453
(b) 4.8046
(c) 5.643
(d) 5.6 × 10³ cm.
Answer:
(b) 4.8046

Question 46.
Volume occupied by one molecule of water (Density = 1g cm^-3) is :
(a) 3.0 × 10^-23cm³
(b) 5.5 × 10²³cm³
(c) 9.0 × 10^-23cm³
(d) 6.02 × 10^-23cm³.
Answer:
(a) 3.0 × 10^-23cm³

2. Fill in the blanks:

1. Deuterium has …………………. atom.
Answer:
one

2. Empirical formula of benzene is ………………..
Answer:
CH

3. Under similar conditions of temperature and pressure equal volume of all gases contain equal number of molecules. This was proposed by …………………….
Answer:
Avogadro

4. Number of moles in 180 gram water will be ………………….
Answer:
10 mol

5. S.I. unit of atomic mass is ……………………..
Answer:
a.m.u(u)

6. Kilogram/metre³ is the unit of …………………
Answer:
Density

7. Five oxides of nitrogen N₂O, NO, N₂O₃, N₂O₄, N₂O₅ obey ………………….. law of chemical combination.
Answer:
Multiple proportion

8. 1 mole CO₂ contain ……………………. carbon atom.
Answer:
6.023 × 10²³

9. Mass of 22.4 litre SO₂ at N.T.R is …………………..
Answer:
64 gram

10. Two gases placed if in two containers of equal volume, under normal temperature
and pressure then the number of molecules in them will be ………………….
Answer:
same

11. On burning 2 mole C₂H₅OH in excess of air …………………… CO is obtained.
Answer:
176 gram

12. Volume of 2 gram H₂ at N.T.P. is …………………
Answer:
22.4 litre

13. At the end of a chemical reaction, the total mass of the reactant remains ………………….
Answer:
Unchanged

14. Simplest formula of a compound with molecular formula of benzene is …………………..
Answer:
CH

15. A compound contains C = 40.6%, H = 6.5%, O = 52.8%. Its empirical formula will
be …………………
Answer:
CH2O

3. Match the following:
I.

‘A’(Prefix)	‘B’ (Multiples)
1. Micro	(a) 106
2. Deca	(b) 109
3. Mega	(c) 10-6
4. Giga	(d) 10-15
5. Femto	(e) 10

Answer:
1. (c) 10^-6
2. (e) 10
3. (a) 10⁶
4. (b) 10⁹
5. (d) 10^-15

II.

‘A’	‘B’
1. Volume of one mole molecules of gas at N.T.P.	(a) Empirical formula × η
2. Molecular formula	(b) Molality
3. Number of mole’s of solute per 1000 gm solvent	(c) Molarity
4. Substance made up of same type of atom	(d) 22.4
5. Number of moles present in 1 litre solution	(e) Element.

Answer:
1. (d) 22.4
2. (a) Empirical formula × η
3. (b) Molality
4. (e) Element.
5. (c) Molarity

4. Answer in one word/sentence:

1. Standard unit of atomic mass is
Answer:
Atomic mass unit (a.m.u.)

2. Gram molecular mass of O₂ is
Answer:
32 gram

3. The ratio by weight of Na and Cl elements in NaCl obtained from various sources is 23 : 35.5. This is proved by which law?
Answer:
Law of constant or definite proportion

4. 16 and 32 gram by weight of oxygen combine separately with 28 gram N₂ to form two oxides NO and N₂O₂. This is proved by which law.
Answer:
Law of multiple proportion

5. Number of molecules 6 023 × 10²³ present in one gram molecular mass of any substance is called.
Answer:
Mole or Avogadro number

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 14 Environmental Chemistry which are most likely to be asked in the exam.

MP Board Class 11th Chemistry Important Questions Chapter 14 Environmental Chemistry

Environmental Chemistry Class 11 Important Questions Very Short Answer Type

Question 1.
Define Environmental chemistry.
Answer:
Environmental chemistry is a branch of science which deals with the study of effects of chemicals on our surroundings (like origin, transport, reactions and effects and fates of chemical species etc.).

Question 2.
Explain Tropospheric pollution.
Answer:
The tropospheric pollution occurs because of the presence of undesirable gaseous and solid particles in the air. In the troposphere mainly the following gaseous and particulate pollutants are present.

• Gaseous air pollutants are oxides of sulphur, nitrogen and carbon dioxide, hydrogen sulphide, ozone and other oxidants.
• Particulate pollutants are dust, fumes, mist, spray, smoke etc.

Question 3.
Make a list of the gases responsible for greenhouse effect.
Answer:
For greenhouse effect mainly CO₂ gas is responsible. Apart from CO₂, other greenhouse gases are methane, water vapour, nitrous oxide, chlorofluorocarbon (CFC) and ozone.

Question 4.
What do you mean by Biochemical oxygen demand BOD?
Answer:
The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water is called Biochemical oxygen demand (BOD). Amount of BOD in the water is a measure of the amount of organic material in the water, in terms of how much oxygen will be required to break it down biologically. For pure and clean water BOD value is less than 5 ppm.

Question 5.
Due to greenhouse effect, earth’s temperature is increasing. Which substances are responsible for greenhouse effect?
Answer:
Greenhouse gases like carbon monoxide, methane, nitrous oxide, ozone and chlorofluorocarbons (CFCs) are responsible for greenhouse effect. These gases absorb the radiations passing through the earth’s surface and increase the earth’s temperature.

Question 6.
Ozone is a poisonous and a strong oxidizing reagent still its presence in the upper part of stratosphere is very important. What will happen if ozone is completely removed from this region?
Answer:
Ozone prevents the harmful ultraviolet rays of the sun to reach the earth’s surface due to which human life is protected by the harmful effects of these ultraviolet rays.
If ozone is completely removed from the upper region of stratosphere, then ultraviolet rays will reach the earth and cause various diseases like skin cancer etc.

Question 7.
What are the sources of oxygen dissolved in air?
Answer:
The sources of oxygen dissolved in water are :

• Photosynthesis,
• Natural Aeration,
• Mechanical Aeration.

Question 8.
Oxygen dissolved in water is very important for aquatic life. Which factor is responsible for the reduction of dissolved oxygen in water?
Answer:
For the reduction of dissolved oxygen in water, discharge of phosphate and nitrate fertilizers, detergents, animal excreta, food, paper and pulp industry, organic waste from industries are responsible. Micro-organisms which oxidize organic substances also use dissolved oxygen. Along with this, photosynthesis stops at night but aquatic plants and animals continuously breathe due to which deficiency of dissolved oxygen in water is caused.

Question 9.
What are Biodegradable and non-biodegradable pollutants?
Answer:
Biodegradable pollutants are substances which are decomposed by organisms.
Example: Excreta, cow dung, fruits and vegetable peels.

Non-biodegradable pollutants are substances which are not decomposed by organisms.
Example: Mercury, Lead, DDT, glass, plastic etc.

Question 10.
What is pollution?
Answer:
In a balanced environment, every component is present in a definite amount and ratio. Sometimes, in the environment amount of one or more component either increases than required or harmful components enter into the environment which are harmful for living organism. This is known as pollution.

Question 11.
What are pollutants?
Answer:
Such substances whose amount increases up to a certain limit in the atmosphere due to which it becomes harmful for plant and animal kingdom are called pollutants. They are as follows:
CO, CO₂, NO, N₂, S₂ etc.

Question 12.
What are contaminants?
Answer:
Some substances do not exist in the environment but are made in artificial form by man. These substances pollute the environment without getting destroyed. They are called contaminants.

Question 13.
Write the names of gases or chemical responsible for the depletion of ozone layer.
Answer:
Nitric oxide, atomic oxygen and chlorofluorocarbon present in the atmosphere is responsible for the depletion of ozone layer.

Question 14.
What are greenhouse gases?
Answer:
CO₂, ozone and water vapour are known as greenhouse gases. They possess the property of absorbing infrared radiations. Thus, they are known as greenhouse gases.

Question 15.
What is polluted air?
Answer:
If air contains such substances which are harmful for health of organisms, then such air is called polluted air. Polluted air mainly contains suspended molecules of CO₂, SO₂, SO₃, CO.

Question 16.
Why is depletion of ozone observed Antarctica?
Answer:
Because in the other part of stratosphere, free chlorine gets converted to chlorine molecules, whereas at Antarctica compounds produced get surrounded by chlorine free radical which cause depletion of ozone.

Question 17.
What is the utility of BOD measurement in a water sample?
Answer:
BOD is measure of pollution caused by organic biodegradable substances present in a given sample of water. Lower value of BOD indicates that less amount of organic waste is present in water.

Question 18.
Oxidation of sulphur dioxide to sulphur trioxide is a slow change in the absence of a catalyst but in presence of atmosphere it oxidizes readily. Why is it so? Also give chemical equation for the conversion of SO₂ to SO₃.
Answer:
In the oxidation of SO₂ to SO₃, particulate matter present in polluted air acts as a catalyst.

Question 19.
How is ozone produced in stratosphere?
Answer:
In stratosphere, ozone is produced by the action of ultraviolet radiations on dioxygen (O₂) molecule. Ultraviolet rays decompose oxygen molecule to oxygen atom. Oxygen atom combine with oxygen molecule to form ozone molecule.

Question 20.
What is chlorosis?
Answer:
In plants, chlorophyll is formed slowly This is due to presence of SO₂. This pollution is called chlorosis.

Question 21.
What is Metathesis?
Answer:
Metathesis is the name of that science, in which utility of scientific methods for a common man is studied.

Question 22.
What are Primary and Secondary Pollutants?
Answer:
Primary Air Pollutants: Air pollutants which directly pollute the air by natural or human activities are called primary pollutants. Like CO, NO, CO_2, SO₂.
Secondary Air Pollutants: Such air pollutants which are prepared by the chemical reactions between pollutants or by the reaction of gases and pollutants in the atmosphere are called secondary pollutants.
Like: SO₃+H₂O → H₂SO₄.

