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Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry

Question 1.
What are polymers?
Answer:
Polymers are high molecular mass macromolecules (10³ – 10⁷ u) and consist of repeating units of monomers.

Question 2.
Write the classification of polymers based on source. Give examples.
OR
What are natural and synthetic polymers? Give two examples of each type.
Answer:
Based on the source polymers are classified as natural, semisynthetic and synthetic polymers.

1. Natural polymers : These polymers are obtained either from plants or animals, e.g., cellulose, jute, linene, rubber, silk.
2. Semisynthetic polymers : The fibres obtained by giving special chemical treatment to natural fibres (cellulose) and further regenerated are called semisynthetic polymers e.g., acetate rayon, viscose rayon, cuprammonium silk.
3. Synthetic polymers : The man made fibres prepared by polymerization of one monomer or copolymerization of
   two or more monomers are called synthetic polymers, e.g., nylon, terylene, polythene, etc.

Question 3.
Write the classification of polymers based on structure. Give examples
OR
Write the reactions involved in the preparation of (1) Polyvinyl chloride (PVC) (2) Polypropylene.
Answer:
Based on structure polymers are classified as linear chain polymers, branched chain polymers and network or cross linked polymers.

(1) Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.

They have high melting points; high densities and high tensile strength.

(2) Branched-chain polymers : These polymers consist of long and straight chain with smaller side chains give rise to branched-chain polymers. They have low density. They have lower melting points and tensile strength. Polypropylene having methyl groups as branches.

(3) Network or cross-linked polymers : These polymers consist of cross-linking of chains by strong covalent bonds leading to a network-like structure. Cross-linking results from polyfunctional monomers, e.g., melamine, bakelite, vulcanization of rubber. These polymers are hard rigid and brittle.

Question 3.
How are polymers classified on the basis of mode of polymerization?
OR
Write the classification of polymers based on mode of polymerization.
Answer:
There are three modes of polymerization depending upon the types of reactions taking place between the monomers.

1. Addition polymerization (or chain growth polymerization)
2. Condensation polymerization (or step growth polymerization)
3. Ring opening polymerization
4. Addition polymerization or chain growth polymerization : It is a process of polymers by the repeated addition of a large number of monomers is called addition polymerization (like alkenes) without loss of any small molecules.
   Example : Formation of polyethylene from ethylene is well known example of addition polymerization. It is a chain growth polymerization.
5. Condensation polymerization or step growth polymerization : The process of formation of polymers from polyfunctional monomers with the elimination of some small molecules such as water, hydrochloric acid, methanol, ammonia is called condensation polymerization.
   
   Example : The formation of terylene, a polyester polymer, from ethylene glycol and terephthalic acid. Condensation polymerization is a step growth polymerization.
6. Ring opening polymerization : The process of formation of polymers from cyclic compounds (like lactams, cyclic ethers, etc.) by ring opening is called ring opening polymerization.
   Example : Polymerization of e-caprolactam.
   Ring opening polymerization proceeds by addition of a single monomer unit to the growing chain molecules. It is a step growth polymerization.

Question 4.
Classify the polymers given below as addition, condensation and ring opening polymers:
PVC, polythene, cyclic ethers, polyester, polyacrylonftrile. polystyrene.
Answer:

• Addition polymers: PVC, polythene. polyacrylonitrile. polystyrene.
• Condensation polymers: Polyester.
• Ring opening polymers : Cyclic ethers

Question 5.
Write the classification of polymers based on intermolecular forces. Give examples.
OR
In which dasses, are the polymers classified on the basis of Inter molecular forces?
Answer:
Molecular forces bind the polymer chains either by hydrogen bonds or Vander Waal’s forces. These forces are called intermolecular forces. On the basis of magnitude of intcrmolccular forces, polymers are further classified as ebstomers, fibres, thermoplastic polymers. thermosetting polymers.

(1) Elastomers: Weak van der Waals type of intermolecular forces of attraction between the polymer chains are observed in cbstomcrs. When polymer is stretched, the polymer chain stretches and when the strain is relieved the chain returns to its odginal position, Thus, polymer shows elasticity and is called elastomers. Elastomers, the elastic polymers, have weak van der Waals type of intermolecular forces which permit the polymer to be stretched. Lilastorners are soft and stretchy and used in making rubber bands. E.g.. neoprene, vulcanized rubber, buna.S, buna-N.

(2) Fibres : It consists of strong intermolecular forces of’ attraction due to hydrogen bonding and strong dipole-dipole forces. These polymers possess high tensile strength. Due to these strong intermolecular forces the fibres are crystalline in nature. They are used in textile industries, strung tyres. etc.. e.g., nylon, terylene.

(3) Thermoplastic polymers: These polymer possess moderately strong intermolecular forces of attraction between those of elastomers and fibres. These polymers arc called thermoplastic because they become soft on heating and hard on cooling. They are either linear or branched chain polymers. They can be remoulded and recycled. E.g. polyethenc, PVC, polystyrene.

(4) Thermosetting polymers: These polymers are cross linked or branched molecules and are rigid polymers. During their formation they have property of being shaped on heating. but they get hardened while hot. Once hardened these become infusible, cannot be softened by heating and therefore, cannot be remoulded and recycled.
This shows extensive cross linking by covalent bonds formed in the moulds during hardening/setting process while hot. E.g. Bakelite, urea formaldehyde resin.

Question 6.
What is meant by the term homopolymer?
Answer:
A polymer made from only one type of repeating units is called homopolymer. in some cases the repeating unit is formed by condensation of two distinct monomers. Examples. Polythene, PVC. Nylon-6.

Question 7.
What is meant by the term copolymer?
Answer:
A polymer made from two or more types of repeating units is called a copolymer. The different monomer units are randomly sequenced in the copolymer, e.g., Terylene, Nylon-6, 6, Buna-S, Buna-N.

Question 8.
Write the classification of polymers on the basis of biodegradability?
OR
(1) What are biodegradable polymers?
(2) What are nonbiodegradable polymers?
Answer:
Based un biodegradability, polymers are classified as biodegradable polymer and nonbiodegradable polymers.

(1) Biodegradable polymers: Polymers that are affected by microbes or disintegrate by themselves afler a certain period of time due to environmental degradation are called biodegradable polymers.

Examples: PHBV i.e., Polyhydroxy butyrate-CO-β-hydroxy valerate Dextron. Nylon-2-nylon-6.

(2) Non biodegradable polymers: Synthetic polymers do not disintegrate by themselves after a certain period or not affected by microbes, are called nonbiodegradhle polymers.

Examples: Bakelite, Nylon, Terylene.

Question 9.
Explain the following terms :
Answer:

1. Branched chain polymer : The polymer consists of long and straight chain with smaller side chains give rise to branched chain polymers, e.g. Polypropylene
2. Addition polymer : The polymer formed by the repeated addition of a large number of monomers (like alkenes) without loss of any small molecules are called addition polymers, e.g. polythene -[-CH₂ – CH₂-]_n. It is a chain growth polymerization.
3. Condensation polymer : The polymers formed by the repeated condensation reaction between polyfunctional monomers with the elimination of some molecules such as water, hydrochloric acid, methanol, ammonia are called condensation polymers, e.g. Nylon-6, 6.
4. Elastomers : Polymers in which the intermolecular forces of attraction between the polymer chains are the weakest. When polymer is stretched, it has ability to stretch and when the strain is relieved it returns to its original position. Thus, polymer shows elasticity and is called elastomers, e.g. natural rubber, neoprene, vulcanized rubber.
5. Homopolymer : A polymer made from only one type of repeating unit of one monomer is called homopolymer e.g. Polythene, PVC, etc.
6. Biodegradable polymer : Polymers which are affected by microbes or disintegrate by themselves after a certain period of time due to environmental degradation are called biodegradable polymers.
   Example : PHSV i.e. Polyhydroxy butyrate -CO-β-hydroxy valerate Dextron.

Question 10.
What is natural rubber?
Answer:
Natural rubber is a high molecular mass linear polymer of isoprene (2-methyl-1, 3-butadiene).

Question 11.
Write a note on natural rubber.
Answer:
Natural rubber is manufactured from rubber latex obtained from the rubber tree in the form of colloidal suspension. Reaction involved in the formation of natural rubber by the process of addition polymerization is as follows :

Natural rubber is -1, 4- polyisoprene. It exhibits elastic properties.

Question 12.
State properties of natural rubber.
Answer:

1. Polyisoprene molecule has cis configuration of the C = C double bond. It consists of various chains held together by weak van der Waals forces and has coiled structure.
2. It can be stretched like a spring and exhibits elastic property.
3. Its molecular mass varies from 130,000 u to 340,000 u.

Question 13.
Write a note on vulcanization of rubber. OR Discuss the main purpose of vulcanization of rubber.
Answer:
The tensile strength, toughness and elasticity of natural rubber can be increased by adding 3 to 5% sulphur and heating at 100-150°C, resulting in cross linking of cis-1, 4-polypropene chains through disulphide bonds, (-S-S-). This process is known as vulcanization of rubber. The physical properties of rubber can be changed by controlling the amount of sulphur in the vulcanization process. Rubber made with 20-30% sulphur is hard, 3 to 10% sulphur is little harder and is used in making tyres and 1 to 3% sulphur is used in making rubber bands.

Question 14.
Write the name and structure of one of the initiators used in free radical polymerisation.
Answer:
The initiator used in free radical polymerisation is acetyl peroxide.

Question 15.
What is LDP? How is it prepared? Give its properties and uses.
Answer:
LDP means low-density polyethene. LDP is obtained by polymerization of ethylene under high pressure (1000 – 2000 atm) and temperature (350 – 570 K) in presence of traces of O₂ or peroxide as initiator.

The mechanism of this reaction involves free radical addition and H-atom abstraction. The latter results in branching. As a result the chains are loosely held and the polymer has low density.

Properties of LDP :

• LDP films are extremely flexible, but tough chemically inert and moisture resistant.
• It is poor conductor of electricity with melting point 110 °C.

Uses of LDP :

• LDP is mainly used in preparation of pipes for agriculture, irrigation, domestic water line connections as well as insulation to electric cables.
• It is also used in submarine cable insulation.
• It is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc.

Question 16.
Whatis HDP ? How is it prepared ? Give its properties and uses ?
Answer:
HDP means high density polyethylene. It is a linear polymer with high density due to close packing.

HDP is obtained by polymerization of ethene in presence of Zieglar-Natta catalyst which is a combination of triethyl aluminium with titanium tetrachloride at a temperature of 333K to 343K and a pressure of 6-7 atm.

Properties of HDP :

• HDP is crystalline, melting point in the range of 144 – 150 °C.
• It is much stiffer than LDP and has high tensile strength and hardness.
• It is more resistant to chemicals than LDP.

Uses of HDP :

• HDP is used in manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, etc.
• It is used to prepare laboratory wares and other objects where high tensile strength and stiffness is required.

Question 17.
How is polyacrylonitrile (PAN) prepared? Give its uses.
Answer:
Acrylonitrile (monomer) on polymerization (addition polymerization) in the presence of peroxide initiator gives polyacrylonitrile.

Uses : Polyacrylonitrile resembles wool and is used as wool substitute and for making orlon or acrilan.

Question 18.
How is nylon-6 prepared?
Write the reaction for the preparation of nylon 6.
Answer:
When epsilon (ε)-caprolactam is heated with water at high temperature it undergoes ring opening polymerization to form the polyamide polymer called nylon-6.

The name nylon-6 is given on the basis of six carbon atoms present in the monomer unit. Nylon-6 has high tensile strength and luster, nylon-6 fibres are used for manufacture of tyre cords, fabrics and ropes.

Question 19.
Draw the structures of polymers formed using the following monomers.

Answer:

Question 20.
How is Novolac prepared?
OR
Write the reaction to prepare Novolac polymer.
Answer:
The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.

Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

Question 21.
How is bakelite prepared?
Answer:
In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.

Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

Question 22.
How is a melamine-formaldehyde polymer (melamine) prepared?
Answer:
The monomers formaldehyde and melamine undergo condensation polymerisation to form cross-linked melamine formaldehyde. It is used in making crockeries, decorative table tops like formica and plastic dinner-ware.

Question 23.
Write the preparation of the following synthetic rubbers and give their uses :
(1) Buna-S or styrene-butadiene rubber (SBR) (2) Neoprene rubber
Answer:
(1) Buna-S rubber : Its trade name is SBR (Styrene-butadiene rubber) Buna-S is a copolymer of styrene and 1, 3-butadiene. When 75 parts of butadiene and 25 parts of styrene subjected to addition polymerization by the action of sodium.

Uses : Buna-S is superior to natural rubber with regard to mechanical strength and has abrasion resistance. Hence, it is used in tyre industry.

(2) Neoprene : Neoprene, a synthetic rubber, is a condensation polymer of chloroprene (2-chloro-l, 3-butadiene), Chloroprene polymerizes rapidly in presence of oxygen.

Vulcanization of neoprene takes place in presence of magnesium oxide.

Uses : Neoprene is resistant to petroleum, vegetable oils, light as well as heat. It is used in making hose pipes for transport of gasoline and making gaskets. It is used for manufacturing insulator cable, jackets, belts for power transmission and conveying.

Question 24.
Write structure of natural rubber and neoprene rubber along with the name and structure of their monomers.
Answer:

Question 25.
Write the preparation of viscose rayon.
Answer:
Viscose rayon is a semisynthetic fibre. It is a regenerated cellulose. The molecular formula of cellulose is (C₆H₁₀O₅)_n. A modified representation of the molecular formula of cellulose Cell-OH is used in the reactions involved in viscose formation.

Cellulose (wood pulp) is treated with the concentrated NaOH to form alkali cellulose. It is then converted to xanthate by treating with CS₂. Further xanthate is mixed with dilute NaOH to form viscose solution which is extruted through spinnerates of spinning machine into acid bath, when regenerated cellulose fibres precipitate, i.e. viscose rayon.

Question 26.
How is PHBV polymer prepared?
Answer:
It is a copolymer. The monomers β-hydroxy butyric acid (3-hydroxy butanoic acid) and β-hydroxy valeric acid (3-hydroxy pentanoic acid) undergo polymerization to form PHBV polymer. It has an ester linkage. Hydroxyl group of one monomer forms ester link by reacting with carboxyl group of the other. Thus PHBV is an aliphatic polyester i.e. poly β-hydroxybutyrate-co-β-hydroxy valerate (PHBV). It is a biodegradable polymer.

Question 27.
Write the name/s of monomer/s, polymer structure and one use of each of the following polymers (trade name) :
Answer:

Question 28.
Write the names of monomers used in preparing following polymers :
(1) Dacron.
Answer:
Monomers : Ethylene glycol and Dmiethyl terephthalate (DMT)

(2) Bakelite.
Answer:
Monomers : o-hydroxy methyl phenol and formaldehyde.