Question 23.
What is photochemical smog?
Answer:
Mixture of smoke and fog is known as smog. When higher concentration of oxidizing agents are present in this smog, it is called photochemical smog. Smoke is an air pollutant, in which various solid particles are suspended in the atmosphere due to incomplete combustion of fuel substances.

Question 24.
What is Acid rain?
Answer:
Due to air pollution, concentration of various gases like CO₂, SO₂, SO₃, NO₂ etc. increases in the atmosphere. These gases dissolve in raindrops according to their solubilities and form acid. These drops fall as rain and are known as acid rain.
CO₂+H₂O → H₂CO₃
SO₃+H₂O → H₂SO₄.

Question 25.
When is CO₂ known as pollutant?
Answer:
Normal amount of CO₂ in the atmosphere is never harmful, but plants prepare their food by it. But, when due to various activity its concentration becomes more than required, then it disturbs the natural balance and become harmful.

Question 26.
Carbon dioxide is considered to be more responsible for greenhouse effect. Why?
Answer:
Water vapour is found to be very close to the earth, and ozone is found at the upper part of the atmosphere. Compared to this CO₂ is evenly found in all places of the atmosphere. Therefore, for greenhouse effect, CO₂ is found to be more responsible because CO₂ possess the tendency to absorb the infrared radiations of the sun and produce greenhouse effect.

Question 27.
Why is acid rain considered as a threat to Tajmahal?
Answer:
Tajmahal is made of marbles. The acid rain contains H2S04 which attacks the marble (CaCO₃).
CaCO₃(s) + H₂SO₄(aq) → CaSO₄(aq) + CO₂+ H₂O(l)
As a result, the marble becomes pitted and weakened mechanically. Therefore, the monument is being slowly eaten away and marble is getting discoloured and lustreless.

Question 28.
A person uses water supplied by municipality. Due to shortage of water, he started using underground water. He felt laxative effect. What could be the cause?
Answer:
Laxative effect can be felt only when the concentration of sulphate present in water is greater than 500 ppm otherwise at moderate concentration sulphate is harmless.

Question 29.
What do you understand by Green chemistry? How is it helpful to control environmental pollution?
Answer:
Green chemistry is a technique to utilize the known knowledge and principles of chemistry and other sciences by which harmful effects of environment can be reduced. Green chemistry is a process of production which can bring minimum pollution or harm to the environment. Co-products produced in a process if not removed in a beneficial way increase pollution in the environment. Such processes are not only harmful for environmental view but are also costly. Along with modified work, use of present knowledge to reduce chemical harm is the basis of Green chemistry.

Question 30.
Why is carbon monoxide more dangerous than carbon dioxide? Explain.
Answer:
Carbon monoxide is a poisonous gas. It combines with haemoglobin present in blood to form carboxyhaemoglobin. It is 300 times more stable than oxyhaemoglobin. When the percentage of carboxyhaemoglobin in the blood becomes 3-4% then ability of the blood to carry oxygen becomes very less.

Due to minimum oxygen, headache, weak eyesight, reduction in awareness and judgement, cardiovascular diseases etc. are produced. Carbon dioxide gas does not combine with haemoglobin. Therefore, it is a less harmful pollutant. Increase in its amount in the environment leads to global warning (increase in earth’s temperature).

Question 31.
What is Fly ash pollutant?
Answer:
Fly ash are small ash particles formed by the combustion of fossil fuels which are carried into the air by gases produced during combustion.

Question 32.
What is contaminated water? Write name of diseases caused due to contaminated water.
Answer:
Water in which due to pollution various harmful products of living beings and micro-organisms are present is called contaminated water. Diseases caused due to contaminated water are :

• Cholera,
• Jaundice,
•  Diarrhoea,
• Skin disease.

Question 33.
What do you understand by Global warming?
Answer:
Increase in earth’s temperature, due to absorption of sun’s energy by various gases like CO₂, methane, chlorofluorocarbon etc. present in large amount in the atmosphere is called Global warming.

Question 34.
What is Ionosphere?
Answer:
A thick layer about 40 km at a height of about 50 to 90 km from the earth’s surface is called Mesosphere. All the gases in this region exist in ionized state, therefore it is called Ionosphere.
O₂ + O⁺ → O₂ + O
O⁺ + N₂ → NO⁺ + N
N₂+O₂ → N₂ + O₂.

Environmental Chemistry Class 11 Important Questions Short Answer Type 

Question 1.
How is the poisonous effect of CO produced on man and animals?
Answer:
CO has poisonous effect on man and plants.
(i) It combines with haemoglobin of the blood more strongly than oxygen. CO + Hb → CO – Hb (Carboxy haemoglobin).
As a result of this amount of haemoglobin available in the blood for the transport of oxygen to the body cells decreases. The normal metabolism is thus, impaired due to less O₂ level. This will cause suffocation and will ultimately lead to death. Carbon monoxide if present in air can cause mental impairment, respiratory problems, muscular weakness and dizziness.

(ii) A high concentration of CO (100 PPM or more) will harmfully affect the plants causing leaf drop; reduction of leaf size and premature aging etc.

Question 2.
Write the harmful effects of sulphur dioxide.
Answer:
Harmful effects of SO₂ :

• SO₂ harms the respiratory tube and lungs due to which various lung diseases, cancer etc. can be caused.
• It stops the growth of plants. Due to this, change of green leaves to yellow is called chlorosis.
• Acid rain due to SO₂ harms the monuments and stones.
• SO₂ gas enhances corrosion.

Question 3.
Write harmful effects of nitrogen oxide.
Answer:
Harmful Effects of Nitrogen oxide :

• Presence of gases like NO, NO₂ etc. more than required cause irritation in eyes.
• Its excess amount may lead to lung and heart diseases.
• Oxides of nitrogen enhance corrosion in metals.
• Reduce rate of photosynthesis.
• Reduce the brightness of dyes.

Question 4.
A farmer was using pesticides on his farm. He used the produce of his farm as food for rearing fishes. He was told that fishes were not fit for human consumption because large amount of pesticides had accumulated in the tissues of fishes. Explain how did it happen?
Answer:
Pesticides transfer from the crops to fish as their food. This way, pesticides through the food chain reach the higher level from the lower level. With time, concentration of pesticides in fish increased to such a level that serious metabolic and body active states were produced. That is why, fishes were not suitable as food for man.

Question 5.
For dry cleaning, in place of tetrachloroethene along with suitable deter-gent liquified carbon dioxide is an alternative. Which loss of the environment can be stopped by the use of tetrachloroethene? Is the use of liquified carbon dioxide with detergent a completely prevention from the view of pollution?
Answer:

• Tetrachloroethene (Cl₂C = CCl₂) is a possible cause of cancer, it also pollutes ground water. Its harmful effects can be controlled by the use of liquified carbon dioxide and suitable detergent.
• Use of liquified CO₂ with suitable detergent is not completely preventive because maximum detergents are non-biodegradable and they also pollute the water. Along with this, liquified CO₂ enter the environment at the end and increase green house effect.

Question 6.
What are the ill effects of greenhouse effect?
Answer:
Main cause of greenhouse effect is the presence of large amount of CO₂ in the atmosphere which absorbs the infrared radiations and increase the temperature of the environment. This way, due to increase in temperature of the environment weather will change. Sun rays gives birth to serious disease like cancer.

Some places will suffer from drought, hot air will flow in some places, some places will have strong storm and somewhere floods may come. Huge icecaps of Arctica and Antarctica will melt, by which sea level will rise and places near sea shores will submerge.

Question 7.
What is smog? Explain its mechanism with its types.
Answer:
Mixture of smoke and fog is known as smog.
Types of Smog:

1. Classical smog: SO₂ gas mixed with fog combines with smoke to form classical smog.
2. Photochemical smog: In a Sunny place, primary air pollutant form secondary pollutant by photochemical reaction. They are oxidizing in nature. Where pollutant gas is more, wind does not blow, there with smoke, smog is formed.

Mechanism :
1. If automobile fuel is used before sunrise, then mainly CO and NO gas are released. With contact of air it forms NO₂ gas.
2NO + O₂ → 2NO₂

2.By the effect of ultraviolet rays active oxygen is released by NO₂.
NO₂ → NO + O
3. This active oxygen converts the hydrocarbon released from the fuel to free radical.
4. These organic free radicals form undesirable compounds by chain reactions and these pollutants form smog.

Question 8.
Discuss the causes of water pollution.
Answer:
Following are the causes of water pollution :

• Waste products from big factories are dumped in sea, rivers or lakes.
• Various fertilizers, manure and insecticides are used in fields.
• All these products mix with the soil of fields, dissolve in rainwater and fall in rivers and pools.
• Atomic electric centre pollute the water.
• Due to increase in population, colonies are formed at the river banks, and waste from the houses pollute the water.
• In rural areas, people take bath, wash clothes in the rivers which pollute water.

Question 9.
How is synthetic greenhouse prepared?
Answer:
Synthetic greenhouse: In nature, coating of CO₂ is forming greenhouse, but synthetic greenhouse can be synthesized by studying its mechanism.

Actually the transparent glass roof and wall of the glasshouse allow sun rays to pass through and strike the surface of the house. The reflected radiation is of longer wavelength than the incident radiation. A significant portion of reflected radiation absorbed by glass. As radiation of longer wavelength (Infrared radiation) generates heat, this causes rise in tem¬perature inside the glasshouse.

Question 10.
You have dug pits in your agricultural area or garden for producing compost. Discuss the process of production of best compost in the light of bad odour, Hies and recycling of waste.
Answer:
For producing compost, pits should be dug in suitable place by which we can be prevented from bad odours and flies. Biodegradable domestic waste like tea leaves, fruits and vegetable waste etc. should be put in the compost pit and should be covered by thin layer of sand. After some time, these waste materials, by the effect of heat and bacteria get converted to compost.

Compost pits should be covered by sand by which flies cannot enter and bad odour is minimum.
Non-biodegradable waste like plastic, glass, metal shavings, polythene bags are sent for recycling. By recycling useful products are obtained from waste products.

Question 11.
How are domestic waste used as manure?
Answer:
Domestic wastes are collected in small containers and sent to disposal area. Here biodegradable waste and non-biodegradable waste are separated from the garbage. Biodegradable waste like vegetables and fruit waste, animal waste etc. are in open ground beneath the soil where it degrades and after some days get converted to manure.