(3) Nylon-6, 8.
Answer:
Monomers : Hexamethylene diamine and Hexamethylene dicarboxylic acid.
H₂N-(CH₂)₆-NH₂ HOOC-(CH₂)₆-COOH

(4) Melamine.
Answer:
Monomers : Melamine and Formaldehyde

(5) Buna-S.
Answer:

(6) Buna-N.
Answer:

(7) Butyl rubber.
Answer:

(8) Teflon.
Answer:
Monomers : F₂C = CF₂ Tetrafluoroethene

(9) Natural rubber.
Answer:
Monomers : 1,3-Butadiene
CH₂ = CH – CH = CH₂

(10) Neoprene.
Answer:
Monomers : Chloroprene or 2-Chloro-l,3-butadiene

Question 29.
Write the structures of monomers used in the preparation of following polymers.

Answer:
The monomers used in the preparation of above polymer are :

Answer:
The monomers used in the preparation of above polymer are :

Answer:
The monomer used in the preparation of above polymer are :

Answer:
The monomer used in the preparation of above polymer is

Question 30.
Following monomers are used to prepare polymers. Predict the structures of polymers:

(1) Ethylene glycol.
Answer:
Ethylene glycol is used in the preparation of polyester (terylene or dacron).

Terylene is polyester fibre formed by the polymerization of terephthalic acid and ethylene glycol.

Terylene is obtained by condensation polymerization of ethylene glycol and terephthalic acid in presence of catalyst zinc acetate and antimony trioxide at high temperature.

Properties :

• Terylene has relatively high melting point (265 °C)
• It is resistant to chemicals and water.

Uses :

• It is used for making wrinkle-free fabrics by blending with cotton (terycot) and wool (terywool), and also as glass reinforcing materials in safety helmets.
• PET is the most common thermoplastic which is another trade name of the polyester polyethylenetereph- thalate.
• It is used for making many articles like bottles, jams, packaging containers.

(2) ε-Caprolactam.
Answer:
ε-caprolactam is used in the preparation of nylon-6.

When epsilon (ε)-caprolactam is heated with water at high temperature it undergoes ring opening polymerization to form the polyamide polymer called nylon-6.

The name nylon-6 is given on the basis of six carbon atoms present in the monomer unit. Nylon-6 has high tensile strength and luster, nylon-6 fibres are used for manufacture of tyre cords, fabrics and ropes.

(3) Ethene.
Answer:
Ethene is used in the preparation of polythene
Polymer:

Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.

They have high melting points; high densities and high tensile strength.

(4) Formaldehyde.
Answer:
Formaldehye is used in the preparation of bakelite.

The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.

Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.

Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

Question 31.
Classify the following polymers as step growth or chain growth polymers :
(1) Nylon-6,6
(2) Terylene
(3) Polyethene (4) PVC.
Answer:
Step growth polymers : Nylon-6,6, terylene
Chain growth polymers : Polythene, PVC.

Question 32.
Classify the following as linear, branched or cross linked polymers :
(1) Bakelite
(2) Starch
(3) Polythene
(4) Nylon.
Answer:
Linear polymers : Polythene, nylon.
Cross-linked polymers : Bakelite, starch.

Question 33.
Classify the following as addition and condensation polymers :
(1) Bakelite
(2) polyvinyl chloride
(3) polythene
(4) terylene.
Answer:
Addition polymers : Polyvinyl chloride, polythene
Condensation polymers : Bakelite, terylene.

Question 34.
Arrange the polymers in increasing order of their intermolecular forces :
Nylon-6,6, Polythene, Buna-S.
Answer:
The increasing order of their intermolecular forces of attraction follows the order :
Buna-S, Polythene, Nylon-6,6.

Question 35.
Classify the following as natural, synthetic and semisynthetic polymers :
Terylene, cuprammonium silk, jute, melamine
Answer:
Natural polymers : Jute
Synthetic polymers : Terylene, melamine
Semisynthetic polymers : Cuprammonium silk

Question 36.
Complete the following statements :
(1) Chemically teflon is …………………………. .
(2) …………………………. is the catalyst used to obtain HDP by polymerisation of ethene.
(3) Viscose rayon is a …………………………. .
Answer:
(1) polytetrafluoroethylene
(2) Zieglar-Natta
(3) semisynthetic fibre (regenerated fibre).

Question 37.
Answer the following in one sentence each.

(1) Name a polymer used for making LCD screen?
Answer:
The polymer used for making LCD screen is Perspex.

(2) Which of the two is a condensation polymer? Bakelite or Polythene?
Answer:
The condensation polymer is bakelite.

(3) Which of the two is a linear polymer? Nylon or Starch.
Answer:
The linear polymer is nylon.

(4) Which of the two is a step growth polymer? Nylon-6,6 or PVC.
Answer:
The step growth polymer is Nylon-6,6.

(5) Write the use of polyacrylamide gel.
Answer:
Polyacrylamide gel is used in electrophoresis.

(6) Write the use of urea formaldehyde resin.
Answer:
Urea formaldehyde resin is used in making unbreakable dinnerware and decorative laminates.

(7) Give an example of semisynthetic polymer.
Answer:
Semisynthetic polymer : Viscose rayon, cuprammonium silk.

(8) Write the monomer unit of teflon.
Answer:
Monomer unit of teflon : Tetrafluoroethene (F₂C = CF₂).

(9) Write the equation for the preparation of polypropylene.
Answer:

(10) Name a synthetic polymer which contains amide linkage.
Answer:
Polymer that contains amide linkage : Nylon-6; Nylon 6,6.

(11) Name a synthetic polymer which contains ester linkage.
Answer:
Polymer that contains ester linkage : Terylene or Dacron.

(12) Name one thermosetting and one thermoplastic polymer.
Answer:
Thermosetting polymer : Bakelite.
Thermoplastic polymer : Polythene, polystyrene.

(13) State the uses of biodegradable polymers.
Answer:
Biodegradable polymers are used as orthopaedic devices, implants, sutures and drug release matrices.

(14) Name a copolymer which is used for making nonbreakable crockeries.
Answer:
The polymer used in making nonbreakable crockeries : Melamine formaldehyde polymer.

(15) Write the structure of monomer used in the preparation of
Answer:
The structure of monomer :

(16) Write the structure of melamine.
Answer:
The monomers formaldehyde and melamine undergo condensation polymerisation to form cross-linked melamine formaldehyde. It is used in making crockeries, decorative table tops like formica and plastic dinner-ware.

(17) What does SBR stand for?
Answer:
SBR stand for styrene(S)butadiene (B) rubber (R).

(18) Draw the structure of repeating unit in nylon-6,10.
Answer:
The structure of repeating unit in nylon-6,10 is

(19) What are the monomers used to prepare nylon given below?

Answer:
Monomers used in the preparation of nylon are

(20) Write the monomers used to prepare nylon given below :

Answer:
Monomers used in the preparation of nylon are

(21) Name the catalyst which is formed from titanium chloride and triethyl aluminium.
Answer:
The catalyst Zieglar Natta is formed from titanium chloride and triethyl aluminium.

(22) Define molecular mass of polymer.
Answer:
Molecular mass of a polymer is an average of the molecular masses of the constituent molecules.

(23) Which factor determines the molecular mass of polymer.
Answer:
Molecular mass of polymer depends upon the degree of polymerization (DP). DP is the number of monomer units in a polymer molecule.

Multiple Choice Questions

Question 38.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which one is the natural polyamide polymer?
(a) Cuprammonium silk
(b) Wool
(c) Perlon-L
(d) Jute
Answer:
(b) Wool

2. The synthetic fibres are prepared from
(a) cellulose
(b) starch
(c) chemical compounds
(d) polymers
Answer:
(c) chemical compounds

3. Which of the following is NOT a vegetable fibre?
(a) Hemp
(b) Jute
(c) Cotton
(d) Keratin
Answer:
(d) Keratin

4. Which of the following fibres are made up of polyamides?
(a) Dacron
(b) Rayon
(c) Nylon
(d) Terylene
Answer:
(c) Nylon

5. An example of natural cellulose fibre is
(a) cotton
(b) wool
(c) silk
(d) rayon
Answer:
(a) cotton

6. Cellulose is the main constituent of
(a) nylon-6
(b) cotton
(c) terylene
(d) wool
Answer:
(b) cotton

7. Of the following, which group contains two cellulosic fibres and one protein fibre?
(a) Cotton, keratin, wool
(b) Linen, keratin, wool
(c) Cotton, linen, rayon
(d) Cotton, keratin, linen
Answer:
(d) Cotton, keratin, linen

8. Which of the following is not a vegetable fibres?
(a) Hemp
(b) Jute
(c) Cotton
(d) Keratin
Answer:
(d) Keratin

9. Which of the following is polyamide?
(a) Teflon
(b) Nylon 6, 6
(c) Terylene
(d) Bakelite
Answer:
(b) Nylon 6, 6

10. The monomer of e-caprolactam is
(a) styrene
(b) amino acid
(c) aminocaproic acid
(d) adipic acid O
Answer:
(c) aminocaproic acid

11. is the formula of _ Jn
(a) Nylon-66 salt
(b) Nylon-66
(c) DMT
(d) Nylon-6
Answer:
(d) Nylon-6

12. Nylon-6 is a
(a) polyester fibre
(b) protein fibre
(c) poly caprolactum fibre
(d) poly amine fibre
Answer:
(c) poly caprolactum fibre

13. The condensation product of e-caprolactum is
(a) teflon
(b) nylon-6
(c) nylon-66
(d) bakelite
Answer:
(b) nylon-6

14. \(\left[\overline{\mathrm{O}} \mathrm{OC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{COO}-\mathrm{N} \mathrm{H}_{3}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{N} \mathrm{H}_{3}\right]\) is the formula of
(a) nylon-6
(b) nylon-6 salt
(c) nylon-66 salt
(d) nylon-66
Answer:
(b) nylon-6 salt

15. The starting material required for the preparation of Nylon-66 is
(a) glycol
(b) α-amino acid
(c) adipic acid and hexamethylene diamine
(d) dimethyl terephthalate and ethylene glycol
Answer:
(c) adipic acid and hexamethylene diamine

16. Terylene is also known as
(a) styrene
(b) butadiene
(c) dacron
(d) teflon
Answer:
(c) dacron

17. Terylene is
(a) vegetable fibre
(b) protein fibre
(c) polyester fibre
(d) polyamide fibre
Answer:
(c) polyester fibre

18. Terylene polymer is formed in
(a) the presence of nitrogen
(b) the presence of hydrogen
(c) the presence of oxygen
(d) the vacuum
Answer:
(d) the vacuum

19. The by-product formed during the preparation of terylene fibre is
(a) glycerol
(b) propylene glycol
(c) ethylene glycol
(d) ethyl alcohol
Answer:
(c) ethylene glycol

20. Nylon polymer cannot be used for making
(a) tyre cords
(b) films
(c) dress materials
(d) fishing nets
Answer:
(b) films

21. Glycol is an important constituent of
(a) nylon-6
(b) nylon-66
(c) terylene
(d) hexamethylene diammonium adipate
Answer:
(c) terylene

22. Terylene is prepared by the process of
(a) halogenation
(b) condensation
(c) esterification
(d) hydrogenation
Answer:
(b) condensation

23. What are the steps during polymerisation to form terylene?
(a) Terephthalic acid is condensed with ethylene glycol at 420 K-460 K.
(b) Ethylene glycol displaces methanol to form dihydroxy diethyl terephthalic acid
(c) Zinc acetate – antimony trioxide is used as catalyst
(d) All of these
Answer:
(d) All of these

24. During polymerisation of nylon salt to nylon-66, the conditions are
(a) room temperature and pressure
(b) temperature 503 K
(c) temperature 553 K in presence of Nitrogen
(d) heating in an autoclave at 373 K
Answer:
(c) temperature 553 K in presence of Nitrogen

25. Which one of the following is a condensation polymer?
(a) Nylon
(b) Polythene
(c) PVC
(d) Teflon
Answer:
(a) Nylon

26. Which one of the following is an addition polymer?
(a) Bakelite
(b) Nylon-6,6
(c) Polystyrene
(d) Terylene
Answer:
(c) Polystyrene

27. A polymer of butadiene and Acrylonitrile is called
(a) Buna-S
(b) Buna-N
(c) Buna-B
(d) Buna-A
Answer:
(b) Buna-N

28. Natural rubber is a polymer of
(a) Styrene
(b) Butadiene
(c) Vinyl chloride
(d) Isoprene
Answer:
(d) Isoprene

29. In which of the following pairs both are copolymers?
(a) PHBV, bakelite
(b) Polythene, terylene
(c) Polyacrylonitrile, nylon-6,6
(d) Polystyrene, melamine
Answer:
(a) PHBV, bakelite

30. The polymer used in paints is
(a) Nylon
(b) Glyptal
(c) Neoprene
(d) Terylene
Answer:
(b) Glyptal

31. Which of the following contains biodegradable polymers only?
(a) Cellulose, dextron, PHBV
(b) Starch, PHBV, PVC
(c) Bakelite, nylon-2-nylon-6, nylon-6,6
(d) Cellulose, starch, terylene
Answer:
(a) Cellulose, dextron, PHBV

32. Thermosetting polymer is
(a) Nylon-6
(b) Nylon-6,6
(c) Bakelite
(d) SBR
Answer:
(c) Bakelite

33. Nylon thread contains the polymer
(a) Polyamide
(b) Polyvinyl
(c) Polyester
(d) Polyethylene
Answer:
(a) Polyamide

34. Polythene, PVC, teflon and neoprene are all
(a) Monomers
(b) Homopolymers
(c) Copolymers
(d) Condensation polymers
Answer:
(b) Homopolymers

35. Which one of the following is NOT a biodegrad¬able polymer?
(a) Starch
(b) Cellulose
(c) Dextron
(d) Decron
Answer:
(d) Decron

36. The polymer used in making blankets (artificial wool) is
(a) Polyester
(b) Polyacrylonitrile
(c) Polythene
(d) Polystyrene
Answer:
(b) Polyacrylonitrile

37. Which one of the following is a linear polymer?
(a) Bakelite
(b) LDP
(c) Nylon
(d) Formaldehyde melamine polymer
Answer:
(c) Nylon

38. Which one of the following is a branched polymer?
(a) PVC
(b) Polyester
(c) Nylon
(d) Polypropylene
Answer:
(d) Polypropylene

39. The polymer used to make non-stick cookware is
(a) Polyethene
(b) Polystyrene
(c) Polytetrafluoroethylene
(d) Polyvinyl chloride
Answer:
(c) Polytetrafluoroethylene

40. The monomer used to prepare orlon is
(a) CH₂ = CH-CN
(b) CH₂ = CHCl
(c) CH₂ = CH-F
(d) CH₂ = CF₂
Answer:
(a) CH₂ = CH-CN

41. Buna-N rubber is a copolymer of

Answer:
(b)

42. Bakelite is a polymer of
(a) Formaldehyde and phenol
(b) Benzaldehyde and phenol
(c) Formaldehyde and benzyl alcohol
(d) Acetaldehyde and phenol
Answer:
(a) Formaldehyde and phenol