Question 12.
Write the ill effects of water pollution and how is it controlled.
Answer:
III effects of water pollution:

• By the use of polluted water, cholera, jaundice, typhoid like contaminated diseases are spread.
• Due to discharged excreta, amounf of oxygen dissolved in water decreases due to which aquatic organisms die.
• Due to soap, detergent waste, polluted water becomes a danger for the life of fish.

Control:

1. Discharged excreta should not be thrown in rivers, pools etc. by chemical reaction should be converted to manure.
2. Atomic identification below the sea should be banned.
3. By the process of recycling cow dung and excreta should be converted to useful material.

Question 13.
What is soil pollution? How can soil pollution be controlled?
Answer:
Soil Pollution: Any undesirable change in physical, chemical or biological properties of soil which has harmful effect on human beings and other organisms or the natural character and utility of soil or earth is destroyed is called soil pollution.
Control:

• Industrial waste should be treated in the factories only and then dumped on earth.
• Use of fertilizers, insecticides and fungicidal medicines should be minimized.
• Cutting of trees should be stopped to prevent soil erosion and destruction of the layer of earth.

Question 14.
Write the ill effects of depletion of ozone layer.
Answer:
Following is the ill effects of ozone layer :

• Entrance of ultraviolet rays directly on earth without obstruction can lead to various diseases like cancer.
• Ultraviolet rays decrease the immunity of body.
• CFC increases greenhouse effect, as a result temperature of earth rises.

Control:

1. Production of CFC should be reduced.
2. Search the option of CFC.

Question 15.
Explain the mechanism of depletion of ozone layer.
Answer:
The mechanism of depletion of ozone layer are the following :
1. Depletion by Nitrogen oxide: Some amount of N₂O is present in the stratosphere. It combines with atomic oxygen to form NO.
N₂O+O → 2NO
NO gas combines with O₃ to form NO2
NO + O₃ → NO₂ + O₂.
NO₂ breaks by atomic oxygen to form NO again
NO₂+O → N0 + O₂
This cycle of NO is the cause of depletion of O₃.

2. Depletion by C.F.C.: It produces free radical by photochemical decomposition.

Formation of free radicals is continuously depleting ozone layer.

Environmental Chemistry Class 11 Important Questions Long Answer Type

Question 1.
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for fish kill.
Answer:
Large amount of phosphate and nitrate salts increases the number of phytoplanktons in water. These phytoplanktons use the dissolved oxygen in water to such a large extent, that it is not available in sufficient amount for the respiration of other organisms. Along with it excessive population of bacteria start the decomposition of organic substance in water like leaves, grass etc. It uses the dissolved oxygen in water. Micro-organisms (like algae) are in such a larger amount that they cover the water level as a sheet and prevents the penetration of sunlight in the water level as a result photosynthesis is obstructed. By all these processes, amount of dissolved oxygen decreases 6 ppm. Thus, growth of fish is stopped and they do not remain alive.

Question 2.
What would have happened, if there would have been no greenhouse gases in the atmosphere?
Answer:
Carbon dioxide, methane, water vapour, CFCs and ozone are green house gases. These gases absorb the developed heat by the objects situated near the earth’s surface and heat it up. This is known as natural green house effect because it controls temperature and makes life better on earth. If there would have been no green gases in the atmosphere, then there would have been no vegetation or life present on earth (because earth would have converted to a cool house).

Question 3.
Into how many parts is the atmosphere divided. Explain the chemical reactions occuring in every region.
Answer:
Atmosphere is divided into four parts :

1. Troposphere,
2. Stratosphere,
3. Mesosphere,
4. Thermosphere.

1.Troposphere: The region upto a height of 8 to 12 km from the earth’s surface is called Troposphere. Its main components are N₂, O₂, CO₂ and H₂O. Fossil fuels bum in oxygen to form CO₂.
CH₄ +2O₂ → CO₂ +2HzO
In the troposphere, compounds are decomposed by various bacteria.
Bacteria compound + O₂ → CO₂ + H₂O
Photosynthesis also occur in this layer.

2. Stratosphere: About 10 to 50 km, layer of 40 km thickness is known as Stratosphere.
(a) Photochemical decomposition of oxygen.

Excited oxygen molecule 02 liberate radiation of 636, 630, 558nm wavelength in visible region. This is known as airglow.

Oxygen molecule gets excited and lose electron by which positively charged oxygen ion is formed.
O₂ +O → O₃
O₃ molecule again decomposes by ultraviolet rays.

O₃ + O → O₂ + O₂

To provide stability to ozone, others like N₂, O₂ are required which absorb the extra energy and stop the decomposition of O₃.
O + O₂+M → O₃ + M
Alternatively, if species like NO or OH are present in stratosphere, then they decompose o3.
O3 + OH → O₂ + HOO
O3 + NO →NO₂ + O₂
NO₂ +O → NO + O₂
N₂ molecule undergo self decomposition by radiation and combine with O₂ to form NO.

O₂ + N → NO + O

3. Mesosphere: About 50 to 90 km height above the earth’s surface a 40 km thick layer is called Mesosphere.
O₂ + O → O₂ + O
O₊ + N₂ → NO⁺ + N
N2⁺+O₂ → N₂ + O₂

4. Thermosphere: This layer of atmosphere starts from a height of 90 km from the earth’s surface. This region till 500 km height is known as Thermosphere.

Question 4.
How is acid rain damaging statues and monuments in India?
Answer:
In India, statues and monuments are made up of marble i.e. CaCO₃. The surrounding air mainly contain oxides of sulphur and nitrogen in large amount. The main reason is the presence of large number of industrial units and energy plants around it. Oxides of sulphur and nitrogen are acidic. SO₂ and NO₂ oxidize with water to form mineral acids which is the main source of acid rain.
2SO_2(g) + O_2(g) + 2H₂O_(l) → 2H₂SO _4(aq)
4NO_2(g) + O_2(g) + 2H₂O_(l) → 4HNO_3(aq)
This acid rain reacts with the marble of statues and monuments and destroy them.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Question 5.
What is smog? How is classical smog different from photochemical smog?
Answer:
The word smog is derived from smoke and fog. This is the most common example of air pollution that occurs in many cities throughout the world. There are two types of smog.

1. Classical smog: It occurs in cool humid climate. It is a mixture of smoke, fog and sulphur dioxide. Chemically it is a reducing mixture.
2. Photochemical smog: Photochemical smog occurs in warm, dry and sunny climate. The main components of photochemical smog result from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories. Chemically it is oxidizing smog as it contains high concentration of oxidizing agents.

Question 6.
How many types of particulate are there? Explain.
Answer:
Particulate: Finely divided solid or liquid particles suspended in air are the cause of pollution. Size of these particles can be from 0.02 Å to 50000Å. Our eyes can see particles of the size about 10000Å.
Various types of particles and their source :

1.Soot: Due to incomplete combustion of carbon-containing fuels minute particles of carbon spread in the environment. This type of combustion generally occur at all places because from wood, coal etc. used as domestic fuels carbon-containing fuels are used in every field.

2. Dust: Sources of dust are present at all places on the earth. These are released in the environment due to natural or human sources during volcanic eruption, storm, during certain industrial processes like crushing, grinding of limestone, cement as fly ash and silica dust. Apart from this, dust is also spread due to running of vehicles.

3. Metal particles: Particles of metals like lead, mercury, chromium, arsenic, zinc, iron, nickel, cadmium etc. spread in the environment by various types of metal treatment.
Though their quantity is very less, still they produce harmful effect even if present in minute quantity.

4. Metal oxide: Generates by combustion of fuels containing metallic compounds.
For example when coal containing iron pyrites is burnt, particulates of Fe₃O₄ are introduced into the air.
3FeS₂+8O₂ → Fe3O₄ + 6SO₂

5. Particles of metal salts and fly ash: Limestone present in coal is left in ash in the form of CaO and can fly with ash in dust. Minute drops of H₂SO₄ situated in the environment react with CaO to form particles of CaSO₄. Ammonia present in air react with H₂SO₄ and form particles of (NH₄)₂SO₄. The flying of ash with air in the form of minute particles producing pollution is called fly ash pollution.

6. Asbestos particles: Asbestos is used in the factories and industries in various forms. Therefore asbestos particles spread in the environment.

7. Mist: We know that water vapour condense into minute droplets in the atmosphere which appear as mist to us. Similarly acids like H₂SO₄ and HNO₃ formed by various activities in the atmosphere also produce mist in the form of minute particles and spread pollution.

8. PAH and TEL: PAH means Polycyclic Aromatic hydrocarbon. TEL means Tetra Ethyl Lead. Due to incomplete combustion of fossil fuels and petroleum, PAH pollutes the environment in the form of minute particles. Tetraethyl lead, Pb (C₂H₅)₄ is added to gasoline as antiknock agent.

During combustion in the engine, it is oxidized to PbO which deposits in different parts of the engine and may cause damage. To avoid this damage, Pb(C₂H₅)₄ is mixed with dichloroethane and dibromo ethane which converts PbO into PbCl₂ and PbBr₂ which are volatile and thus come along with the exhaust gases and introduce into the atmosphere.
Pb(C₂H₅)₄ + O₂ + C₂H₄C₁₂ + C₂H₄Br₂ → CO₂ + H₂O + PbCl₂ + PbBr₂
Thus, engine is protected but air gets polluted.

Question 7.
What is Acid rain. Explain in detail its effect on the environment.
Answer:
All pollutant gases spread as pollutant particles in the form of smoke by the burning of fossil fuels and other fuels. Due to high temperature of industries and other engines, oxides of nitrogen spread in the atmosphere by the combination of N₂ and O₂. These gases mix in raindrops form acid and then fall on the earth as acid raindrops. This is called acid rain.
CO₂ + H₂O → H₂CO₃
SO₂ + H₂O → H₂SO₃

Effects of Acid rain :

• Acid rain neutralizes the base present in the soil, as a result of which soil becomes acidic. This affects the bacteria, plants and nitrogen fixation.
• Due to acid rain, water of rivers, lakes become acidic which has harmful effect on organisms.
• Acid rain leads to depleting forest resources and vegetation.
• Statues and monuments are losing their brightness because the metals present in them get corroded. Due to acid rain marble of Tajmahal is depleting.
• Acid rain harms leaves of aquatic plants and reduces the rate of photosynthesis.
• Acid rain has harmful effect on skin, lungs and neck.