43. The process involving heating of natural rubber with sulphur is known as
(a) Sulphonation
(b) Vulcanisation
(c) Galvanisation
(d) Calcination
Answer:
(b) Vulcanisation

44. A polymer of bisphenol and phosgene is called
(a) Polyamide
(b) Glyptal
(c) Polycarbonate
(d) Polystyrene
Answer:
(c) Polycarbonate

45. Thermocol is a homopolymer of
(a) terephthalic acid
(b) acrylonitrile
(c) methyl a-cyanoacrylate
(d) styrene
Answer:
(d) styrene

46. The polymer is used to prepare shatter resistant glass is called
(a) Perspex/acrylic glass
(b) Soda glass
(c) Buna N
(d) Polyacrylamide
Answer:
(a) Perspex/acrylic glass

47. A polymer used in making shoe soles is
(a) Glyptal
(b) Buna-N
(c) Buna-S
(d) Poly carbonate
Answer:
(b) Buna-N

48. The Zieglar-Natta catalyst is used in the preparation of
(a) LDPE
(b) PHBV
(c) PAN
(d) HDPE
Answer:
(d) HDPE

49. Which of the following is natural rubber?
(a) cis-1, 4-polyisoprene
(b) neoprene
(c) Trans-1. 4-polyisoprene
(d) Butyl rubber
Answer:
(a) cis-1, 4-polyisoprene

50. Which one from the following is the Terylene polymer?

Answer:
(b)

51. Equivalent amount of Hexamethylene diamine and adipic acid on complete neutralization produces :

Answer:
(a)

52. Polyhydroxy butyrate-CO-β-hydroxy valerate represents
(a) Dextron
(b) Nylon-6, 6
(c) Nylon-2-nylon-6
(d) PHBV
Answer:
(d) PHBV

53. Among the following, the biodegradable polymer is
(a) PVC
(b) Nylon-6, 6
(c) Nylon-2-nylon-6
(d) Neoprene
Answer:
(c) Nylon-2-nylon-6

54. The monomers of Nylon-2-nylon-6 are
(a) glycine and ω-amino caproic acid
(b) lactic acid and glycolic acid
(c) glycolic acid and co-amino caproic acid
(d) isobutylene and isoprene
Answer:
(a) glycine and ω-amino caproic acid

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Negotiable Instrument which can be discounted with the bank _____________
(a) Cheque
(b) Bills of Exchange
(c) Bank Draft
Answer:
(b) Bills of Exchange

Question 2.
Bank overdraft is a _____________ term facility provided by bank.
(a) medium
(b) short
(c) long
Answer:
(b) short

Question 3.
A loan for a period of more than 5 years is called as _____________ loans.
(a) short term
(b) long term
(c) medium-term
Answer:
(b) long term

Question 4.
Purchase and sale of securities is _____________ function of Bank.
(a) primary
(b) utility
(c) agency
Answer:
(c) agency

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) Current Account	(1) Time Deposits
(b) Financial Arrangement	(2) ATM, Credit Card and Debit Card
(c) Agency Function	(3) Demand Deposits
(d) Utility Function	(4) Collection of Dividend and Interest
(e) Cash Credit	(5) Bank Overdraft
(f) Bill of Exchange	(6) Advance against the stock of raw material
(g) Letter of Credit	(7) Non-negotiable instrument
	(8) Bank Draft
	(9) Negotiable Instrument
	(10) International trade transaction

Answer:

Group ‘A’	Group ‘B’
(a) Current Account	(3) Demand Deposits
(b) Financial Arrangement	(5) Bank Overdraft
(c) Agency Function	(4) Collection of Dividend and Interest
(d) Utility Function	(2) ATM, Credit Card, and Debit Card
(e) Cash Credit	(6) Advance against the stock of raw material
(f) Bill of Exchange	(9) Negotiable Instrument
(g) Letter of Credit	(10) International trade transaction

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
Deposits generating saving habits among the people.
Answer:
Saving Deposits

Question 2.
Deposits are not repayable on demand.
Answer:
Time Deposits

Question 3.
Deposits repayable on demand.
Answer:
Demand Deposits

Question 4.
The loan was provided for a period of less than one year.
Answer:
Short Term Loan

Question 5.
The loan is provided for a period between 1 to 5 years.
Answer:
Medium-Term Loan

Question 6.
The loan was provided for a period of more than 5 years.
Answer:
Long Term Loan

1D. State whether the following statements are True or False

Question 1.
Letter of credit is issued by banks for domestic trade transactions.
Answer:
False

Question 2.
Bill of Exchange is a negotiable instrument.
Answer:
True

Question 3.
The bank acts as a trustee, attorney, and executor of the will.
Answer:
True

Question 4.
RTGS is an agency function of the Bank.
Answer:
False

Question 5.
The bank plays the role of Depository Participant (DP).
Answer:
True

1E. Find the odd one.

Question 1.
Loan, Deposit, Cash credit, Overdraft facility.
Answer:
Deposit

Question 2.
Cheque, Withdrawal slip, Pay in slip.
Answer:
Pay in slip

Question 3.
Travellers Cheque, Safe Deposit Vault, NEFT, Transfer of Money.
Answer:
Transfer of Money

Question 4.
Collection of Dividend, Collection of cheque, Buying, and Selling of Securities, Payment of Electricity Bill.
Answer:
Buying and Selling of Securities

1F. Complete the sentences.

Question 1.
The appointment of bankers of a company is made by the _____________
Answer:
Board of Directors

Question 2.
Banker is a dealer in _____________
Answer:
Money

Question 3.
Bank account in which money is deposited at regular interval is a _____________
Answer:
Recurring Deposit Account

Question 4.
The instrument which can be discounted with the bank is a _____________
Answer:
Bills of Exchange

Question 5.
Businessman, Companies etc, usually open a _____________ account with a bank.
Answer:
Current

Question 6.
There is no restriction on the withdrawals from _____________ account.
Answer:
Current

Question 7.
Interest is not paid on the _____________ account.
Answer:
Current

Question 8.
_____________ account is suitable for salaried people.
Answer:
Saving

Question 9.
Stop payment letter is sent, when the cheque is _____________
Answer:
lost/misplaced

Question 10.
Withdrawals are not permitted from the _____________ accounts.
Answer:
Fixed Deposits

Question 11.
The facility given by the bank to draw more money than the actual balance in the credit is called _____________
Answer:
Overdraft facility

Question 12.
A deposit which is kept for a fixed period in a bank is a _____________
Answer:
Fixed Deposit

Question 13.
A type of bank account which is generally opened by a businessman for his business transactions is a _____________
Answer:
Current account

Question 14.
A type of account in which the interest is paid at higher rate is a _____________
Answer:
Fixed Deposit Account

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) Loans not more than one year	……………………….
(2) ……………………..	Cash Credit
(3) Collection of cheques	………………………..
(4) ……………………..	Letter of Credit

(Interest charged on, actual amount withdrawn, Agency function, Utility function, Short Term loans)
Answer:

Group ‘A’	Group ‘B’
(1) Loans not more than one year	Short Term loans
(2) Interest charged on actual amount withdrawn	Cash Credit
(3) Collection of cheques	Agency function
(4) Utility function	Letter of Credit

1H. Answer in one sentence.

Question 1.
What is a demand deposit?
Answer:
Deposits that are repayable on demand are called demand deposits. There are 2 types of demand deposits i.e Saving Deposits and Current Deposits.

Question 2.
What is a time deposit?
Answer:
Deposits that are repayable after a specific period of time are called time deposits. There are 2 types of time deposits i.e. Fixed Deposits and Recurring Deposits.

Question 3.
What do you mean by loan?
Answer:
Any amount granted or lent for a specific period of time against personal security, gold or silver, or other movable or immovable assets is called a loan.

Question 4.
What is the Bill of Exchange?
Answer:
It is a written unconditional order by the seller to the buyer, to pay a certain sum of money on a future fixed date for payment of goods or services received.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
Short-term loans are of more than 5 years.
Answer:
Long-term loans are of more than 5 years.

Question 2.
Bills of Exchange are the non-negotiable instrument.
Answer:
Bills of exchange is a negotiable instrument.

Question 3.
The collection of cheques and bills is a utility function of the bank.
Answer:
The collection of cheques and bills is the agency function of the bank.

Question 4.
The rate of interest is high for saving deposits.
Answer:
The rate of interest is high for fixed deposits.

Question 5.
Deposits that are repayable after the specific period is Demand Deposit.
Answer:
Deposits that are repayable after the specific period are Time Deposits.

2. Answer in brief.

Question 1.
Explain the types of term loans.
Answer:
A loan granted for a specific time period against personal security, gold or silver, and movable and immovable assets is called a term loan.
Types of term loans:

• Short-term loan: A loan repayable within 1 year is called a short-term loan. It is generally taken by businessmen to meet their working capital requirements.
• Medium-term loan: A loan repayable within 2 years to 5 years period is called a medium-term loan.
• Long-term loan: A loan repayable after 5 years is called a long-term loan. It is taken by businessmen for the growth and development of the business.

Question 2.
Explain the types of Advances.
Answer:
Advance is a credit facility provided by the bank to its customers. Advances are for a shorter period than loans.
Type of Advances:

• Overdraft: It is an arrangement in which a customer i.e. Current A/c holder, is allowed to overdrew money in excess of the credit balance in his account. It is allowed against collateral securities like – shares, F.D.R., Government securities, etc.
• Cash credit: In this mode of advance, a separate bank account called ‘Cash Credit Account is opened in the name of the borrower. He can withdraw from this account as and when required. It is allowed against security like – stock of raw materials, finished goods, etc.
• Discounting of Bills: Discounting of bills means encashing the Bills of Exchange before its due date with the banker. The bank charges a certain amount of interest for this service which is called a discounting rate.

3. Justify the following statements.

Question 1.
In cash credit, the customer’s account is credited by a bank with the sanctioned amount.
Answer:
Cash credit is another kind of credit facility given by the bank to its customers including businessmen, companies, etc.

• A separate bank account known as a “Cash Credit Account” is required to be opened in the name of the borrower.
• The bank credits the account as per the sanctioned cash limit from which the customer can utilize funds whenever required for cash credit.
• Generally, the security of tangible assets like goods, finished stock is required to be kept.
• The interest is charged only on the actual amount utilized by the customer.
• Cash credit arrangement is for a longer period as compared to overdraft.
• This system of lending is prevalent in India only.
• Thus, in cash credit, the customer’s account is credited by a bank with the sanctioned amount.

Question 2.
Bank correspondence should be brief and to the point.
Answer:

• The bank is a financial institution.
• A company secretary has to conduct bank correspondence as per the instruction of the Board.
• Bank correspondence needs careful and cautious drafting.
• The company secretary has to use his knowledge, skill, and experience while conducting bank correspondence,
• The company secretary has to conduct bank correspondence promptly and accurately.
• Mistakes and delays in bank correspondence may bring financial loss to the company.
• Bank correspondence should be always brief, compact, and precise.
• Unnecessary or irrelevant information should be avoided in bank correspondence.
• Thus, bank correspondence should be brief and to the point.

Question 3.
No interest is paid by the bank on the current account.
Answer:

• The Current account is normally opened by businessmen, firms, or companies.
• A current account is a running account and in practice it never becomes time-barred.
• This account is opened with a minimum deposit.
• There is no limit on the amount or number of withdrawals.
• Interest is not payable on this account.
• Overdraft facility is given only to current depositors after following the prescribed bank procedure.
• Hence, interest is paid only in the case of recurring, fixed, and saving accounts and not in the case of the current account.
• Thus, no interest is paid by the bank on the current account.

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 3 Ionic Equilibria Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 3 Ionic Equilibria

Question 1.
What are electrolytes?
Answer:
Electrolytes: The substances which in their aqueous solutions (or in any polar solvents) dissociate or ionize forming positively charged ions (cations) and negatively charged ions (anions) are called electrolytes. For example, NaCl, HCl, etc.

Question 2.
What is an ionic equilibrium?
Answer:
Ionic equilibrium: The equilibrium between ions and unionized molecules of an electrolyte in solution is called an ionic equilibrium.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+}\)

Question 3.
What are the types of electrolytes?
Answer:
There are two types of electrolytes as follows :
(a) Strong electrolyte The electrolytes which ionise completely or almost completely are called strong electrolytes. For example, NaCl, HCl, H₂SO₄, etc.
(b) Weak electrolytes: The electrolytes which dissociate to a less extent are called weak electrolytes. For example, CH₃COOH, NH₄OH, etc.

Question 4.
Define degree of dissociation.
Answer:
Degree of dissocsavIt is defined as a fraction of total number of moles of an electrolyte that dissociate into its ions at equilibrium.
It is denoted by a and represented by,

OR α = \(\frac{\text { Per cent dissociation }}{100}\)
∴ Per cent dissociation = α × 100

Question 5.
Define acid and base. Give examples.
Answer:
Acid : A hydrogen containing substance which gives H⁺ ions in aqueous solution is called an acid. For example, HCl, CH₃COOH, etc.
\(\mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)

Base : A substance that contains OH group and produces hydroxide ions (OH^–) in aqueous solution is called a base.
\(\mathrm{NaOH}_{(\mathrm{aq})} \longrightarrow \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)

Question 6.
What are the limitations of Arrhenius acid-base theory ?
Answer:
Limitations of Arrhenius theory :

1. This theory is applicable only for aqueous solutions and not for non-aqueous solutions.
2. It fails to explain the acidic nature of non-hydrogen compounds like BF₃, AlCl₂, FeCl₃, etc.
3. It fails to explain the basic nature of non-hydroxy compounds like NH₃, amines, Na₂CO₃, KCN, aniline, etc. in their aqueous solutions.
4. It does not explain role of solvent or existence of H₃O⁺ in an aqueous solution of an acid.

Question 7.
Explain neutralisation reaction according to Arrhenius theory.
Answer:
Neutralisation reaction : According to Arrhenius theory neutralisation is a reaction between an acid and a base in their aqueous solutions produciijg salt and unionised water.
\(\mathrm{HCl}_{(\mathrm{aq})}+\mathrm{NaOH}_{(\mathrm{aq})} \rightarrow \mathrm{NaCl}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
Since strong acid, strong base and salt dissociate completely, the above reaction is represented as,

Hence, according to Arrhenius theory neutralisation reaction is defined as a reaction between H⁺ ions and OH^– ions forming unionised water molecules.

Question 8.
Explain Bronsted-Lowry theory of acids and bases.
Answer:
Acid : According to Bronsted-Lowry theory acid is a substance that donates a proton (H⁺) to another substance.
Base : According to this theory base is a substance that accepts a proton (H⁺) from another substance. For example,

Since HCl donates a proton it is an acid while NH₃ accepts a proton it is a base.

Question 9.
For each of the following reactions, identify the Lowry-Bronsted conjugate acid-base pairs :

Answer:

In this Acid-1 (CH₃COOH) and Base-1 (CH₃COO^–) is one acid base conjugate pair while Base-2 (NH₃) and Acid-2 \(\mathrm{NH}_{4}^{+}\) is another conjugate pair.

Question 10.
Mention a conjugate acid and a conjugate base for each of the following :
(a) H₂O (b) \(\mathrm{HSO}_{4}^{-}\) (C) Br^– (d) H₂CO₃ (e) \(\mathbf{H}_{2} \mathbf{P O}_{4}^{-}\) (f) \(\mathbf{N H}_{4}^{+}\)
Answer:

Substance	Conjugate acid	Conjugate base
(a) H2O	H3O+	OH–
(b) \(\mathrm{HSO}_{4}^{-}\)	H2SO4	\(\mathrm{SO}_{4}^{-2}\)
(c) Br–	HBr	–
(d) H2CO3	–	\(\mathrm{HCO}_{3}^{-}\)
(e) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)	H3PO4	\(\mathrm{HPO}_{4}^{-2}\)
(f) \(\mathrm{NH}_{4}^{+}\)	–	NH3

Question 11.
Identify conjugate acid-base pairs in the following :


Answer:

Question 12.
Define acids and bases on the basis of Lewis concept. Give examples.
Answer:
Lewis concept of an acid and a base is based on the electronic theory.
Acid : It is defined as any species (molecule or ion) that can accept a pair of electrons. E.g. BF₃, AlCl₃ and all electron deficient species like cations (K⁺, Ag⁺) and molecules having incomplete octet, like BeF₂, BF₃.

Base : It is defined as any species (molecule or ion) that can donate a pair of electrons. E.g. NH₃, C₂H₅NH₂ and all electron rich species like anions (Cl^–, OH^–) and all molecules with lone pair of electrons.

Question 13.
Explain : (A) BF₃ is a Lewis acid, (B) NH₃ is a Lewis base.
Answer:
(A) According to Lewis theory, an acid is a substance which can accept a pair of electrons.
In BF₃ molecule, the octet of B is incomplete, hence it needs two electrons or a pair of electrons to complete its octet. Hence BF₃ acts as a Lewis acid.

(B) According to Lewis theory, a base is a substance which can donate a pair of electrons.
In NH₃ molecule, nitrogen atom has one lone pair of elctrons to donate.

The reaction between BF3 and NH3 can be represented as,

Question 14.
Classify the following into Lewis acids and bases :
CN^–, Cl^–, S^2-, Cu⁺⁺, H₂O, OH^–, BF₃, Ag⁺.
Answer:
(1) Lewis acid : Cu⁺⁺, BF₃, Ag⁺
(2) Lewis bases : CN^–, Cl^–, S^2-, OH^–.

Question 15.
Explain amphoteric nature of water.
Answer:
(1) Since water acts as an acid as well as a base, it is amphoteric in nature.
(2) H₂O has a tendency to donate a proton forming OH^– as well as has a tendency to accept a proton forming H₃O⁺.


Therefore H₂O is amphoteric in nature.

Question 16.
How are acids and bases classified on the basis of extent of their dissociation?
Answer:
On the basis of extent of dissociation acids and bases are classified as follows :
(1) Strong acids and strong bases : The acids and bases which dissociate to a greater extent or almost completely are called strong acids and strong bases.
\(\mathrm{HCl}_{\text {(aq) }} \longrightarrow \mathrm{H}_{\text {(aq) }}^{+}+\mathrm{Cl}_{\text {(aq) }}^{-}\)
\(\mathrm{NaOH}_{(\mathrm{aq})} \longrightarrow \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)

(2) Weak acids and weak bases : The acids and bases which dissociate partially are called weak acids and weak bases. There exists an equilibrium between undissociated molecules and ions in solution.

Question 17.
Give examples of (a) strong acids and strong bases, (b) weak acids and bases.
Answer:
(a) Strong acids : HCl, H₂SO₄
Strong bases : NaOH, KOH
(b) Weak acids : HCOOH, CH₃COOH
Weak bases : NH₄OH, C₂H₅NH₂

Question 18.
Define and explain dissociation constant of a weak acid.
Answer:
Dissociation constant of a weak acid : It is defined as the equilibrium constant for dissociation equilibrium of a weak acid and denoted by K_a.
Explanation : Consider an aqueous solution of a weak acid HA.
\(\mathrm{HA}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-}\)
The equilibrium constant called dissociation constant Ka is represented as,
K_a = \(\frac{\left[\mathrm{H}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)

Question 19.
Derive the expression of Ostwald’s dilution law in case of a weak acid (HA).
OR
Derive the relationship between degree of dissociation and dissociation constant of a weak acid.
Answer:
Expression of Ostwald’s dilution law in case of a weak acid : Consider the dissociation of a weak acid HA. Let V dm³ of a solution contain one mole of weak acid HA. Then the concentration of a solution is, C = \(\frac{1}{V}\) mol dm^3-3. Let α be the degree of dissociation of HA.

Applying the law of mass action to this dissociation equilibrium, we have,

As the acid is weak, a is very small as compared to unity,
∴ (1 – α) ≈ 1.

This is an expression for Ostwald’s dilution law. This shows that the degree of dissociation of a weak acid is directly proportional to the volume of solution containing one mole of acid or inversely proportional to square root of its concentration.

Question 20.
Derive the relationship between degree of dissociation and dissociation constant of a weak base.
OR
Derive the expression of Ostwald’s dilution law in case of a weak base.
Answer:
Expression of Ostwald’s dilution law in case of a weak base : Consider the dissociation of a weak base BOH. Let V dm³ of a solution contain one mole of weak base BOH. Then the concentration of a solution is, C = \(\frac{1}{V}\) mol dm^-3. Let α be the degree of dissociation of BOH.

Consentration at equilibrium (mol dm^-3) \(\frac{(1-\alpha)}{V} \quad \frac{\alpha}{V} \quad \frac{\alpha}{V}\)
Applying the law of mass action to this dissociation equilibrium, we have,

This is an expression of Ostwald’s dilution law. Thus, the degree of dissociation of a weak base is directly proportional to the square root of the volume of the solution containing one mole of a base or inversely proportional to the square root of its concentration.

Solved Examples 3.4

Question 21.
Solve the following:

(1) At 298 K, 0.01 M formic acid solution is 1.2% dissociated. Calculate the dissociation constant of formic acid.
Solution :
Given : Concentration of HCOOH = C = 0.01 M
Percent dissociation = 1.2
Dissociation constant = K_a = ?
∴ Degree of dissociation = \(\frac{\text { Percent dissociation }}{100}\)
α = \(\frac{1.2}{100}\)
= 1.2 × 10^-2
K_a = Cα²
= 0.01 × (1.2 × 10^-2)²
= 1.44 × 10^-6
Ans. Dissociation constant = K_a = 1.44 × 10^-6

(2) The degree of dissociation of ammonium hydroxide is 0.0232 in 0.5 M solution. What will be the dissociation constant of ammonium hydroxide ?
Solution :
Given : Degree of dissociation = α = 0.0232
Concentration of NH₄OH = C = 0.5 M
Dissociation constant = K_b = ?
K_b = Cα²
= 0.5 × (0.0232)²
= 2.692 × 10^-4
Ans. Dissociation constant of NH₄OH
= K_b = 2.692 × 10^-4.

(3) Calculate the hydrogen ion concentration in 0.1 M acetic acid solution when the acetic acid is 2% dissociated in the solution.
Solution : The dissociation of acetic acid is represented below :
CH₃COOH ⇌ CH₃ – COO^– + H⁺
Given : Dissociation = 2%, C = 0.1 M
[H₃O⁺] = ?
The concentration of hydrogen ion, [H⁺], is given by the following formula :
[H₃O⁺] = αC
α = Degree of dissociation = \(\frac{\text { Per cent dissociation }}{100}\)
= \(\frac{2}{100}\) = 0.02
[H₃O⁺] = Hydrogen ion concentration = ?
C = Molar concentration of acetic acid
= 0.1 M = 0.1 mol dm^-3
∴ [H₃O⁺] = C × α = 0.1 × 0.02
= 0.002 = 2.0 × 10^-3 mol dm^-3
Ans. Hydrogen ion concentration = 2.0 × 10^-3 mol dm^-3.

(4) Calculate the percentage dissociation of 0.01 M NH₄OH solution. The K_b for NH₄OH is 1.75 × 10^-5.
Solution :
Given : Concentration of NH₄OH = C = 0.01 M
Dissociation constant of NH₄OH
= K_b= 1.75 × 10^-5
Percentage dissociation = ?
K_b = Cα²

∴ Per cent dissociation = α × 100
= 4.183 × 10^-2 × 100
= 4.183
Ans. Dissociation of NH₄OH = 4.183%

(5) 0.01 mole of a weak base is dissolved in 8 dm³ of water. The dissociation constant of the base is 4.0 × 10^-10. Calculate the degree of dissociation of the base in the solution.
Solution :
Given : V = 8 dm³; n = 0.01 mole; K_b = 4 × 10^-10
α = ?
The degree of dissociation and dissociation constant of a weak base are related to each other by the following formula :
K_b = α²C OR α = \(\sqrt{\frac{K_{b}}{C}}\)
K_b = Dissociation constant of the base = 4.0 × 10^-10
α = Degree of dissociation of the base = ?
C = Molar concentration of the base
= 0.01 mole in 8 dm³

Ans. Degree of dissociation of the base = 5.656 × 10^-4

(6) The dissociation constant of benzoic acid (C₆H₅COOH) is 6.6 × 10^-5. Calculate the hydrogen ion concentration of a solution containing 1.22 g of benzoic acid in 2000 mL of water.
Solution :
Given : K_a = 6.6 × 10^-5; V = 2000 mL;
W = 1.22 g; [H⁺] = ?
Molar mass of benzoic acid (C₆H₅COOH) = 122
The concentration of the solution is 1.22 g benzoic acid in 2000 ml (2 dm³) of solution.
1.22 g = \(\frac{1.22}{122}\) = 0.01 mol
∴ Molar concentration of benzoic acid = \(\frac{0.01}{2}\)
= 0.005 mol dm^-3
The dissociation constant and degree of dissociation of a weak acid are,
K_a = α²C OR α = \(\sqrt{\frac{K_{\mathrm{a}}}{C}}\)
α = Degree of dissociation of benzoic acid = ?
C = 0.005 mol dm^-3

Since benzoic acid is a monobasic acid, [H⁺] = aC
∴ [H⁺] = 0.1149 × 0.005
= 5.745 × 10^-4 mol dm^-3
Ans. Hydrogen ion concentration
= 5.745 × 10^-4 mol dm^-3

(7) The degree of dissociation of acetic acid in its 0.1 M solution is 0.0132 at 25 °C. Calculate the degree of dissociation in its 0.01 M solution.
Solution :
Given : C = 0.1 M, α = 0.0132, C’ = 0.01 M,
α = 0.0132, α’= ?
K_a = α²C
C = Molar concentration of acetic acid = 0.1 M
α = Degree of dissociation in 0.1 M solution = 0.0132
∴ α = \(\sqrt{\frac{K_{\mathrm{a}}}{C}}\)
If the concentration is C’, α’ = \(\sqrt{\frac{K_{\mathrm{a}}}{C^{\prime}}}\)

Ans. Degree of dissociation in 0.01 M solution = 4.175 × 10^-2

Question 22.
Explain autoionisation of water. Derive a relation for ionic product of water.
Answer:
Pure water ionises to a very less extent. The ionisation equilibrium is represented as follows,
\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
The equilibrium constant K for the above equilibrium is represented as,
K = \(\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}}\)
∴ K × [H₂O]² = [H₃O⁺] × [OH^–]
Since K and active mass of pure water [H₂O] are constant we can write,
K × [H₂O] = K_w,
∴ K_w= [H₃O⁺] × [OH^–]
where K_w is called ionic product of water. At 25 °C,
K_w= 1 × 10^-14.

Question 23.
Define ionic product of water.
Answer:
Ionic product of water : It is defined as the product of molar concentrations of hydronium ions (or hydrogen ions) and hydroxyl ions at equilibrium in pure water at constant temperature.
It is represented as,
K_w = [H₃O⁺] × [OH^–]
At 25 °C, K_w= 1 × 10^-14.

Question 24.
Define the following :
(i) pH (2) pOH. (2 marks)
Ans.
(1) pH : The negative logarithm, to the base 10, of the molar concentration of hydrogen ions, H⁺ is known as the pH of a solution.
PH = -log₁₀ [H⁺]
(2) pOH : The negative logarithm, to the base 10, of the molar concentration of hydroxyl ions, OH^– is known as the pOH of a solution.
pOH = -log₁₀ [OH^–]

Question 25.
What are approximate concentrations of H₃O⁺ and OH^– in, (a) pure water or neutral solution, (b) acidic solution and (c) basic solution ? Also mention pH values.
Answer:

Question 26.
Write a note on pH scale.
Answer:
Most of the chemical reactions and industrial processes are carried out in aqueous solutions, hence there is a need to know concentration of H⁺ and OH^– ions in the solution.
Sorensen developed a convenient scale to represent the acidic, basic or neutral nature of the solution.
The pH scale is used to express the concentration of H⁺ and OH^– along with pH and pOH of the solution.
According to Sorensen,
pH = -log₁₀ [H+], pOH = -log₁₀ [OH^–]
pH + pOH = 14.

Acids, basic and neutral solutions.

Question 27.
Solve the following :

(1) At 50 °C the value of ionic product of water is 5.5 × 10^-14. What are the concentrations of [H₃O⁺] and OH^– in a neutral solution at 50°C temperature ?
Solution :
Given : at 50 °C
Ionic product of water = K_w = 5.5 × 10^-14
[H₃O⁺] = ? OH^– = ?
Water at any temperature will be neutral.
Hence, [H₃O⁺] = [OH^–] = x mol dm^-3
[H₃O⁺] × [OH^–] = K_w
x × x = 5.5 × 10^-14
∴ x = 2.345 × 10^-7
∴ [H⁺] = [OH^–] = 2.345 × 10^-7 M
Ans. Concentrations : [H₃O⁺] = [OH^–]
= 2.345 × 10^-7 M

(2) The concentration of H+ ion in lemon juice is 2.5 × 10^-3 M. Calculate the OH^– ion concentration and classify the solution as acidic, basic or neutral.
Solution :
Given : [H₃O⁺] = 2.5 × 10^-3 M, K_w = 1 × 10^-4
[OH^–] = ?
By ionic product of water,
[H₃O⁺] × [OH^–] = K_w
∴ [OH^–] = \(\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}\)
= \(\frac{1 \times 10^{-14}}{2.5 \times 10^{-3}}\)
Ans. Concentration of OH^–
= [OH^–] = 4 × 10^-12 M
Hence the solution of lemon juice is acidic.

(3) Calculate pH and pOH of 0.02 M HCl solution.
Solution :
Given : C = 6.02 M HCl; pH = ? pOH = ?
\(\begin{aligned}
\mathrm{HCl}_{(\mathrm{aq})} \longrightarrow & \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \\
& 0.02 \mathrm{M}
\end{aligned}\)
[H⁺] = [H₃O⁺] = 0.02 M
PH= -log₁₀ [H₃O⁺]
= -log₁₀ 0.02
= -(\(\overline{2} .3010\))
= 2 – 0.3010 = 1.699
pH + pOH = 14
∴ pOH = 14 – pH
= 14 – 1.699
= 12.3010
Ans. pH = 1.6990; pOH = 12.3010.