Control of Acid rain :

1. Rivers and lakes cannot be prevented from acid rain but by mixing lime, its effect can be reduced.
2. By pouring lime in the fields, effect of acid rain can be reduced.
3. By using coal with low content of sulphur, unleaded petrol process of combustion of fuel in plants can be controlled and effect of acid rain can be destroyed.
4. Measures to control the effect of gases causing acid pollution and converting unburnt hydrocarbons to ineffective compound are being tried.

Question 8.
Explain the process of pollution of oxides of sulphur and nitrogen. How can they be controlled?
Answer:
Oxide of sulphur (SO_x): Among the oxides of sulphur, SO₂ is the main pollutant. It gets partially oxidised to SO₃ by air. SO₃ forms H₂SO₄ in presence of moisture this way SO₂ is a primary pollutant and SO₃ and H₂SO₄ is a secondary pollutant.
Process of pollution : (i) Fossil fuels and petroleum contain 0-3% to 3% of S which form S02 on burning.
S+O₂ → SO₂

(ii) SO₂ gas is released by industrial extraction of metals.
2FeS₂ + 5O₂ → 2FeO + 4SO₂
2PbS + 3O₂ → 2PbO + 2SO₂
2ZnS + 3O₂ → 2Zn0 + 2SO₂

(iii) During biological activities of various bacteria H₂S gas is released which oxidizes by air to form SO₂.
H₂S+\(\frac{3}{2}\) O₂ → SO₂+H₂O

(iii) Volcanic gases contain sufficient amount of SO₂

Control:

• To control SO₂, impurity of S should be removed from fuels.
• To remove sulphur from the engine by physical methods it is washed properly by making powder.
• In chemical methods, fossil fuels are carbonated.
• Liquation and vaporization are other methods of sulphur control.
• Environmental SO₂ can be controlled by scrubbing.

Oxides of nitrogen (NO_x): Nitric oxide is primary pollutant whereas N₂O, NO₂ or other oxides are secondary pollutant which are formed in the environment by various chemical activities.
Source:

• Due to high temperature produced by lightning in the atmosphere N₂ and O₂ combine to form NO.
• Production of NO due to high temperature in automobile engine.
• Formation of NO and other oxides in various industries.
• Formation of NO by natural and biological activities and due to fire in jungle.

Control:

• In the combustion of fuel, on keeping the amount of airless, the equilibrium
  for the manufacture of NO can be shifted to backward which can control the formation of NO.
  N₂ + O₂ ⇌ double arow 2NO
• Oxides of nitrogen released from automobile engines can be reduced by catalyst like Pt to N₃ or NH₃.
• Oxides of N₂ are acidic, therefore can be removed by basic solution.
• Scrubbing of nitrogen oxides can be done by acids.

Question 9.
Have you seen soil pollution in the nearby area? What measures will you take to control soil pollution ?
Answer:
Yes, it can be controlled by the following methods :
1. insecticides and pesticides which are used to protect our crops spread soil pollution. Therefore its use should be limited.

2. After the second world war, DDT was used in agriculture to control insects, pests, weeds and various crop diseases. Due to its adverse effect, its use was banned in India. Pesticides like Aldrin and Dieldrin are actually toxic. These are insoluble and non-biodegradable in water due to which they cause serious metabolic and physiological disorders.

Nowadays organophosphates and carbamates are used as pesticides. They are more biodegradable compounds but they are serious toxins and are harmful to human beings. Thus, fertilizers, detergents, pesticides, polymers like chemicals are used only under extreme necessity.

3. Biodegradable domestic waste should be dumped in pits.
4. Non-biodegradable waste should be recycled.
5. Use of polythene should be avoided.
6. Domestic waste, biological waste and chemical waste should be burnt. As a result of incineration volume of waste products is reduced.

Question 10.
Sometime ago formation of polar stratosphere clouds was reported over Antarctica. Why these formed? What happens when such clouds break up by warmth of sunlight?
Answer:
In summer season, nitrogen dioxide and methane react with chlorine monoxide and chlorine atoms and form chlorine sink which prevent ozone depletion to a large extent.
In winter season, special type of clouds, polar stratosphere clouds are formed over Antarctica. These clouds provide a type of surface by which chlorine nitrate hydrolysis to form hypochlorous acid. It reacts with hydrogen chloride to form molecular chlorine.

In spring, when sunlight return to Antarctica, heat of sun disperse the clouds and HOCl and Cl₂ decompose by sunlight.

This way, the chloride radical produced start the chain reaction for the depletion of ozone.

Question 11.
Describe at least 7 medicines obtained from plants.
Answer:
1. Neem: It is the plant of family Malice.
Uses :

• It keeps the environment pure,
• Neem leaves are insecticides,
• Neem oil is used in the treatment of wound.

2. Amla: Botanical name: Emblica Officinalis.
Uses :

• Used to prepare Triphala powder,
• Powder of seeds of Amla are used for washing hairs.

3. Tulsi: Botanical name: Ocimum sanctum.
Uses :

• It is grown in houses and temples to keep the environment clean,
• Its leaves are used for the treatment of fever and cough,
• It has antiseptic property.

4. Mahua: Botanical name: Madhuca indica.
Uses :

• Leaves, seeds and flowers are used for the treatment of skin disease,
• Its flowers are used to prepare wine.

5. Arandi: Botanical name: Ricinus communis.
Uses :

• Its oil is used to prepare soap, candles etc.
• Oil is used as medicine.

6. Harra : Botanical name : Terminalia chebula.
Uses :

• In dyeing leather,
• In the manufacture of medicines,
• Powder of fruit is used for the treatment of asthma.

7. Khus: Botanical name: Velivera ziga includes.
Uses :

• Oil is used in cosmetics, perfume and making beverage,
• It is insect repellant.
• It is used in the treatment of arthritis, backache, sprain etc.

Question 12.
How will you use Green chemistry for the following :
1. To control photochemical smog.
2. To avoid use of halogenated solvents in dry-cleaning and that of chlorine in bleaching.
3. To reduce use of synthetic detergents.
4. To reduce the consumption of petrol and diesel.
Answer:
1. Some plants like Pinus, Pyrus can metabolise nitric oxide (NO). Thus, their growth can control photochemical smog.

2. In dry-cleaning, use of halogenated solvent is replaced by a process, where liquified CO₂ with a suitable detergent is used. Use of H₂O₂ in place of Cl₂ for bleaching clothes gives better results and makes use of lesser amount of water.

3. Soap is 100% biodegradable. Thus, it can be used in place of detergents. Nowadays biodegradable detergents are also available. Thus, these are used in place of non-biodegradable detergents.

4. CNG (Compressed Natural Gas) should be used because it causes least pollution. Along with this, use of electrical automobiles can also reduce the consumption of petrol and diesel.

Question 13.
Differentiate between London classical smog and photochemical smog.
Answer:
Differentiate between London classical smog and Photochemical smog

Classical smog	Photochemical smog
1. This type of smog was first observed in London in 1952.	This type of smog was first observed in Los Angeles in 1950.
2. It is formed by the reaction of moist SO2 with H2SO4, which collects as particulate matter.	It is formed by photochemical reaction between NO2 and hydrocarbon.
3. It is formed by smoke and fog.	It does not contain smoke and fog.
4. It is formed in the morning in winter.	It is formed in warm sunny climate at day time.
5. It causes lung diseases.	It causes irritation in eyes.
6. It shows reducing character.	It shows oxidizing character.

Question 14.
(a) What are the main sources of soil pollution?
(b) How can Environmental pollution be controlled?
Answer:
(a) Main sources of soil pollution are :

• Industrial waste.
• Insecticides, pesticides and herbicides.
• Fertilizer, DDT, BHC, NaClO₃, Na₃ASO₃ etc.
• Radioactive products.

(b) Environmental pollution includes domestic waste, industrial waste etc. which can be controlled by the following methods :

1. By the recycling of glass, plastic, iron, polythene, paper etc.
2. By burning.
3. By management of waste.
4. By the use of Green chemistry.
5. By sewage treatment.
6. By awareness.
7. By planting more trees.
8. By adopting control tools and techniques.
9. By installing high chimneys fitted with equipments to discharge pollutants.

Mechanical Properties of Fluids Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Source of non-pollution energy :
(a) Fossil fuel
(b) Sun
(c) Gasoline
(d) Nuclear energy.
Answer:
(b) Sun

Question 2.
Which radiations manufacture O₃ :
(a) Ultraviolet
(b) Visible
(c) Infrared
(d) Radio waves.
Answer:
(a) Ultraviolet

Question 3.
Which radiations provide greenhouse effect:
(a) Infrared
(b) Visible
(c) Ultraviolet
(d) X-rays.
Answer:
(a) Infrared

Question 4.
PAN is responsible for :
(a) Depletion of ozone layer
(b) Smog
(c) Acid rain
(d) Poisonous food.
Answer:
(b) Smog

Question 5.
Which is not an air pollutant:
(a)H₂
(b) H₂S
(c) NO_x
(d) O₃.
Answer:
(a) H₂

Question 6.
Air pollutant released from Jet aeroplanes in the form of an air pollutant:
(a) Photochemical oxidant
(b) Photochemical reductant
(c) Aerosol
(d) Physical pollutant.
Answer:
(c) Aerosol

Question 7.
Is not present in acid rain :
(a) H₂SO₄
(b) HNO₃
(c) H₂SO₃
(d) CH₃COOH.
Answer:
(d) CH₃COOH.

Question 8.
Is responsible for disease of lungs :
(a) O₂
(b) N₂
(c) CO₂
(d) SO₂.
Answer:
(d) SO₂.

Question 9.
03 is manufactured in :
(a) Troposphere
(b) Stratosphere
(c) Mesosphere
(d) Thermosphere.
Answer:
(b) Stratosphere

Question 10.
For acid rain ‘sink’ is :
(a) Leaves
(b) Reservoir
(c) Limestone
(d) CO₂.
Answer:
(c) Limestone

Question 11.
Primary pollutant is :
(a) SO₂
(b) NO₂
(c) N₂O
(d) NO.
Answer:
(d) NO.

Question 12.
Most dangerous is :
(a) Smoke
(b) Dust
(c) Smog
(d) NO.
Answer:
(c) Smog

Question 13.
Which of the following easily combine with haemoglobin:
(a) CO
(b) NO
(c) O₂,
(d) CO₂.
Answer:
(b) NO

Question 14.
Aromatic compound which is found in the form of particles :
(a) Benzene
(b) Toluene
(c) Nitrobenzene
(d) Polyring hydrocarbon.
Answer:
(d) Polyring hydrocarbon.

Question 15.
Following disease is caused due to depletion of ozone :
(a) Blood cancer
(b) Lung cancer
(c) Skin cancer
(d) Chest cancer.
Answer:
(c) Skin cancer

Question 16.
Photochemical smog is formed :
(a) In summer mornings
(b) In winter days
(c) In rainy days in the morning
(d) In day time in rains.
Answer:
(b) In winter days

Question 17.
Which of the following compound does not form photochemical smog:
(a) NO
(b) O₃
(c) C_xH_y
(d) SO₂.
Answer:
(d) SO₂.

Question 18.
Main pollutant hydrocarbon is :
(a) Methane
(b) Ethane
(c) Propane
(d) Butane.
Answer:
(a) Methane

Question 19.
Which of the following acid is in abundance in acid rain :
(a) HNO₃
(b) H₂SO₄
(c) HCl
(d) H₂CO₃.
Answer:
(b) H₂SO₄

Question 20.
Which of the following is not a pollutant:
(a) NO₂
(b) CO₂
(c) O₃
(d) (C_xH_y).
Answer:
(b) C0₂

Question 21.
Which of the following is not a greenhouse gas :
(a) CO₂
(b) CH₄
(e) CFCs
(d) O₂.
Answer:
(d) O₂.

Question 22.
Region near earth’s surface is :
(a) Thermosphere
(b) Mesosphere
(c) Stratosphere
(d) Troposphere.
Answer:
(c) Stratosphere

Question 23.
Main source of CO pollutant is :
(a) Industrial process
(b) Means of transport
(c) Earthquake
(d) Volcano.
Answer:
(b) Means of transport

Question 24.
Marble is the main sink of:
(a) Metallic pollutant
(b) NH₃ pollutant
(c) Acid pollutant
(d) None.
Answer:
(c) Acid pollutant

Question 25.
Which of the following is not a Nobel Prize winner in 2005 :
(a) Y. Chadwin
(b) R.H. Grooms
(c) R.R. Shark
(d) Polythene.
Answer:
(d) Polythene.

Question 26.
Is correct for photochemical smog :
(a) It is reducing in nature
(b) Is formed in winter season
(c) Is a mixture of smog and fog
(d) Cause irritation in eyes.
Answer:
(d) Cause irritation in eyes.

2. Fill in the blanks:

1. The air pollutant released by jet aeroplanes in the form of fluro carbon is …………… .
Answer:
aerosol

2. D.D.T. is …………… poisonous pollutant as compared to B.H.C.
Answer:
more
3. 03 is formed in the …………… level of atmosphere.
Answer:
stratosphere

4. A definite tolerable level of pollutants in the environment is expressed by …………… .
Answer:
TLV

5. Maximum air pollutants are present in …………… level of the atmosphere.
Answer:
troposphere

6. Ozone layer prevent us from …………… rays.
Answer:
ultraviolet

7. Oxides of …………… and …………… cause acid rain.
Answer:
nitrogen, sulphur

8. …………… is the main cause of ozone layer depletion.
Answer:
C.F.C
9. …………… gas is responsible for green house effect.
Answer:
CO₂
10. Components which produce pollution are called …………… .
Answer:
pollutant

11. …………… causes harmful effect on lungs.
Answer:
Photochemical smog

12. SO₂ pollutant is responsible for the disease of …………… .
Answer:
Lungs

13. …………… is responsible for ultraviolet light.
Answer:
Skin cancer

14. …………… also cause harm to ozone layer.
Answer:
Methane.

3. Match the following:

I.

‘A’	‘B’
1. Ozone Hole	(a) Fly ash
2. Particulate	(b) CFC
3. Greenhouse effect	(c) Smog
4. PAN	(d) Infrared radiations
5. Fluorosis	(e) Bone disease.

Answer:
1. (b) CFC
2. (a) Fly ash
3. (d) Infrared radiations
4. (c) Smog
5. (e) Bone disease.

II.

‘A’	‘B’
1. Smog	(a) Water pollution
2. Main component of acid rain	(b) Depletion of ozone layer
3. COD or BOD	(c) Aerosol
4. Chlorofluorocarbon	(d) SO2

Answer:
1. (c) Aerosol
2. (d) SO₂
3.  (a) Water pollution
4. (b) Depletion of ozone layer.

4. Answer in one word/sentence :

1. Write the definition of pollutants.
Answer:
Chemicals which affect man, animal and plant kingdom are called pollutants.

2. Write the names of two air pollutants.
Answer:
SO₂, SO₃.

3. Name two pollutants which depletes ozone layer.
Answer:

1. Cycle of nitric oxide (NO) and
2. C.F.C. (Chlorofluorocarbon) in which CFC is main.

4. In which region ozone is found?
Answer:
Ozone is found in stratosphere region.

5. Tell two health problems caused by SO₂.
Answer:

1. SO₂ affects the respiratory canal and lungs due to which various health diseases are caused like cancer also.
2. Acid rain caused due to SO₂ produces boils on the skin.

6. What is Acid rain?
Answer:
Various gaseous pollutant present in the atmosphere like SO₂, SO₃, NO₂, NO dissolve in raindrop.
These drops fall with rain and are called acid rain.
SO₂ + H₂O → H₂SO₃
SO₃+H₂O → H₂SO₄.

7. Name any two greenhouse gases.
Answer:
CFC and CO₂.

8. What is PAN?
Answer:
Peroxy acyl nitrate which is photochemical smog.

9. What is CFC?
Answer:
It is chlorofluorocarbon which is the main cause of depletion of ozone layer.

10. What is green chemistry?
Answer:
A technique to check pollution in which such chemical reactions are suggested which do not cause pollution and if pollution spreads then it can be destroyed. This is called green chemistry.

11. What is TLV?
Answer:
A definite value of pollutants can be tolerated. This is expressed by TLV. TLV means ‘Threshold Limit Value’.

12. What are particulate pollutants?
Answer:
The pollutants mixed up with air in liquid or solid state in such a manner that the remain suspended for a long time are called particulates.

13. What is specimasion?
Answer:
Many pollutants can be made from an element. The method of determining which of the product is more dangerous is called specimasion.

14. What is sink?
Answer:
Sink is that in which the substance totally gets consumed and then also there is no effect on sink. ‘

15. Name the biggest sink of the earth.
Answer:
The biggest sink of the earth is the sea.

16. What is greenhouse effect?
Answer:
Greenhouse effect: The heating of atmosphere due to absorption of infrared radiation by carbon dioxide and other gases is called greenhouse effect.

17. Explain the mechanism of acid rain.
Answer:
Due to combustion of fuels various pollutant particles of all gas pollutants in the form of smoke spread. Due to very high temperature of industries and other fuels N2 and 02 combine and spread in the atmosphere as oxides of nitrogen. These various gases fall on the earth as rain drops which is called acid rain.

18. Gas leaked in 1984 in Bhopal gas tragedy is.
Answer:
CH₃-N = C= O Methyl isocyanate

19. Name the main component responsible for depletion of ozone layer.
Answer:
Chlorofluorocarbon.

20. Name the person who started “Chipko Andolan” for forest conservation.
Answer:
Sunderlal Bahuguna.

21. Write name of disease caused by water pollution.
Answer:
Water pollution causes gastrointestinal diseases such as cholera, typhoid, dysentry, gastroenteritis, polio, hepatitis etc.

22. Name the important medicinal plant used to remove environmental pollution and also useful in skin cancer.
Answer:
Neem (Azaderecta indica).

23. Which gas is the cause of greenhouse effect?
Answer:
Carbon dioxide (CO₂).

24. Which is the main sink of CO pollutant?
Answer:
Organisms present on earth.

25. Write the name of four methods used in Green chemistry.
Answer:

1. Use of sunlight,
2. Micro-oven,
3. Micro-waves,
4. Use of Sound waves and Enzyme.

Students get through the MP Board Class 11th Physics Important Questions Chapter 9 Mechanical Properties of Solids which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 9 Mechanical Properties of Solids

Mechanical Properties of Solids Class 11 Important Questions Very Short Answer Type

Question 1.
What do you mean by deforming force?
Answer:
External force applied on a body is known as deforming force.

Question 2.
What is deformation?
Answer:
If an application of deforming force, a change in the shape or size of a body takes place, then it is said to be deformed.

Question 3.
What is meant by elasticity?
Answer:
It is the property of a body by virtue of which, it regains its original length, volume and shape when the deforming force is removed.

Question 4.
What is stress and its unit?
Answer:
The internal force of restitution per unit area of a deformed body is called stress.
Stress = \(\frac{\text { Force }}{\text { Area }}\)
Unit: SI unit of stress is Nm^-2.

Question 5.
Define plasticity.
Answer:
Plasticity is the property of remaining deformed even after the removal of deforming forces.

Question 6.
What do you mean by elastic limit?
Answer:
The maximum value of deforming force, by which if the force is applied on a body, the body looses its elastic property, is called elastic limit.

Question 7.
What do you mean by brittleness?
Answer:
Some substances are neither elastic nor plastic but they break easily into pieces, this property of a body is called brittleness and the bodies are called brittle substances.