(4) The pH of a solution is 6.06. Calculate its [H₃O⁺] ion concentration.
Solution :
Given : pH = 6.06, [H₃O⁺] = ?
PH = -log₁₀ [H₃O⁺]
∴ log₁₀ [H₃O⁺] = -pH
∴ [H₃O⁺] = Antilog – pH
= Antilog – 6.06
= Antilog \(\overline{7} .94\)
= 8.714 × 10^-7 M
Ans. [H₃O⁺] = 8.714 × 10^-7 M.

(5) Calculate number H⁺ ions present in 1 mL of 0.01 M H₂SO₄ solution.
Solution :
Given : C = 0.01 M H₂SO₄; Y = 1 mL
Number of H⁺ ions = ?
\(\begin{aligned}
\mathrm{H}_{2} \mathrm{SO}_{4(\mathrm{aq})} \longrightarrow & 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \\
& 0.01 \times 2 \mathrm{M}
\end{aligned}\)
∵ 1000 mL solution contains 0.02 mol H⁺
∴ 1 mL solution contains \(\frac{0.02}{1000}\) mol
= 2 × 10^-5 mol H⁺
∴ Number of H⁺ ions = 2 × 10^-5 × 6.022 × 10²³
= 1.204 × 10¹⁹
Ans. Number of H⁺ ions = 1.204 × 10¹⁹

(6) The pH of a 0.1 M monoacidic base is 11.11. What is the percent dissociation of base ?
Solution :
Given : pH = 11.11; per cent Dissociation of base = ?
c = 0.1 M
\(\begin{gathered}
\mathrm{BOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{B}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \\
c(l-\alpha) \quad c \alpha \quad c \alpha
\end{gathered}\)
pH + pOH = 14
∴ pOH = 14 – pH= 14 – 11.11 = 2.89
POH = -log₁₀ [OH^–]
∴ [OH^–] = Antilog – pOH
= Antilog – 2.89
= Antilog \(\overline{3} .11\)
= 1.29 × 10^-3 M
∵ [OH^–] = cα
∴ α = \(\frac{\left[\mathrm{OH}^{-}\right]}{c}=\frac{1.29 \times 10^{-3}}{0.1}\) = 1.29 × 10^-2
∴ Per cent dissociation = α × 100
= 1.29 × 10^-2 × 100 = 1.29
Ans. Per cent dissociation = 1.29.

(7) Calculate the pH of dedmolar solution of sulphuric add aqueous solution. Assuming complete Ionization of sulphuric add.
Solution:
Given : Concentration of H₂SO₄ = decimolar
= 0.1 M
H₂SO₄ → 2H⁺ + \(\mathrm{SO}_{4}^{-2}\)
∴ [H⁺] = 2 × 0.1
= 0.2 M
PH = -log₁₀ [H⁺]
= -log₁₀ 0.2 M
= \(-[\overline{1} .3010]\)
= 1 – 0.3010 = 0.6990
Ans. pH of H₂SO₄ solution = 0.6990.

(8) The pH of a 0.2 M solution of ammonia is 10.78. Calculate (i) OH^– ions concentration (ii) the degree of dissociation (iii) the dissociation constant.
Solution :
Given: pH = 10.78, C = 0.02 M, [OH^–] = ? K_b = ?
As NH₃ is a base,
pOH = 14 – pH
pH = 10.78
∴ pOH = 14 – 10.78 = 3.22
pOH = -log₁₀ [OH^–]
∴ 3.22 = -log₁₀ [OH^–]
∴ -3.22 = log₁₀ [OH^–]
[OH^–] = antilog (-3.22)
∴ [OH^–] = antilog \((\overline{4} .78)\)
= 6.026 × 10^-4 M
As NH₄OH is a monoacidic base, [OH^–]

(9) NH₄OH is 4.3% ionised at 298 K in 0.01 M solution. Calculate the ionization constant and pH of NH₄OH.
Solution :
Given : Per cent dissociation = 4.3,
C = 0.01 M, K_b = ?, pH = ?
The degree of dissociation and dissociation constant of NH₄OH are related to each other by the formula :
K_b = α²C
K_b = Dissociation constant of NH₄OH = ?
α = Degree of dissociation of NH₄OH = 4.3%
= 4.3 × 10^-2
C = Molar concentration of NH₄OH = 0.01 M
∴ K_b = (4.3 × 10^-2)² × 0.01
= 18.49 × 10^-4 × 10^-2
∴ K_b = 1.849 × 10^-5
Since NH₄OH is a monoacidic base,
[OH^–] = αC
= 4.3 × 10^-2 × 0.01
= 4.3 × 10^-4 mol dm^-3
pOH = -log₁₀ [OH-]
= -log₁₀ 4.3 × 10^-4
= -[0.6335 – 4] = 3.3665
pH + pOH = 14
pH = 14 – 3.3665 = 10.6335
Ans. K_b = 1.849 × 10^-5, pH = 10.6335

Question 28.
Define hydrolysis.
Answer:
Hydrolysis : A reaction in which the cations or anions or both the ions of a salt react with water to produce acidity or basicity or sometimes neutrality is called hydrolysis.

Question 29.
What are the types of the salts? Give examples.
Answer:
A salt is formed by the reaction between equivalent amounts of an acid and a base. According to the nature of an acid and a base, there are four types of the salts as follows :
(1) Salt of a strong acid and a strong base :

(2 ) Salt of a weak acid and a strong base :

(3) Salt of a strong acid and a weak base :

(4) Salt of a weak acid and a weak base : CH₃COONH₄ :

Question 30.
A salt of strong acid and strong base does not undergo hydrolysis. Explain.
OR
An aqueous solution of sodium chloride is neutral. Explain.
Answer:
(1) Sodium chloride is a salt of strong acid HCl and strong base NaOH.
(2) In water, it reacts forming HCl and NaOH.
(3) As both are strong, they dissociate almost completely to liberate H⁺ and OH^– ions, respectively.
(4) H⁺ and OH^– ions combine together to form weakly dissociating H₂O. As there are no free H⁺ ions and OH^– ions, the solution is neutral and the salt does not undergo hydrolysis.
NaCl + H₂O ⇌ NaOH + HCl
Ionic equation :
Na⁺ + Cl^– +H₂O ⇌ Na⁺ + OH^– + H⁺ + Cl^–
H₂O ⇌ H⁺ + OH^–
Since the solution contains equal number of H⁺ and OH^– ions, it is neutral.
Hence the salt of strong acid and strong base does not undergo hydrolysis.

Question 31.
Explain the hydrolysis of the salt of strong acid and weak base.
OR
A solution of CuSO₄ reacts acidic. Explain.
Answer:
(1) Consider a salt of strong acid and weak base, like CuSO₄ obtained from strong acid H₂SO₄ and weak base Cu(OH)₂.
(2) When it is dissolved in water it undergoes hydrolysis as follows :

(3) Since [H₃O⁺] > [OH^–], the solution is acidic in nature.

Question 32.
Explain the hydrolysis of a salt of weak acid and strong base.
OR
A solution of sodium acetate, CH₃COONa reacts basic explain.
Answer:
(1) Consider a salt of weak acid and strong base like CH₃COONa. In aqueous solution it undergoes hydrolysis as follows :

(2) Since the base NaOH is strong, it dissociates completely while acid CH₃COOH being weak dissociates partially.
(3) Hence [OH^–] > [H₃O⁺] in the solution and the solution reacts basic.

Question 33.
Explain the hydrolysis of the salt of weak acid and weak base.
Answer:
(1) Consider a salt BA of weak acid (HA) and weak base (BOH).
(2) In aqueous solution it undergoes hydrolysis as follows :

(3) The nature of the solution will depend upon relative strength of weak acid and weak base, hence will depend upon their dissociation constants K_a and K_b.

(i) A salt of weak acid and weak base for which K_a > K_b :
Consider hydrolysis of NH₄F.

Since K_a (7.2 × 10^-4) for HF is greater than K_b (1.8 × 10^-5) for NH₄OH, the acid dissociates partially more than the base, hence, [H₃O⁺] > [OH^–] and the solution reacts acidic after hydrolysis.

(ii) A salt of weak acid and weak base for which K_a < K_b :
Consider hydrolysis of NH₄CN.

Since K_a (4 × 10^-10) for HF is less than K_b (1.8 × 10^-5) for NH₄OH, the base dissociates more than acid and hence [H₃O⁺] < [OH^–] and the solution reacts basic after hydrolysis.

(iii) A salt of weak acid and weak base for which K_a = K_b:
Consider hydrolysis of CH₃COONH₄.

Since K_a = K_b, the weak acid CH₃COOH and weak base NH₄OH dissociate to the same extent, hence, [H₃O⁺] = [OH^–] and the solution reacts neutral after hydrolysis.

Question 34.
Define buffer solution.
OR
What is buffer solution?
Answer:
Buffer solution : It is defined as a solution which resists the change in pH even after the addition of a small amount of a strong acid or a strong base or on dilution or on addition of water.

Question 35.
What are the types of buffer solutions ?
Answer:
These are two types of buffer solutions :
(A) Acidic buffer solution : it is a solution containing a weak acid e.g. (CH₃COOH) and its salt of a strong base. e.g. (CH₃COONa).
pH of an acidic buffer is given by following Henderson Hasselbalch equation,
pH = \(\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]}\)
where pK_a = -log₁₀ K_a
and K_a is the dissociation constant of weak acid.

(B) Basic buffer solutions : It is a solution containing a weak base (e.g. NH₄OH) and its salt of strong acid, (e.g. NH₄Cl).
pOH of a basic buffer is given by Henderson Hassebalch equation,
pOH = \(\mathrm{p} K_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]}\)
where pK_b = -log10 K_b
and K_b is the dissociation constant of a weak base.

Question 36.
Explain a buffer action of an acidic buffer.
Answer:
Mechanism of action of an acidic buffer :
(1) An acidic buffer is a mixture of a weak acid and its salt with a strong base. The weak acid dissociates feebly, but the salt dissociates almost completely. Moreover, due to the common ions, largely supplied by the salt, dissociation of the weak acid is further suppressed.
(CH₃COOH + CH₃COONa) :
CH₃COONa_(aq) → CH₃COO_(aq) + \(\mathrm{Na}_{(\mathrm{aq})}^{+}\) (Complete)

(2) When a small quantity of strong acid (H⁺) is added to this mixture, hydrogen ions combine with acetate ions to form undissociated acetic acid. Thus, addition of an acid does not change the pH of the buffer.
\(\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-} \rightleftharpoons \mathrm{CH}_{3}-\mathrm{COOH}_{(\mathrm{aq})}\)
This removal of added H⁺ is called reserved basicity.

(3) When a small quantity of a strong base (OH^–) is added, the hydroxide ions react with the acid producing the corresponding anions and water. Thus, the concentrations of H⁺ and OH^– in the solution do not change and the pH remains constant.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})}+\mathrm{OH}_{(\mathrm{aq})}^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
This removal of added OH^– is called reserved acidity.

Question 37.
Explain a buffer action of a basic buffer.
Answer:
Mechanism of action of a basic buffer :
(1) A basic buffer solution is a solution containing a weak base and its salt with a strong acid. The weak base dissociates feebly, but the salt dissociates completely. Moreover, due to the presence of the common ion, largely supplied by the salt, the dissociation of the base is further suppressed. (NH₄OH + NH₄Cl) :
\(\mathrm{NH}_{4} \mathrm{Cl}_{(\mathrm{aq})} \rightarrow \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\) (Complete)

(2) When a small quantity of a strong acid is added to the solution, the hydrogen ions combine with the base producing corresponding cations and water. Thus, the addition of an acid does not change the pH of the buffer.
\(\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
This removal of added H⁺ is called reserved basicity.

(3) When a small quantity of a strong base is added, the hydroxide ions combine with \(\mathrm{NH}_{4}^{+}\) ions to form undissociated NH₄OH. As a result, the hydrogen or hydroxyl ion concentration does not change. Thus, the pH of the solution does not change.
\(\mathrm{OH}_{(\mathrm{aq})}^{-}+\mathrm{NH}_{4}^{+} \rightleftharpoons \mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}\)
This removal of added OH^– is called reserved acidity.

Question 38.
What are properties of a buffer solution ?
OR
What are the advantages of a buffer solution ?
Answer:
Properties (or advantages) of a buffer solution :

1. The pH of a buffer solution is maintained appreciably constant.
2. By addition of a small amount of an acid or a base pH does not change.
3. On dilution with water, pH of the solution doesn’t change.

Question 39.
What are the applications of a buffer solution ?
Answer:
Buffer solutions have many applications as follows :
(1) In a biochemical system : Blood in our body has pH 7.36 – 7.42 due to (\(\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2} \mathrm{CO}_{3}\)) and little change of 0.2 pH unit may be fatal. For example, saline solution used in intravenous injection contains a buffer solution maintaining pH of the blood in the required range.

(2) Agriculture : The properties of soil depend upon its pH. The salts present in soil such as phosphates, carbonates, bicarbonates and organic acids impart definite pH to the soil. Depending on pH the fertilizers are selected.

(3) Industry : In many industries, buffer solutions are used to carry out chemical processes very effectively, such as the industries of paper, dye, paints, drugs, ink, etc.

(4) Medicines : Many medicines particularly in the liquid state have a good stability and optimum activity at a definite pH, for which buffer solutions are used. For example penciline preparations are carried out in the presence of a buffer of sodium citrate. A buffer solution of magnesium citrate is prepared by adding citric acid to Mg(OH)₂.

(5) Analytical chemistry : In a qualitative analysis, the precipitation of groups, the chemical tests for detection of ions, etc. are carried out at a definite pH. For example, precipitation of cations of IIIA are carried in the presence of a basic buffer of pH 8 – 10 obtained by using NH₄OH and NH₄Cl.

Solved Examples 3.8

Question 40.
Solve the following :

(1) Calculate the pH of a buffer solution containing 0.1 M CH₃COOH and 0.05 M CH₃COONa. Dissociation constant of CH₃COOH is 1.8 × 10^-5 at 25 °C.
Solution :
Given : [CH₃COOH] = 0.1 M,
[CH₃COONa] = 0.05 M; K_a = 1.8 × 10^-5; pH = ?
pK_a = -log₁₀ K_a
= -log₁₀ 1.8 × 10^-5
= -(\(\overline{5} \cdot 2553\))
= 5 – 0.2553
= 4.7447
pH = \(\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}\)
= \(4.7447+\log _{10} \frac{0.05}{0.1}\)
= 4.7447 + \((\overline{1} .6990)\)
= 4.7447 + (-1 + 0.6990)
= 4.7447 – 0.3010
= 4.4437
Ans. pH = 4.4437

(2) A buffer solution contains 0.3 M NH₄OH and 0.4 M NH₄Cl. If K_b for NH₄OH is 1.8 × 10^-5, calculate pH of the solution.
Solution :
Given : [NH₄OH] = 0.3 M; [NH₄Cl] = 0.4 M
K_b =1.8 × 10^-5; pH = ?
pK_b = -log₁₀ K_b
= -log₁₀ 1.8 × 10^-5
= \(-(\overline{5} .2553)\)
= 4.7447
pOH = \(\mathrm{p} K_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\mathrm{Base}]}\)
= \(4.7447+\log _{10} \frac{0.4}{0.3}\)
= 4.7447 + log₁₀ 1.333
= 4.7447 + 0.1248
= 4.8695
∵ pH + pOH = 14
∴ pH = 14 – pOH
= 14 – 4.8695
= 9.1305
Ans. pH = 9.1305.