Question 8.
What is difference between stress and pressure?
Answer:
Stress is internal force while pressure is external force.

Question 9.
How many types of stress are there?
Answer:
There are three types of stress :

1. Tensile stress or longitudinal stress,
2. Normal stress,
3. Tangential or shearing stress

Question 10.
What is strain? Write its unit.
Answer:
The ratio of change in length, volume or shape and its original length, volume or shape is called strain.
As it is the ratio of same quantities, therefore it has no unit.

Question 11.
What is longitudinal stress?
Answer:
It is the restoring force developed per unit cross-sectional area of a body when the length of the body increases in the direction of deforming force.

Question 12.
Define tangential stress.
Answer:
When the deforming force is applied along the plane of the area of cross-section of the body then the stress produced is known as tangential stress.

Question 13.
State the types of strain.
Answer:
There are three types of strain :

1. Longitudinal strain,
2. Shearing strain and
3. Volume strain.

Question 14.
What is longitudinal strain?
Answer:
Within elastic limit, it is the ratio of change in length to original length.
i.e., Longitudinal strain = \(\frac{\text { Change in length }}{\text { Initial length }}\)

Question 15.
Define volume strain.
Answer:
The change in unit volume due to deforming force is called volume strain.
Or
Within elastic limit, the ratio of change in volume to the initial volume is called volume strain, i.e.,
Volume strain = \(\frac{\text { Change in volume }}{\text { Initial volume }}\)

Question 16.
Define shearing strain.
Answer:
When a deforming force is applied on a body along its surface in such a way that the volume of the body remains constant, but the shape changes, then the body is said to be sheared and the strain produced is called shearing strain. Within elastic limit, it is measured by the ratio of the relative displacement of one plane to its distance from fixed plane, i.e.,
Shearing strain, θ = \(\frac{\Delta l}{l}\)
Δ l is relative displacement between two layers and l is the distance between the layers.

Question 17.
State Hooke’s law.
Answer:
Within elastic limit the stress is directly proportional to the strain. i.e.,
Stress ∝ Strain.

Question 18.
What are the types of modulus of elasticity?
Answer:
There are three types of modulus of elasticity :

1. Young’s modulus,
2. Bulk modulus,
3. Modulus of rigidity.

Question 19.
Define Young’s modulus and give its unit.
Answer:
Within elastic limit the ratio of longitudinal stress to the longitudinal strain is called Young’s moduluSIe.,
Young s modulus = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)
Unit: Its SI unit is Nm^-2.

Question 20.
What is compressibility?
Answer:
‘The reciprocal of bulk modulus is called compressibility.

Question 21.
Define coefficient of rigidity.
Answer:
Within elastic limit, the ratio of tangential stress to the shearing strain is called coefficient of regidity.

Question 22.
Write the following in form of increasing elasticity of copper, iron, glass and rubber.
Answer:
Rubber < Glass < Copper < Iron.

Question 23.
If an ivory ball and an iron ball are dropped from the same height on a hard floor. Which one will rebounce higher?
Answer:
The ivory ball will rebounce higher because its elasticity is greater than that of iron ball.

Question 24.
Springs are made of steel and not of copper. Why?
Answer:
The Young modulus of steel is more than that of copper. If the same deforming force is applied on the spring of steel and spring of copper then the steel regain its initial position faster than that of copper spring. So the spring are made of steel.

Question 25.
What do you mean by restoring force?
Answer:
The force due to which the body regain its original shape and size is called restoring force.

Question 26.
State the condition for deforming force and restoring force to be equal and opposite.
Answer:
When the strain produced in the body is within the elastic limit then the deform¬ing force and restoring force are equal and opposite to each other.

Question 27.
On increasing temperature, Young modulus of elasticity increases or de¬creases.
Answer:
Decreases.

Question 28.
Why the work is done in stretching a wire?
Answer:
When a wire is stretched, the inter reaction force is developed within the wire and hence work is done against this force.

Question 29.
What happens with work to stretch a wire?
Answer:
The work done to stretch a wire is stored as elastic potential energy within wire.

Question 30.
If the length of the wire is halved, then what will be the effect in its Young modulus of elasticity?
Answer:
No effect on Young modulus of elasticity as it depends on the mass of the mate¬rial of the body not in its length.

Question 31.
If the length of wire is cut to its half then :
(i) What will be the effect on increase in length?
Answer:
Increase in length will get halved with respect to initial length.

(ii) Effect on maximum weight it can resist?
Answer:
No effect.

Question 32.
If a wire is replaced by another wire of same length and material but of double diameter, then :
(i) What will be increase in length for same given weight?
Answer:
Increase in length will remain one-fourth.

(ii) Effect on maximum weight that it can resist?
Answer:
Maximum weight it can resist will become four times.

Question 33.
Why the suspension wires are made up of quartz or phospher bronze?
Answer:
This is because the elastic after effect is negligible in quartz and phospher bronze.

Question 34.
In Searl’s apparatus, two similar wires of same material are taken, why?
Answer:
The temperature of room changes, hence the effect of temperature will be same on both the wires. Thus, the change in temperature does not cause any error.

Question 35.
In Searl’s experiment the readings are taken both by loading and unload¬ing the weights, why?
Answer:
There are two reasons :

1. It ensures that the experiment is performed within elastic limit.
2. The errors produced due to elastic fatigue, torsion and backlash ‘are removed.

Question 36.
Why the readings are taken after some time in the experiments of elasticity?
Answer:
To remove the error caused by elastic after effect.

Question 37.
What is elastic hysteresis?
Answer:
On applying deforming force on a body the strain is produced in it on removal of deforming force some time body do not regain its original position completely. Some strain remains left in that body. It is called elastic hysteresis.

Question 38.
Write down the effect of impurities on elasticity.
Answer:
In the presence of impurities the elasticity of the bodies is increased.

Question 39.
What is the effect of temperature on elasticity?
Answer:
With increase in temperature, usually the elasticity of the bodies decreases and with decrease in temperature, the elasticity increases. But the effect of temperature is negligible in the elasticity of’Invar steel’.

Mechanical Properties of Solids Class 11 Important Questions Short Answer Type

Question 1.
State the difference between elastic and plastic bodies.
Answer:
Difference between Elastic and Plastic bodies :

Elastic bodies	Plastic bodies
1. The bodies which regain their original position after the withdrawn of deforming force, are called elastic bodies.	1. The bodies which get permanently deformed by the deforming force, are called plastic bodies.
2. The change is temporary.	2. The change is permanent.
3. Examples : Gold, iron etc.	3. Examples : Melting wax, wet clay, plasticine.

Question 2.
Steel is more elastic than rubber. Explain with reason.
or
Why steel is more elastic than rubber?
Answer:
Let two wires of steel and rubber of equal length L and equal area of cross-section A are taken. Both are subjected by a force F, so that the elongation produced in steel is l_s and in rubber l_r.
∴ Modulus of elasticity of steel, Y_s = \(\frac{F \cdot L}{A \cdot l_{s}}\) …(1)
and Modulus of elasticity of rubber, Y_r = \(\frac{F \cdot L}{A \cdot l_{r}}\) …(2)
Now, dividing eqn. (1) by eqn. (2),
\(\frac{Y_{s}}{Y_{r}}=\frac{l_{r}}{l_{s}}\)
∵ l_r > l_s
∴ Y_s > Y_r
Hence, it is clear that steel is more elastic than rubber.

Question 3.
What do you mean by Young’s modulus of elasticity. Write its unit and dimensional formula. Derive the formula for Young’s modulus and hence define it.
Answer:
Within elastic limit, the ratio of longitudinal stress to the longitudinal strain is called Young’s modulus.
i.e., Y = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)

SI unit is Nm^-2 and dimensional formula is [ML^-1T^-2].
Formula: Let a mass M is suspended by a wire of length L and hence the elongation produced is l. If the radius of the wire is r, then
Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{M g}{\pi r^{2}}\)
and strain = \(\frac{l}{L}\)
But Y = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)
∴ Y = \(\frac{M g / \pi r^{2}}{l / L}\)
Or
Y = \(=\frac{M g L}{\pi r^{2} l}\)
Within elastic limit, Young’s modulus of elasticity is numerically equal to that force, which extends the length of wire of unit cross-sectional area by the original length i.e., the final length becomes double the initial length.

Question 4.
What do you mean by bulk modulus? Establish the formula for bulk modulus.
Answer:

Bulk modulus : Within elastic limit, the normal stress is directly proportional to the volume strain.
∴ K = \(\frac{\text { Normal stress }}{\text { Volume strain }}\)
Formula : Let initial volume of a gas is V and pressure P. If the pressure is increased to P + p, so that its volume becomes V- v, hence the decrease in volume is v.
Since, force acting per unit area is called pressure. Hence,
Normal stress = Increase in pressure = (P + p)- P = p
Volume strain = – \(\frac{v}{V}\)
By Hooke’s law, within elastic limit,
Bulk modulus of elasticity = \(\frac{\text { Normal stress }}{\text { Volume strain }}\)
K = \(\frac{p}{-\frac{v}{V}}=\frac{-p V}{v}\)
-ve sign shows that, with increase in pressure, volume decreases.

Question 5.
What do you mean by coefficient of rigidity? Derive formula for it.
Answer:
Coefficient of rigidity : Within elastic limit the ratio of tangential stress and shearing strain is called modulus of rigidity.
i.e., η = \(\frac{\text { Tangential stress }}{\text { Shearing strain }}\)

Formula derivation : Let ABCDEFGH be a cuboid (of side l). A tangential force F is applied on the surface ADEF, If the area of surface is A and angular displacement due to force F is θ, then
Tangential stress = \(\frac{F}{A}\)
If the relative displacement between the layers is
AA’ = X, then
Shearingstrain, θ = \(\frac{A A^{\prime}}{A B}=\frac{x}{l}\)
Modulus of rigidity = \(\frac{\text { Tangential stress }}{\text { Shearing strain }}\)
∴ η = \(\frac{F / A}{x / l}=\frac{F l}{x A}\)

Question 6.
Why are girders I shaped?
Answer:

Girders of iron are made of I shapped.
Let l be the length of a rod, b be the breadth and d be its thickness. If weight W be suspended at its middle point the bending (δ) is given by
δ = \(\frac{\mathrm{W} l^{3}}{b d^{3} \mathrm{Y}}\) …(1)
For the strength of girder, should be smaller, δ will be smaller if
(a) Y is large.
(b) From eqn. (1) it is clear that δ∝\(\frac{1}{b}\) and δ∝\(\frac{1}{d^{3}}\) Therefore depth d should be greater than breadth b.
Therefore, girders of made of I shape.

Question 7.
Define Poisson’s ratio. What is its unit? What is its value?
Answer:
When a deforming force F is applied to a wire, then its length increases, but its radius or breadth decreases. This strain which occurs along the width of the body is called lateral strain. Within elastic limit, the lateral strain is proportional to longitudinal strain.
i.e., Lateral strain ∝ Longitudinal strain
or
Lateral strain = σ × Longitudinal strain
or
Poisson’s ratio,
Where D = Initial diameter of wire, d = Change in diameter, L = Original length of wire, l = Change in length.
It has no unit. Theoretical values lies from – 1 to 0.5.

Question 8.
Derive an equation for the force acting on the clamps, when the wire gets cooled. f’
Answer:

Let a wire of length L and area of cross-section A be clamped between the two ends. If it is being cooled, then its temperature decreases by Δ t°C, as a result it gets strinked and so it tries to contract the clamps. This force has to be found out. If Y is the Young’s moduliis of elasticity and a is the coefficient of linear expansion, then
α = \(\frac{\text { Change in length }(l)}{\text { Original length }(L) \times \text { Change in temp. }(\Delta t)}\)
Hence, αLΔt = l or \(\frac{l}{L}\) = αΔt
∴ Strain = αΔt …(1)
But, Young s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ Stress = Y × Strain= YαΔt …(2)
Moreover, Stress = \(\frac{\text { Force }}{\text { Area }}\)
or
Force = Stress × Area = YαΔt × A
∴ F =YAαΔt
This is the required relation.

Question 9.
Figure shows the strain-stress curve for a given material. What are:
(a) Young’s modulus and
(b) Approximate yield strength for his material?

Answer:
From the given graph for a stress of 150 × 10⁶ Nm^-2, the strain is 0.002.
(a) Young’s modulus of the material (Y) is given by
∴ Y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{150 \times 10^{6}}{0.002}=\frac{150 \times 10^{6}}{2 \times 10^{-3}}\)
= 75 ×10⁹ Nm^-2
= 7.5 × 10¹⁰Nm^-2.

(b) Yield strength of a material is defined as the maximum stress it can sustain.
∴ From graph, the approximate yield strength of the given material = 300 × 10⁶ Nm2 = 3 × 10⁸ Nm^-2. Actually, it is slightly less than 3 × 10⁹ Nm^-2.

Question 10.
The stress-strain graphs for materials A and if are shown in figure (a) and (b):

The graphs are drawn to the same scale :
(a) Which of the materials has greater Young’s modulus?
(b) Which of the two is the stronger material?
Answer:
(a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence Young’s modulus = \(\left(\frac{\text { Stress }}{\text { Strain }}\right)\) is greater forA than that of B.

(b) Strength of a material is determined by the amount of stress (load) required to cause breaking or fracture of material corresponding the breaking point.
∴ Material A is stronger than B as it can withstand more load without breaking than the material B corresponding to point D.

Question 11.
Prove that energy stored per unit volume in a stretched string is, given by:
U = \(\frac{1}{2}\) × Stress × Strain
= \(\frac{1}{2}\) × Young’s modulus × (Strain )²
Derive the formula for the potential energy of a stretched string.
Answer:
Consider a wire of length L and it is extended by a force F due to which its extension takes place by l.
If A is the area of cross-section of the wire and Fis the Young’s modulus of elasticity,
then
Y = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\) = \(\frac{F / A}{l / L}\) = \(\frac{F L}{A l} .\)
Or
YAl = Fl or F = \(\frac{Y A l}{L}\) …(1)

When, the wire is stretched by a force F, then work is done by the deforming force
which is stored in the wire in the form of potential energy.
Small amount of work done to stretch by dl i.e.,
dW = Fdl= \(\frac{Y A l}{L}\)dl
∴ The total amount of work done will be

Where, \(\frac{l}{L}\)is the longitudinal strain
Work done per unit volume is called elastic potential energy per unit volume (U).
∴ U= \(\frac{1}{2}\) × Y × (Strain)², [from eqn. (3)]
But, Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ U= \(\frac{1}{2}\) × \(\frac{\text { Stress }}{\text { Strain }}\) × (Strain)²
or
U= \(\frac{1}{2}\) × Stress × Strain. Proved.

Mechanical Properties of Solids Class 11 Important Numerical Questions

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^-5 m²stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 × 10^-5
m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Solution:
Here, for steel wire,
Length of wire, l₁ = 4.7 m
Area of cross-section, a₁ = 3.0 × 10^-5 m²
Stretching, Δl₁, = Δl(say)
Stretching force on steel, F₁ = F
For copper wire,
Length of wire,l₂ = 3.5 m
Area of cross-section, a₂ = 4.0 × 10^-5 m²
Stretching, Δl₂ = Δl(given);
Stretching force on copper, F₂ = F
Let Y₁ and Y₂ be the Young’s modulus of steel and copper wire respectively.

Dividing eqn. (1) by eqn. (2), we get

= \(\frac{18 \cdot 8}{10 \cdot 5}\) = 1.79 = 1.8:1

Question 2.
Two wires of diameter 0.25 cm. One made of steel and the other made of brass are loaded as shown in fig. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. If Young’s modulus of steel and brass are 2.0 × 10¹¹ Pa and 0.91 × 10¹¹ Pa, then compare the elongation of the steel and the brass wires.
Solution:

Using, Y = \(\frac{M g L}{\pi r^{2} l}\)
l = \(\frac{M g L}{\pi r^{2} Y}\)
For brass wire,
M= 6 kg, L = 1.0 m, Y = 0.91 × 10¹¹N/m²
and 2r = 0.25 cm = 0.25 × 10^-2m
r = 0.125 × 10^-2m
Putting the values in formula,
l = \(\frac{6 \times 9 \cdot 8 \times 1 \cdot 0}{3 \cdot 14 \times\left(0 \cdot 125 \times 10^{-2}\right)^{2} \times 0 \cdot 91 \times 10^{1}}\)
= 1.3 × 10^-4 m.
For steel wire,
M = 6 + 4 = 10 kg,L = 1.5 m, Y = 2.0 × 10¹¹ N/m²and 2r = 0.25cm
∴ r = 0.125 × 10^-2m
Putting the values in formula,
l = \(\frac{10 \times 9 \cdot 8 \times 1 \cdot 5}{3 \cdot 14 \times\left(0 \cdot 125 \times 10^{-2}\right)^{2} \times 2 \cdot 0 \times 10^{1}}\)
= 1.5 × 10^-4m

Question 3.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44500 N. force, producing only elastic deformation. Calculate the resulting strain. Y for copper = 1.1 × 10¹¹ Nm^-2.
Solution:
Here, Y = 1.1 × 10¹¹Nm^-2
A = Area of cross-section
= 15.2 mm × 19.1 mm
= 15.2 × 10^-3m × 19.1 × 10^-3 m
Force, F = 44500 N
Resulting Strain = Longitudinal Strain = ?
∴ Y = \(\frac{\text { Stress }}{\text { Strain }}\)
or
Strain \(=\frac{\text { Stress }}{Y}=\frac{F}{A Y}\)
or
Longitudinal Strain = \(\frac{44500}{15 \cdot 2 \times 19 \cdot 1 \times 10^{-6} \times 1 \cdot 1 \times 10^{1}}\)
= 139.34 × 10^-3 m = 0.139.

Question 4.
A steel cable with a radius of 1.5 cm supports a chair lift at a ski area. If the maximum stress is not to exceed 10⁸ Nm^-2, what is the maximum load the cable can support?
Solution:
Here, radius of steel cable r = 1.5cm = 1.5 × 10^-2 m
Max. Stress = 10⁸ Nm^-2
∴ Area of cross-section of cable A = πr^-2 = π(1.5 × 10^-2)²
Maximum load the cable can withstand = Maximum force ? We know that,
Maximum stress = \(\frac{\text { Maximum force }}{\text { Area of cross-section }}\)
or
Maximum force = Maximum stress × Area of cross-section
= 10⁸ × π × (1.5 × 10^-2)²
= 3.142 × 2.25 × 10⁸ × 10^-4 N
or
Maximum load the cable can withstand = 7.1 × 10⁴N.

Question 5.
A 4.0 metre long copper wire of area of cross-section 1.2 cm² is stretched by a force of 4.8 × 10³N. If the Young’s modulus of copper is Y = 1.2 × 10¹¹ N/m², then calculate:
(i) Stress,
(ii) Strain,
(iii) Increase in length of wire.
Solution:
Given : L=4.0 metre, A = 1.2 cm² = 1.2 × 10^-4m², F = 4.8 × 10³ N, Y= 1.2 × 10¹¹ N/m²

(i) Stress : The stress on wire will be = \(\frac{F}{A}=\frac{4 \cdot 8 \times 10^{3}}{1 \cdot 2 \times 10^{-4}}\)
∴ Stress = 4 × 10⁷ N/m².

(ii) Strain: The Young’s modulus of wire is Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ Strain = \(\frac{\text { Stress }}{\mathrm{Y}}=\frac{4 \times 10^{7}}{1 \cdot 2 \times 10^{11}}\)
∴ Strain = 3.3 × 10^-4

(iii) Increase in Length of wire : From the formula of Young’s modulus
Y = \(\frac{\mathrm{FL}}{\mathrm{A} . l}\)
∴ l = \(\frac{\mathrm{FL}}{\mathrm{AY}}\)
\(=\frac{4 \cdot 8 \times 10^{3} \times 4}{1 \cdot 2 \times 10^{11} \times 1 \cdot 2 \times 10^{-4}}\)
= 13.2 × 10^-4 = 1.32 mm.