(3) 0.2 dm³ acidic buffer solution contains 1.18 g acetic acid and 2.46 g sodium acetate. If K_a for acetic acid is 1.8 × 10^-5 at 25 °C, find pH of the solution.
Solution :

(4) A basic buffer solution contains 0.3 M NH₄OH and 0.2 M (NH₄)₂SO₄. If K_b for NH₄OH at a certain temperature is 2 × 10^-5, what is the pH of the solution ?
Solution :
Given : [NH₄OH] = 0.3 M
[NH₄⁺ ] = 2 × 0.2 = 0.4 M

pH + pOH = 14
∴ pH = 14 – pOH = 14 – 4.8239 = 9.1761
Ans. pH = 9.1761

Question 41.
Define solubility. How is it expressed ?
Answer:
Solubility : It is defined as the maximum amount of a substance in moles, that can be dissolved at constant temperature to give one litre of its saturated solution.
It is expressed in moles per litre or moles per decimeter cube of a saturated solution at given temperature.

Question 42.
Derive a relationship between solubility and solubility product.
Answer:
Consider a saturated solution of a spraingly soluble electrolyte (or salt) AxBy at a given constant temperature. Let S mol dm^-3 be the solubility of AxBy.
A following heterogeneous ionic equilibrium exists.

By a law of mass action, the equilibrium constant K will be represented as,

Since the active mass (concentration) of pure solid, A_xB_y(s) is treated as constant, [A_xB_y(s)] = K’
K × [A_xB_y(s)] = K × K’ = K_(sp)
Therefore, K_sp = [A^y+]^x × [B^x-]^y
where K_sp is called solubility product of A_xB_y.
At equilibrium the concentrations are,
[A^y+] = xS mol dm^-3
[B^x-]^y = yS mol dm^-3
∴ K_sp = [A^y+]^x × [B^x-]^y
= (xS)^x × (yS)^y
∴ K_sp = x^x.y^y.(S)^x+y ……….(1)
Hence solubility S is given by,
S = \(\left(\frac{K_{(\mathrm{sp})}}{x^{x} \cdot y^{y}}\right)^{\frac{1}{x+y}} \mathrm{~mol} \mathrm{dm}^{-3}\) …………(2)
The above equations, (1) and (2) give the relationship between solubility and solubility product.
Here x and y represent number of cations and anions respectively from the electrolyte.

Question 43.
Write expression for solubility and solubility product of following sparingly soluble salts : (1) AgBr (2) PbI₂ (3) Al(OH)₃
Answer:
In general, for a sparingly soluble salt A_xB_y,
K_sp = x^x.y^y.(S)^x+y

Question 44.
What is ionic product?
Answer:
Ionic product (IP) : It is defined as the product of concentrations in mol dm^-3 of ions of an electrolyte in the solution and denoted by IP.
In a saturated solution,
IP = K_sp where K_sp is the solubility product of the electrolyte.

Solved Examples 3.9

Question 45.
Solve the following :

(1) The solubility of AgBr in water is 1.28 × 10^-5 mol/dm³ at 298 K. Calculate the solubility product of AgBr at the same temperature.
Solution :
Given : S = 1.28 × 10^-5 mol dm^-3; K_sp = ?
AgBr dissociates as,
\(\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Br}_{(\mathrm{aq})}^{-}\)
K_sp = [Ag⁺] [Br^–]
As the solubility of AgBr in water is 1.28 × 10^-5 moles/dm³,
[Ag⁺] = [Br^–] = 1.28 × 10^-5 mol dm³
∴ K_sp = [1.28 × 10^-5] [1.28 × 10^-5]
= 1.638 × 10^-10
Ans. Solubility product of AgBr = 1.638 × 10^-10.

(2) The solubility of lead sulphate is 3.03 × 10^-5 kg/dm³. Calculate its solubility product. [Molecular mass of PbSO₄ = 303]
Solution :
Given : S = 3.03 × 10^-5 kg dm^-3, K_sp = ?
Lead sulphate dissociates as
\(\begin{aligned}
&\mathrm{PbSO}_{4} \rightleftharpoons \mathrm{Pb}^{2+}+\mathrm{SO}_{4}^{2-} \\
&\text { (solid) } \quad \text { (aq) } \quad \text { (aq) }
\end{aligned}\)
Molecular weight of PbSO₄ = 303
= 303 × 10^-3 kg
The solubility of PbSO₄ is 3.03 × 10^-5 kg/dm³.
Solubility in mol dm^-3
Weight of PbSO₄ per dm³ Molecular weight 3.03 × 10^-5

(3) The solubility product of AgBr is 3.3 × 10^-12 at 298 K. What concentration of Br^– ion is needed to precipitate AgBr from solution of 0.01 M Ag⁺?
Solution :
Given : K_sp = 3.3 × 10; [Ag⁺] = 1 × 10^-2 M;
[Br^–] = ?
AgBr dissociates as
\(\begin{aligned}
&\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-} \\
&\text {(solid) } \quad \text { (aq) } \quad \text { (aq) }
\end{aligned}\)
K_sp = [Ag⁺] [Br^–]
K_sp = Solubility product = 3.3 × 10^-12
[Ag⁺] = Concentration of Ag⁺
= 0.01 = 1.0 × 10^-2 M
[Br^–] = Concentration of Br^– = ?
∴ 3.3 × 10^-12 = 1.0 × 10^-2 × [Br^–]
∴ [Br^–] = 3.3 × 10^-10 mol/dm³
Ans. The concentration of Br^– required for precipitation of AgBr should be greater than 3.3 × 10^-10 mol/dm³.

(4) The solubility product of magnesium hydroxide is 1.4 × 10^-11. Calculate the solubility of magnesium hydroxide.
Solution :
Given : K_sp =1.4 × 10^-11; S = ?
Magnesium hydroxide dissociates as shown below :
Mg(OH)₂ ⇌ Mg²⁺ + 2(OH)^–
K_sp = [Mg²⁺] [OH^–]²
Let the solubility of Mg(OH)₂ be S mol dm^-3.
∴ [Mg²⁺] = Concentration of Mg²⁺ ions
= S mol dm^-3
∴ [OH^–] = Concentration of OH^– ions
= 2S mol dm^-3
∴ K_sp = S × (25)² = 4S³
K_sp = 1.4 × 10^-11
∴ 1.4 × 10^-11 =4S³
1.4 × 10^-11 = 4S³.

S = 1.518 × 10^-4 mol dm^-3
Ans. Solubility of Mg(OH)₂
= 1.518 × 10^-4 mol dm^-3

(5) The solubility of silver chloride is 1.562 × 10^-10 mol dm^-3 at 298 K. Find its solubility in g dm^-3 at the same temperature.
Solution :
Given :
Solubility of AgCl = S = 1.562 × 10^-10 mol dm^-3
Solubility of AgCl in g dm^-3 = ?
Molar mass of AgCl = M = 143.5 g mol^-1
Solubility in gram per dm³
= solubility in mol dm^-3 × molar mass
= 1.562 × 10^-10 × 143.5
= 2.241 × 10^-5 g dm-3
Ans. Solubility of AgCl = 2.241 × 10^-8 g dm^-3

(6) The solubility of PbSO₄ in water is 0.038 g dm^-3 at room temperature. Calculate its solubility and solubility product at the same temperature. (Atomic weights : Pb = 207.3, S = 32, O = 16)
Solution :
Given : Solubility of PbSO₄ = 0.038 g/dm^-3
Molar mass of PbSO₄ = 303.3 g mol^-1
Solubility in mol dm^-3 = ?
K_sp =?
Solubility in mol dm^-3 = \(\frac{0.038}{303.3}\)
= 1.253 × 10^-4 mol dm^-3

Ans. S = 1.253 × 10^-4 mol dm^-3;
K_sp = 1.57 × 10^-8

(7) The solubility product of PbS at 298 K is 4.2 × 10^-28, The concentration of Pb⁺⁺ ion is 0.001 M. Calculate S^2- ion concentration at which PbS just gets precipitated.
Solution :
Given :
Solubility product of PbS = K_sp = 4.2 × 10^-28
Concentration of Pb⁺⁺ = [Pb⁺⁺] = 0.001 M
Concentration of S^{– –} = [S^{– –}] = ?
For PbS,
\(\mathrm{PbS}_{(\mathrm{s})} \rightleftharpoons \mathrm{Pb}^{++}+\mathrm{S}^{–}\)
∴ K_sp = [Pb⁺⁺] × [S^{– –}]
∴ [S–] = \(\frac{K_{\mathrm{sp}}}{\left[\mathrm{Pb}^{++}\right]}\)
= \(\frac{4.2 \times 10^{-28}}{0.001}\)
= 4.2 × 10^-25 M
To precipitate Pb⁺⁺ as PbS, ionic product must be greater than 4.2 × 10^-28.
Hence, [S^{– –}] > 4.2 × 10^-25 M.
Ans. Concentration of S^{– –} required > 4.2 × 10^-25 M

(8) At 298 K, the solubility of silver sulphate is 1.85 × 10^-2 mol dm^-3. Calculate the solubility product of silver sulphate.
Solution :
Given : Silver sulphate dissociates as follows :

Solubility of Ag₂SO₄ = 1.85 × 10^-2 mol dm^-3
K_sp = Solubility product of Ag₂SO₄ = ?
[Ag⁺] = Concentration of Ag⁺ ion
= 2 × 1.85 × 10^-2 = 3.70 × 10^-2 mol dm^-3
[latex]\mathrm{SO}_{4}^{2-}[/latex] = Concentration of \(\mathrm{SO}_{4}^{2-}\)
= 1.85 × 10^-2 mol dm^-3
∴ K_sp = (3.70 × 10^-2)² × (1.85 × 10^-2)
= 13.69 × 10^-4 × 1.85 × 10^-2
= 25.33 × 10^-6
= 2.533 × 10^-5
Ans. Solubility product of Ag₂SO₄
= 2.533 × 10^-5

Question 46.
What is common ion?
Answer:
Common ion : An ion common to two electrolytes is called common ion. This is generally applicable to a mixture of a strong and a weak electrolyte. For example, a solution containing weak electrolyte CH₃COOH and strong electrolyte salt CH₃COONa.
CH₃COONa → CH₃COO^– + Na⁺;
CH₃COOH ⇌ CH₃COO^– + H⁺
Hence CH₃COOH and CH₃COONa have a common ion CH₃COO^–.

Question 47.
Define the term common ion effect.
Answer:
Common ion effect : The suppression of the degree of dissociation of a weak electrolyte by the addition of a strong electrolyte having an ion in common with the weak electrolyte is called common ion effect. For example, CH₃COOH and CH₃COONa have common ion CH₃COO^–.

Question 48.
Explain common ion effect with suitable example.
Answer:
A weak electrolyte dissociates partially in aqueous solution to produce cations and anions. Equilibrium exists between ions thus formed and the undissociated molecules.
BA ⇌ B⁺ +A^–
For such an equilibrium, the dissociation constant K is defined as
K = \(\frac{\left[\mathrm{B}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{BA}]}\)
K is constant for the weak electrolyte at a given temperature.
Now, if another electrolyte BC or DA is added to the solution BA, having a common ion either B⁺ or A^–, then the concentration of either B⁺ or A^– is increased. However, as K is always constant, the increase in the concentration of any one of the ions shifts the equilibrium to left. In other words, the dissociation of BA is suppressed. This is called common ion effect. For example, the dissociation of a weak acid CH₃COOH is suppressed by adding CH₃COONa having common ion CH₃COO^–.
CH₃COOH ⇌ CH₃COO^– + H⁺
CH₃COONa → CH₃COO^– + Na⁺

Question 49.
Explain the common ion effect on dissociation of a weak acid.
Answer:
(1) Consider the dissociation or ionisation of a weak acid, CH₃COOH in its solution.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+}\)
The dissociation constant Ka for CH₃COOH will be,
K_a = \(\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] \times\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}\)
K_a is constant for CH₃COOH at constant temperature.

(2) If a strong electrolyte like salt CH₃COONa is added to the solution of CH₃COOH, then on dissociation it gives a common ion CH₃COO^–.
CH₃COONa → CH₃COO^– + Na⁺

(3) Due to common ion CH₃COO^–, overall concentration of CH₃COO^– in the solution is increased, which increases the ratio,
\(\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] \times\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}\). In order to keep this ratio constant, the concentration of H⁺ is decreased, by shifting the equilibrium to the left hand side according to Le Chatelier’s principle.

(4) Thus the ionisation of a weak acid is suppressed by a common ion.

Question 50.
Explain the effect of common ion on the dissociation of weak base.
Answer:
(1) Consider the dissociation or ionisation of a weak base, NH₄OH in its dilute solution. \(\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
The dissociation constant K_b for NH₄OH will be,
\(K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{4} \mathrm{OH}\right]}\)

(2) If a strong electrolyte like salt NH₄Cl is added to the solution of NH₄OH, then it gives common ion \(\mathrm{NH}_{4}^{+}\).
NH₄Cl → \(\mathrm{NH}_{4}^{+}\) + Cl^–

(3) Due to common ion \(\mathrm{NH}_{4}^{+}\), overall concentration of \(\mathrm{NH}_{4}^{+}\) is increased, which increases the ratio \(\left[\mathrm{NH}_{4}^{+}\right]\) × [OH^–]/[NH₄OH],
In order to keep this ratio constant, the equilibrium is shifted to the left hand side which satisfies Le Chatelier’s principle.

(4) Thus the ionisation of a weak base is suppressed by a common ion.

Multiple Choice Questions

Question 51.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. According to Lowry-Bronsted concept, base is a substance which acts as –
(a) a proton donor
(b) an electron donor
(c) a proton acceptor
(d) an electron acceptor
Answer:
(c) a proton acceptor

2. BF₃ is a
(a) Lewis acid
(b) Lewis base
(c) amphoteric compound
(d) Electrolyte only
Answer:
(a) Lewis acid

3. Which of the following is a conjugate acid-base pair ?
(a) HCl, NaOH
(b) KCN, HCN
(c) NH₄Cl, NH₄OH
(d) H₂SO₄, \(\mathrm{HSO}_{4}^{-}\)
Answer:
(d) H₂SO₄, \(\mathrm{HSO}_{4}^{-}\)

4. The conjugate acid of \(\mathrm{NH}_{2}^{-}\) is
(a) NH₃
(b) NH₂OH
(c) \(\mathrm{NH}_{4}^{+}\)
(d) N₂H₄
Answer:
(a) NH₃

5. Which of the following molecules is not a Lewis base?
(a) H₂O
(b) BF₃
(c) NH₃
(d) CO
Answer:
(b) BF₃

6. In the following reaction
\(\mathrm{HC}_{2} \mathrm{O}_{4(\mathrm{aq})}^{-}+\mathrm{PO}_{4}^{3-} \rightleftharpoons \mathrm{HPO}_{4}^{2-}+\mathrm{C}_{2} \mathrm{CO}_{4}^{2-}\) Which of two are Lowry-Bronsted bases ?
(a) \(\mathrm{HC}_{2} \mathrm{C}_{4}^{-} \text {and } \mathrm{PO}_{4}^{3-}\)
(b) \(\mathrm{HPO}_{4}^{2-} \text { and } \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)
(c) \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \text {and } \mathrm{HPO}_{4}^{2-}\)
(d) \(\mathrm{PO}_{4}^{3-} \text { and } \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)
Answer:
(d) \(\mathrm{PO}_{4}^{3-} \text { and } \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)

7. According to the Arrhenius theory,
(a) an acid is a proton donor
(b) an acid is an electron pair acceptor
(c) a hydrogen ion exists freely in an aqueous solution
(d) a hydrogen ion is always hydrated to form a hydrogen ion
Answer:
(c) a hydrogen ion exists freely in an aqueous solution

8. The species which will behave both as a conjugate acid and base is
(a) NH₄OH
(b) \(\mathrm{CO}_{3}^{–}\)
(c) \(\mathrm{HSO}_{4}^{-}\)
(d) H₂SO₄
Answer:
(c) \(\mathrm{HSO}_{4}^{-}\)

9. According to the Lewis theory, an acid is
(a) nucleophile
(b) an electrophile
(c) a proton acceptor
(d) an electron donor
Answer:
(b) an electrophile

10. If a 0.1 M solution of HCN is 0.01% dissociated, the dissociation constant for HCN is,
(a) 10^-3
(b) 10⁺³
(c) 10^-7
(d) 10^-9
Answer:
(d) 10^-9

11. The pH of decimolar solution KOH is
(a) 1
(b) 4
(c) 10
(d) 13
Answer:
(d) 13

12. Ostwald’s dilution law is applicable in case of dilute solution of
(a) HCl
(b) H₂SO₄
(c) NaOH
(d) CH₃COOH
Answer:
(d) CH₃COOH

13. The degree of dissociation of a 0.1 M monobasic acid is 0.4%. Its dissociation constant is
(a) 0.4 × 10^-4
(b) 4.0 × 10^-4
(c) 1.6 × 10^-6
(d) 0.8 × 10^-5
Answer:
(c) 1.6 × 10^-6

14. The ionic product of water will increase, if
(a) Pressure is decreased
(b) H⁺ ions are added
(c) OH^– ions are added
(d) Temperature is increased
Answer:
(d) Temperature is increased

15. The [OH^–] for a weak base of dissociation constant K_b and concentration C is nearly equal to
(a) \(\sqrt{\frac{K_{\mathrm{b}}}{C}}\)
(b) KbC
(c) \(\sqrt{K_{\mathrm{b}} C}\)
(d) \(\frac{C}{K_{\mathrm{b}}}\)
Answer:
(c) \(\sqrt{K_{\mathrm{b}} C}\)

16. The [H⁺] for a weak acid of dissociation constant K_a and concentration C is nearly equal to
(a) \(\sqrt{\frac{K_{\mathrm{a}}}{C}}\)
(b) \(\sqrt{K_{\mathrm{a}} C}\)
(c) \(\frac{K_{\mathrm{a}}}{\sqrt{c}}\)
(d) \(\frac{c}{K_{\mathrm{a}}}\)
Answer:
(b) \(\sqrt{K_{\mathrm{a}} C}\)

17. Which of the following solution with same concentration will have highest pH
(a) Al(OH)₃
(b) K₂CO₃
(c) NH₄OH
(d) NaOH
Answer:
(d) NaOH

18. 10 ml of 0.1 M H₂SO₄ is mixed with 20 ml of 0.1 M KOH, the pH of resulting solution will be
(a) 0
(b) 7
(c) 2
(d) 9
Answer:
(b) 7

19. The gastric juice in our stomach contains enough hydrochloric acid to make the hydrogen ion concentration 0.01 mol/dm³. The pH of gastric juice is-
(a) 0.01
(b) 1
(c) 2
(d) 14
Answer:
(c) 2

20. If the hydrogen ion concentration of an acid is decreased ten times, its pH will be
(a) increased by one
(b) decreased by one
(c) remains unchanged
(d) increase by 10
Answer:
(a) increased by one

21. Which of the following metal sulphide is precipitated in an acidic medium ?
(a) NiS
(b) CoS
(c) CuS
(d) MnS
Answer:
(c) CuS

22. The relationship between the solubility and solubility product for silver carbonate is
(a) K_sp = s²
(b) \(\sqrt{K_{\mathrm{sp}}}\) = 4s²
(c) K_sp = 27S⁴
(d) K_sp = 4s³
Answer:
(d) K_sp = 4s³

23. If ‘S’ is solubility in mol dm^-3 and K_sp is solubility product of BA₂ type of salt, then relation between them is
(a) S = \(\sqrt{K_{\mathrm{sp}}}\)
(b) K_sp = 4S³
(c) K_sp = S³
(d) S = K_sp
Answer:
(b) K_sp = 4S³

24. The addition of solid sodium carbonate to pure water results in
(a) an increase in H⁺ ion concentration
(b) an increase in pH
(c) no change in pH
(d) a decrease in OH^– concentration
Answer:
(b) an increase in pH

25. Which of the following salt, when dissolved in water will hydrolyse ?
(a) NaCl
(b) NH₄Cl
(c) KCl
(d) Na₂SO₄
Answer:
(b) NH₄Cl

26. A solution of blue vitriol is acidic in nature because
(a) CuSO₄ reacts with water
(b) Cu²⁺ ions reacts with water
(c) SO₄^2- ions reacts with water
(d) CuSO₄ removes OH^– ions from water
Answer:
(a) CuSO₄ reacts with water

27. What is the nature of the solution of salt FeCl₃ ?
(a) Acidic
(b) Basic
(c) Neutral
(d) Amphoteric
Answer:
(a) Acidic

28. Which of the following salts does not hydrolyse in water ?
(a) Sodium acetate
(b) Sodium carbonate
(c) Sodium nitrate
(d) Sodium cyanide
Answer:
(c) Sodium nitrate

29. An aqueous solution of magnesium chloride changes blue litmus red due to
(a) the formation of Cl^– ions
(b) the formation Mg²⁺ ions
(c) reaction of Cl^– ions with water
(d) hydrolysis of the salt
Answer:
(d) hydrolysis of the salt

30. An aqueous solution of which of the following salts is basic ?
(a) CH₃COONa
(b) NH₄Cl
(c) KNO₃
(d) CuSO₄
Answer:
(a) CH₃COONa

31. The number of moles of hydroxide ions (OH^–) produced from 2 moles of Na₂CO₃ is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

32. The POH value for solution is 4, its hydrogen ion concentration will be
(a) 10^-4
(b) 10^-10
(c) 10¹⁰
(d) 10⁴
Answer:
(b) 10^-10

33. If an acid is diluted
(a) pH increases
(b) pH decreases
(c) no change occurs
(d) can vary depending on an acid
Answer:
(a) pH increases

34. pH of a solution is 13. H⁺ ions present in 1 cm³ of the solution is
(a) 6.023 × 10¹⁰
(b) 6.023 × 10⁷
(c) 6.023 × 10^-10
(d) 6.023 × 10^-7
Answer:
(b) 6.023 × 10⁷

35. pH of blood is maintained constant by mechanism of
(a) common ion effect
(b) buffer
(c) solubility
(d) all of these
Answer:
(b) buffer

36. The pH of 0.05 M solution of dibasic acid is
(a) +1
(b) -1
(c) +2
(d) -2
Answer:
(a) +1

37. The pH of a 0.63% nitric acid solution is (Equivalent weight of nitric acid is 63)
(a) 6
(b) 7
(c) 1
(d) 9
Answer:
(c) 1

38. 100 ml of 0.01 M solution of NaOH is diluted to 1 dm³. What is the pH of the dilute solution?
(a) 12
(b) 11
(c) 2
(d) 3
Answer:
(b) 11

39. If the H⁺ ion concentration in a solution is 0.01 M, the pOH of the solution is
(a) 12
(b) 10^-10
(c) 2
(d) 14
Answer:
(a) 12

40. If the pH value of a solution is zero, the solution is
(a) a strong acid
(b) a very weak acid
(c) neutral
(d) a base
Answer:
(a) a strong acid

41. The pH of a solution is 5, when the hydroxyl ion concentration is
(a) 10^-5 mol/dm³
(b) 10^-7 mol/dm³
(c) 10^-9 mol/dm³
(d) 10^-14 mol/dm³
Answer:
(c) 10^-9 mol/dm³

42. The pH of human blood in a normal person is approximately
(a) 4.7
(b) 6.04
(c) 7.40
(d) 8.74
Answer:
(c) 7.40

43. If molarity of NaOH is 3.162 × 10^-3 M, its pH is
(a) 8.5
(b) 9.5
(c) 10.5
(d) 11.5
Answer:
(d) 11.5

44. The common ion effect is based on
(a) Sorensen’s principle
(b) Le Chatelier’s principle
(c) Heisenberg’s principle
(d) Freundlich’s principle
Answer:
(b) Le Chatelier’s principle

45. The ion that cannot be precipitated by both HCl and H2S is
(a) Pb²⁺
(b) Cu²⁺
(c) Ag⁺
(d) Ca²⁺
Answer:
(d) Ca²⁺

46. The correct representation for solubility product of SnS2 is
(a) [Sn⁴⁺] [S^2-]²
(b) [Sn⁴⁺] [S^2-]
(c) [Sn⁴⁺] [2S^2-]
(d) [Sn⁴⁺] [2S^2-]
Answer:
(a) [Sn⁴⁺] [S^2-]²

47. The solubility product of a salt BA at room temperature is 1.21 × 10^-6. Its molar solubility is
(a) 1.21 × 10^-3 M
(b) 1.1 × 10^-4 M
(c) 1.1 × 10^-3 M
(d) 1.21 × 10^-2 M
Answer:
(c) 1.1 × 10^-3 M

48. Among the following hydroxides, the one which has the lowest value of solubility product at temperature 298 K is,
(a) Mg(OH)₂
(b) Ca(OH)₂
(c) Ba(OH)₂
(d) Be(OH)₂
Answer:
(d) Be(OH)₂

49. A solution becomes unsaturated when
(a) ionic product = solubility product
(b) ionic product < solubility product
(c) ionic product > solubility product
(d) ionic product ≥ solubility product
Answer:
(b) ionic product < solubility product

50. The solubility product of Fe(OH)₃ is
(a) [latex]\mathrm{F}_{\mathrm{e}}^{2+}[/latex] [OH^–]³
(b) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex] [OH^–]²
(c) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex] [OH^–]³
(d) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex]³ [OH^–]³
Answer:
(c) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex] [OH^–]³

51. The solubility product of PbS in 4.2 × 10^-28 at 300 K. The sulphide ions concentration required to precipitate PbS from a solution containing 0.001 M of lead ion is
(a) ≥ 2.1 × 10^-14 mol/dm³
(b) ≥ 4.2 × 10^-14 mol/dm³
(c) ≥ 4.2 × 10^-25 mol/dm³
(d) ≤ 4.2 × 10^-28 mol/dm³
Answer:
(c) ≥ 4.2 × 10^-25 mol/dm³

52. 0.025 M CH₃COOH is dissociated 9.5%. Hence the pH of the solution is
(a) 2.6244
(b) 3.128
(c) 2.988
(d) 2.267
Answer:
(a) 2.6244

53. 0.1 M HCN is dissociated 0.01%. The dissociation constant of HCN is
(a) 1.1 × 10^-6
(b) 1 × 10^-8
(c) 1 × 10^-9
(d) 1 × 10^-7
Answer:
(c) 1 × 10^-9

54. The solubility of product of a sparingly soluble salt AB₂ is 3.2 × 10^-11. If solubility in mol dm^-3 is
(a) 4 × 10^-4
(b) 3.2 × 10^-4
(c) 1 × 10^-5
(d) 2 × 10^-4
Answer:
(d) 2 × 10^-4

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 3 Kingdom Plantae Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 3 Kingdom Plantae

Question 1.
What is the basis of classification of kingdom Plantae?
Answer:
Kingdom plantae is classified on the basis of characteristics like absence or presence of seeds, vascular tissues, differentiation of plant body, etc.

Question 2.
What are Phanerogams and Cryptogams?
Answer:
1. Phanerogams are seed producing plants. These plants produce special reproductive structures that are visible.
2. Cryptogams are spore producing plants. These plants do not produce seed and flowers. They reproduce sexually by gametes, however their sex organs are concealed.

Question 3.
Write a short note on Chlorophyceae.
Answer:

1. Chlorophyceae includes green algae.
2. These are mostly fresh water (few brackish water and marine).
3. Plant body is unicellular, colonial or filamentous.
4. Cell wall contains cellulose.
5. Chloroplasts are of various shapes like discoid, plate-like, reticulate, cup-shaped, ribbon-shaped or spiral with chlorophyll a and b.
6. Reserved food is in the form of starch.
7. Pyrenoids are located in the chloroplast.
8. Green algae like Chlorella are rich in protein, hence used as food even by space travelers, e.g. Chlamydomonas, Spirogyra, Chara, Volvox, Ulothrix, etc.

Question 4.
Observe the given figure of Chara and identify the parts labelledas.

Answer:
X: Oogonium (contains egg)
Y: Antheridium (contains sperms)

Question 5.
Internet my friend (Textbook page no. 20)
Make a list of green algae with their characteristic shape of chloroplast.
Answer:
Green algae with their characteristic shapes of chloroplast:

1. Chlamydomonas – Cup-shaped
2. Spirogyra – Spiral or ribbon-shaped
3. Oedogonium – Reticulate
4. Zygnema – Stellate or Star-shaped

[Students are expected to search for more information regarding green algae with their characteristic shape of chloroplast from internet.]

Question 6.
Write the characteristics of Phaeophyceae.
Answer:
Characteristics of Phaeophyceae (Brown algae):

1. These algae are mostly marine, rarely fresh water.
2. Plant body is simple branched, filamentous (e.g. Ectocarpus) or profusely branched (e.g. Petalonia).
3. Cell wall has cellulose, fucans and algin.
4. Photosynthetic pigments like chlorophyll-a, chlorophyll-c and fucoxanthin are present.
5. Mannitol, laminarin are stored food materials. Body is usually differentiated into holdfast, stalk called stipe and leaf-like photosynthetic organ called frond.
6. Many species of marine algae are used as food. e.g. Laminaria, Sargassum.
7. Some species are used for the production of hydrocolloids (water holding substances), e.g. Ectocarpus, Fucus, etc.