Question 6.
Two wires made of same material, length of first wire is half the length of second wire and its diameter is double the diameter of second wire. If same weight is suspended in both wires then find the ratio of increase in length.
Solution:

Question 7.
On stretching a gold wire, Its length Increases by 1%. Find out linear strain of it, if its coefficient of Young modulus is 8 × 10¹⁰ N/m². Find out the stress produced.
Solution:
According to question,
l = 1%, Y = 8 × 10¹⁰ N/m²
∴ Linear strain = \(\frac{\text { Increase in length }}{\text { Original length }}\)
= \(\frac{1}{100}\)
= 0.01
Stress = Y × Strain = (8 × 10¹⁰) × (0.01)
=8 ×10⁸N/m².

Question 8.
2 kg weight elongates a wire of 3 m length by 1 mm. The diameter of wire is 1 mm. Determine Young’s modulus of the material of the wire.
Solution:
Given : M = 2 kg, L = 3 m, l = 1 mm = 10^-3 m, r= \(\frac{1}{2}\)mm= 0.5 x 10^-3m, g = 9.8 ms^-2. ,
We have, Y = \(\frac{M g L}{\pi r^{2} l}\)
or
Y = \(\frac{2 \times 9 \cdot 8 \times 3}{3 \cdot 14 \times\left(0 \cdot 5 \times 10^{-3}\right)^{2} \times 10^{-3}}\)
=  \(\frac{58 \cdot 8 \times 10^{6} \times 10^{3}}{3 \cdot 14 \times 0 \cdot 25}\)
∴ Y = \(\frac{588 \times 10^{12}}{314 \times 25}\)
= 7.49 × 10¹⁰Nm^-2

Question 9.
The diameter of a wire is \(\frac{2}{\sqrt{\pi}}\) cm. The force required to double the length is 2 × 10¹²dyne. Calculate Young’s modulus of the material of the wire.
Solution:
Given: r = \(\frac{1}{2}\) × \(\frac{2}{\sqrt{\pi}}\)cm = \(\frac{1}{\sqrt{\pi}}\) 10^-2 m, L = l
and Mg = 2 × 10¹²dyne = 2 × 10¹² × 10^-5N = 2 × 10⁷N
Now, we have Y = \(\frac{M g L}{\pi r^{2} l}\)

∴ Y = \(\frac{2 \times 10^{7}}{10^{-4}}\) = 2 × 10¹¹ Nm^-2

Question 10.
Calculate the force required to elongate 1 m long the 2 mm thick wire by 1 mm. (Y = 2 x 10¹¹ Nm^-2)
Solution:
Given: L = 1m,r = \(\frac{2}{2}\) = 1mm =10^-3m = 10^-3m, l = 1mm 10^-3m Y = 2 × 10¹¹Nm^-2
Frmula, Y = \(\frac{M g L}{\pi r^{2} l}\)
or
Mg = \(\frac{Y \pi r^{2} l}{L}\)
\(=\frac{2 \times 10^{11} \times 3 \cdot 14 \times\left(10^{-3}\right)^{2} \times 10^{-3}}{1}\)
= 6.28 × 10¹¹ × 10^-9
= 6.28 × 10² = 628 N.

Question 11.
Calculate the work done in extending a wire of length lm and of cross-section 1mm2 through 2mm. (Given y = 2 × 10¹¹Nm^-2)
Solution:
Given: L= 1m, A = nr² = 1mm² = (1 × 10^-3m)²= 10^-6m²
l = 2mm = 2 × 10^-3m and Y = 2 × 10^-11 Nm^-2
Strain = \(\frac{l}{L}=\frac{2 \times 10^{-3}}{1}\) = 2 × 10^-3
∴ Volume of wire = A.L = 10^-6 × 1 = 10^-6m³
Work done = \(\frac{1}{2}\) × Y × (Strain)² × Volume
= \(\frac{1}{2}\) × 2 × 10¹¹ × (2 ×10^-3)¹¹× 10^-6
= 0.4 J.

Question12.
A rubber ball while taking to the bottom of a lake 200 m deep its volume de-creases by 0.1%. Calculate the Bulk modulus of rubber. Density of water is 10³ kg/ m³ and g = 10 m/sec².
Solution:
Given : Depth h = 200m, Density of water = 10³ kg/m³ Acceleration due to gravity g = 10 m/sec²,
Volume strain = 0.1% = –\(\frac{0 \cdot 1}{100}\)
∴ Bulk modulus K = \(\frac{\frac{-p}{\Delta v}}{\frac{V}{V}}\)
or
K = \(\frac{-200 \times 10^{3} \times 10}{\frac{-0 \cdot 1}{100}}\)
= 2 × 10⁹N/m²

Question 13.
A weight of 20 kg is suspended in a wire. Area of cross-section of it is 1mm². The length of the wire in extended state is 6 m. On withdrawing the weight its length become 5.995 m. Find out Young modulus of elasticity.
Solution:
Given : M = 20 kg, A = 1mm² = 10^-6m², L = 6m,
Change in length = 6 – 5.995 = 0.005 m
∴ Y = \(\frac{M g L}{A l}\)
or
Y \(=\frac{20 \times 9 \cdot 8 \times 6}{10^{-6} \times 0-005}\)
= 2.35 × 10¹¹N/m².

Question 14.
On iron wire of radius 0.5 mm is heated up to 300°C and then it is fixed between two clamp. When the temperature reduces to 30°. What will be the force acting upon the clamp. (α = 1.2 × 10^-5/°C and Y= 1.4 × 10¹² dyne/cm²)
Solution:
Given : r = 0.5mm = 0.05 cm, α = 1.2 × 10×^-5/°C, Y = 1.4 × 10¹²dyne/cm²
ΔT = 300 – 30 = 270° C
From F = YΔ α. AT, we get
F = Y πr² α ΔT
∴ F = 1.4 × 10¹² × 314 × (0.05)² × 1.2 × 10^-5 × 270
F = 3.56 × 10⁷ dyne.

Mechanical Properties of Solids Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
A wire of radius r and length L on which a mass M is suspended so that its increase in length is T. Young modulus will be:
(a) \(\frac{\mathrm{MgL}}{\pi r^{2} l}\)
(b) \(\frac{\mathrm{Mgl}}{\pi r^{2} \mathrm{~L}}\)
(c) \(\frac{\pi r^{2} L}{M g l}\)
(d) \(\frac{\pi r^{2} l}{\mathrm{MgL}}\)
Answer:
(a) \(\frac{\mathrm{MgL}}{\pi r^{2} l}\)

Question 2.
Which substance is more elastic:
(a) Glass
(b) Steel
(c) Plastic
(d) Rubber.
Answer:
(b) Steel

Question 3.
Magnitude of Poisson’s ratio is in between :
(a) – 1 and \(\frac{1}{2}\)
(b) –\(\frac{3}{4}\) and –\(\frac{1}{2}\)
(c) \(\frac{1}{2}\) and 1
(d) 1 and 2
Answer:
(a) – 1 and \(\frac{1}{2}\)

Question 4.
Work done per unit volume for a wire is :
(a) Stress × Strain
(b) \(\frac{1}{2}\)× Stress × Strain
(e) 2(Stress × Strain)
(d) Stress/Strain.
Answer:
(b) \(\frac{1}{2}\)× Stress × Strain

Question 5.
Four wire made of same material, equal weight is suspended on them, increase in length will be maximum for wire :
(a) Length 50 cm and diameter 0.5 mm
(b) Length 100 cm and diameter 1 mm
(c) Length 200 cm and diameter 2 mm
(d) Length 300 cm and diameter 3 mm.
Answer:
(a) Length 50 cm and diameter 0.5 mm

Question 6.
10 newton force is needed to break a wire of radius one millimetre. What is force needed to break a wire of radius 3 mm :
(a) 90N
(b) \(\frac{10}{3}\)N
(c) \(\frac{10}{9}\)N
(d) 30N
Answer:
(a) 90N

Question 7.
In general modulus of rigidity is than modulus of elasticity:
(a) Less
(b) More
(c) Equal
(d) None of these.
Answer:
(a) Less

Question 8.
A wire of length L, Area of cross-section A, Young modulus of elasticity Y and coefficient of linear expansion a is heated up to t°C. Force experience on wire will be :
(a) γAαt
(b) γAαLt
(c) tALα
(d) \(\frac{t \alpha \mathrm{L}}{\mathrm{A}}\)
Answer:
(a) γAαt

2. Fill in the blanks:

1. SI unit of stress is …………………
Answer:
Newton/metre²

2. On increasing temperature, coefficient of elasticity ………………….
Answer:
decreases

3. Within elastic limit ……………….. is directly proportional to strain.
Answer:
stress

4. Rubber is ………………. elastic than steel.
Answer:
less

5. For an ideal rigid body Young modulus is ………………..
Answer:
infinity

3. Match the following:

Column ‘A’	Column ‘B’
1. Hooke’s law	(a) Proportional to area of cross-section
2. Normal stress	(b) Zero
3. Breaking stress	(c) Shearing strain
4. Young modulus of elastic body	(d) Applicable within elastic limit
5. Tangential strain	(e) Volume strain.

Answer:
1. (d) Applicable within elastic limit
2. (e) Volume strain
3. (a) Proportional to area of cross-section
4. (b) Zero
5. (c) Shearing strain

4. Write true or false:

1. Hooke’s law is applicable within elastic limit.
Answer:
True

2. Diamond can be considered as rigid body.
Answer:
True

3. Product of strain and stress is equal to stored energy.
Answer:
False

4. Young modulus of elasticity is dimensionless.
Answer:
False

5. Breaking stress depends on body of material.
Answer:
True

6. Young modulus of elasticity is valid for solid only.
Answer:
True

7. Rubber is more elastic than steel.
Answer:
False

8. Hookes law is defined within elastic limit.
Answer:
False

9. Quartz does not exhibit elastic after effect practically.
Answer:
True

10. It is difficult to twist a small rod in comparison to long rod.
Answer:
True

11. A hollow rod of same length and mass is more strong than a solid rod.
Answer:
True