Question 7.
Identify the given figure of a algae and explain the characteristics of its class with the help of following points:
Habitat, Plant body, photosynthetic pigments, cell wall, stored food

Answer:
The given figure is of Gracillaria. It belongs to class Rhodophyceae (Red algae).
Characteristics of Rhodophyceae:

1. Habitat: These are found in marine as well as fresh water on the surface, deep sea and brackish water.
2. Plant body: Plant body is thalloid.
3. Photosynthetic pigments: Cells contain chlorophyll-a, chlorophyll-d and phycoerythrin.
4. Cell wall: Cell wall is made up of cellulose and pectin glued with other carbohydrates.
5. Stored food: Stored food is in the form of Floridean starch.

Question 8.
What is the commercial use of red algae?
Answer:
Red algae like Gelidium and Gracilaria are used to obtain agar-agar which is used as solidifying agent in tissue culture medium.

Question 9.
Differentiate between red algae and brown algae.
Answer:
1. Photosynthetic pigments are chlorophyll-a, chlorophyll-d and phycoerythrin. Photosynthetic pigments are chlorophyll – a, chlorophyll-c and fucoxanthin.
2. Reserve food is Floridean starch. Reserve food is mannitol and laminarin.
e.g. Porphyra, Gracilaria, Gelidium, Polysiphonia, etc. Ectocarpus, Sargassum, Fucus, Laminaria, etc.

Question 10.
How rhizoids in liverworts differ from that of mosses?
Answer:
Rhizoids are unicellular in liverworts while they are multicellular in mosses.

Question 11.
Explain the thallus structure in lower members of Bryophyta. Give its two examples.
Answer:
1. Liverworts (Hepaticeae) are known as lower members of Bryophyta.
2. Gametophyte possesses flat plant body called thallus.
The thallus is green, dorsiventral, prostrate with unicellular rhizoids.
Examples: Riccia, Marchantia.

Question 12.
What are Hornworts? Give one example.
Answer:
Hornworts (Anthocerotae) are bryophytes which have flattened thallus that produces hornlike structures called as sporophytes. e.g. Anthoceros. In liverworts, asexual reproduction occurs by fragmentation of thalli or with the help of specialized structures called as gemmae. These are green, multicellular, asexual buds which grow in receptacles called gemma cup located on thalli. These gemmae detach from the thallus and germinate to form new individual.

Question 13.
Explain alternation of generation in life cycle of Bryophyta.
Answer:

1. Life cycle of Bryophytes shows sporophytic and gametophytic stages.
2. They alternate with each other to complete their life cycle.
3. Gametophyte is haploid, thalloid or leafy and dominant, (photosynthetic, independent thalloid or erect phase)
4. Sporophyte is short lived, multicellular and depends totally or partially on gametophyte for nutrition and anchorage.

Question 14.
Explain in detail the two stages of gametophytic phase in life cycle of Mosses.
Answer:

1. Gametophytic phase of the life cycle of Mosses (Musci) includes two stages namely; protonema stage and leafy stage.
2. The protonema is prostrate green, branched and filamentous (it is also called juvenile gametophyte). It bears many buds.
3. Leafy stage is produced from each bud.
4. Vegetative reproduction takes place by fragmentation and budding in secondary protonema.
5. The leafy stage has erected, slender stem like (Cauloid) main axis bearing spiral leaf like structures (Phylloid).
6. It is fixed in soil by multicellular branched rhizoids.
7. Leafy stage bears sex organs.

Question 15.
1. Name the two groups of Bryophytes.
2. Give the role of rhizoids in Bryophytes.
Answer:
1. Liverworts and mosses
2. Rhizoids absorb water and minerals and also help in fixation of thallus to the substratum.

Question 16.
Write economic importance of Bryophytes.
Answer:
Economic importance of Bryophytes:

1. Some mosses provide food for herbivorous mammals, birds, etc.
2. Species of Sphagnum, a moss; provides peat used as fuel.
3. Mosses are also used as packing material for transport of living materials because they have significant water holding capacity.
4. Mosses along with lichens are the first living beings to grow on rocks. They decompose rocks to form soil and make them suitable for growth of higher plants.
5. Dense layers of mosses help in prevention of soil erosion, thus act as soil binders.

Question 17.
Which group of plant is known as first vascular and true land plants? Write their characteristics in detail.
Answer:

1. Pteridophytes are known as first vascular and true land plants.
2. Habitat: Pteridophytes grow in moist and shady places, e.g. Ferns, Horsetail. Some are aquatic (Azolla, Marsilea), xerophytic (Equisetum) and epiphytic (Lycopodium).
3. Plant body: It is differentiated into root, stem and leaves.
4. Primary root: The primary root is short lived and is soon replaced by adventitious roots.
5. Stem: The stem may be aerial or underground.
6. Leaves: This group contains plants with pinnate (feather – like) leaves. Leaves may be scaly (e.g. Equisetum), simple and sessile (e.g. Lycopodium), small (microphylls e.g. Selaginella) or large (macrophylls) and pinnately compound (e.g. Nephrolepis l Ferns).
7. Vascular tissues: In these members xylem consists of only tracheids and phloem consists of only sieve cells.
8. Secondary growth: Secondary growth is not seen in pteridophytes due to absence of cambium.
9. Alternation of generations: Pteriodphytes show heteromorphic alternation of generations in which the sporophyte is diploid, dominant, autotrophic and independent. Gametophyte is haploid multicellular, generally autotrophic and short lived.

Question 18.
Match the columns.

Column I	Column II
1. Psilopsida	(a) Selaginella
2. Lycopsida	(b) Equisetum
3. Sphenopsida	(c) Adiantum
	(d) Psilotum

Answer:

Column I	Column II
1. Psilopsida	(d) Psilotum
2. Lycopsida	(a) Selaginella
3.Sphenopsida	(b) Equisetum

Question 19.
Write economic importance of Pteridophytes.
Answer:
1. Pteridophytes are used for medicinal purpose and as soil binders.
2. Many varieties are grown as ornamental plants.

Question 20.
Compare the gametophyte and sporophyte of Bryophytes with that of Pteridophytes.
Answer:

	Bryophytes	Pteridophytes
Gametophyte	It is haploid, dominant, photosynthetic, independent, thalloid or erect.	It is haploid, multicellular, generally autotrophic and short lived.
Sporophyte	It is short lived, multicellular and depends totally or partially on gametophyte for nutrition and anchorage.	It is dominant, independent and | vascular plant body.    i

Question 21.
Explain the given figure.

Answer:
1. The given figure represents megasporophyll of Cycas.
2. Megasporophyll of Cycas:
Megasporophylls are usually arranged in compact structures called female cones or female strobili. Megasporophyll contains megasporangia (ovule) which produce megaspores.
[Students are expected to collect more information about coralloid roots, scale leaf and megasporophyll of Cycas.]

Question 22.
Give the economic importance of Cycas and Pinus.
Answer:
1. Cycas is grown as an ornamental plant.
2. Pinus is used as source of pine wood, turpentine oil and pine resin.

Question 23.
Name the following:

Question 1.
Smallest gymnosperm
Answer:
Zamiapygmaea

Question 2.
The plant known as the ‘Coast red wood of California’.
Answer:
Sequoia sempervirens

Question 24.
Ginlcgo biloba is called as living fossil. Why?
Answer:
Ginkgo biloba is called as living fossil, because this plant is found in living as well as fossil form and the number of fossil forms is much more than the living forms.

Question 25.
Which of the following nuts will not be enclosed in fruits?
Betel nut/ Areca nut, pine nut, walnut, almond, cashew nut, nutmeg.
Answer:
1. Pine nuts are edible seeds of pines which are not enclosed in a fruit. It belongs to class gymnospermae thus, seeds are not enclosed within the fruit.
2. Nuts like betel nut/ areca nut, walnut, almond, cashew nut, nutmeg will be enclosed in fruits. It is because these plants belong to class angiospermae in which seeds are enclosed within the fruit.

Question 26.
Name various groups of vascular plants. Give one characteristic feature of each group.
Answer:
There are 3 groups of vascular plants:
1. Pteridophytes
2. Gymnosperms
3. Angiosperms
Characteristics of Pteridophytes: Pteridophytes are the only cryptogams with vascular tissue. Characteristics of Gymnosperms: Gymnosperms are the plants which possess naked seeds and also known as phanerogams without ovary.
Characteristics of Angiosperms: Angiosperms are the flowering plants in which the seeds remain enclosed within the fruits. Double fertilization is the unique feature of angiosperms. [Any one feature]

Question 27.
Classify the given plants into their respective groups and complete the given table.
Equisetum, Chara, Marchantia, Ginkgo biloba, Riccia, Spirogyra, Adiantum, Sorghum
Answer:

Chlorophyceae	Liverworts	Pteridophyta	Gymnosperms	Monocotyledonae
Chara, Spirogyra	Riccia, Marchantia	Equisetum, Adiantum	Ginkgo biloba	Sorghum

Question 28.
Match the columns.

Column I	Column II
1. Bryophyta	(a) 70 genera and 1000 living species
2. Pteridophyta	(b) 32 genera and 80 species
3. Gymnospermae	(c) 960 genera and 25000 species
	(d) 400 genera and 11000 species

Answer:

Column I	Column II
1. Bryophyta	(c) 960 genera and 25000 species
2. Pteridophyta	(d) 400 genera and 11000 species
3. Gymnospermae	(a) 70 genera and 1000 living species

[Source: Textbook of Biology, standard XI, First Edition; 2019, page no. 21,22,23.]

Question 29.
Identify the plants in the given figure and match the columns.


Answer:
1. c – 1
2. d – 2
3. a – 4
4. b – 3

Question 30.
Write a short note on Haplontic life cycle.
Answer:
1. In haplontic life cycle mitosis occurs in haploid cells.
2. It results in the formation of a single celled haploid or a multicellular haploid organism.
3. These forms produce the gametes through mitosis.
4. Zygote is formed after fertilization. This cell is the only diploid cell in the entire life cycle of the organism.
5. Thus, the same zygotic cell later undergoes meiosis.
6. This type of life cycle observed in some algae and fungi.
[Note: Haplontic life cycle is observed in many algae]

Question 31.
Observe the given figure and explain in detail.

Answer:

1. The given figure indicates diplontic life cycle.
2. Here, mitotic division occurs only in diploid cells.
3. Gametes formed through meiosis are haploid in nature.
4. The diploid zygote formed after fertilization divides mitotically.
5. In this process, production of multicellular diploid organism or the production of many diploid single cells takes place.
6. Animals show diplontic life cycle.

[Note: Diplontic type of life cycle is commonly observed in animals and all seed-bearing plants i.e. gymnosperms and angiosperms.]

Question 32.
Explain the term: Haplo-diplontic life cycle.
Answer:
1. In haplo-diplontic life cycle, mitosis occur in both diploid and haploid cells.
2. These organisms undergo through a phase in which they are multicellular and haploid (the gametophyte), and a phase in which they are multicellular and diploid (the sporophyte).
3. It is observed in land plants and in many algae.
[Note: It is commonly observed in bryophytes and pteridophytes.]

Question 33.
Fill in the blanks.
1. In haplo-diplontic life cycle, mitosis occurs in cells.
2. In diplontic life cycle, mitosis occurs in cells.
3. In haplontic life cycle, mitosis occurs in cells.
Answer:
1. diploid and haploid cells
2. diploid cells
3. haploid cells

Question 34.
Practical/Project:

Question 1.
Visit any nursery or botanical garden. Observe some older leaves of fern plant. You can observe some brown spots on back side of the leaflets as shown in the picture given below. Collect more information about it.
Answer:
1. The brown spots on the back side of older leaves of fern are sori.
2. They reproduce asexually by spores produced within sporangia, which are present in sori. These sori are located along the posterior surface of leaflets.

Question 35.
Read the given points.
1. A plant shows thalloid body.
2. A plant shows presence of rhizoids instead of true roots.
3. A plant needs external water for fertilization.
4. Vascular tissues are absent.
Identify the division of the plant described above.
Answer:
The plant belongs to division Bryophyta.

Question 36.
If a person wants to obtain agar for tissue culture, which plant group he should search?
Answer:
A person should search Rhodophyceae. It is because, ‘agar’ which is used as solidifying agent in tissue culture is obtained from red algae-Gelidium and Gracilaria.

Question 37.
Vinaya while playing in garden observed a pond with a green coloured covering which was floating on the surface of water? Next day she asked her teacher about the same. What her teacher must have told her?
Answer:
Vinaya’s teacher must have told her that the green coloured covering floating on the surface of pond water can be green algae like Spirogyra, Chlorella, Chlamydomonas, etc.

Question 38.
Identify the following:

1. These plants belong to thallophyta and grow upto 100 meters in height.
2. Plants used to obtain a product which is used a solidifying agent in preparation of ice-creams and jellies.
3. Gymnosperm which has girth of about 125 feet.
4. Xerophytic fern which belongs to sphenopsida.
5. Unicellular motile alga which belongs to Chlorophyceae and shows cup-shaped chloroplast.

Answer:

1. Kelps
2. Gelidium, Gracilaria
3. Taxodium mucronatum
4. Equisetum
5. Chlamydomonas

Question 39.
Quick Review:

Question 40.
Exercise:

Question 1.
Name the group of spores producing plants in
which sex organs are concealed.
Answer:
Cryptogams are spore producing plants. These plants do not produce seed and flowers. They reproduce sexually by gametes, however their sex organs are concealed.

Question 2.
Name the two divisions of phanerogams.
Answer:
v – phanerogamae

Question 3.
Complete the given flow chart.
Answer:
Quick Review

Question 4.
Define phanerogams.
Answer:
Phanerogams are seed producing plants. These plants produce special reproductive structures that are visible.

Question 5.
Write any two examples of phaeophyceae.
Answer:
Examples of phaeophyceae

Question 6.
Enlist the accessory pigments of algae.
Answer:
Various types of photosynthetic pigments are found in algae.
1. The accessory pigments are chlorophyll-b, chlorophyll-c, chlorophyll-d, carotenes, xanthophylls and phycobilins. Phycobilins are of two types, i.e. phycocyanin and phycoerythrin.
[Students are expected to collect more information about pigments found in algae from internet.]

Question 7.
Bryophytes are the amphibians of the plant kingdom. Justify
Answer:
Members of Bryophyta are mostly terrestrial plants which depend on water for fertilization and completion of their life cycle. Hence, they are called ‘amphibians of Plant Kingdom’.

Question 8.
Distinguish between Rhodophyceae and phaeophyceae with respect to photosynthetic pigments and reserve food.
Answer:
1. Photosynthetic pigments are chlorophyll-a, chlorophyll-d and phycoerythrin. Photosynthetic pigments are chlorophyll-a, chlorophyll-c and fucoxanthin.
2. Reserve food is Floridean starch. Reserve food is mannitol and laminarin.
e.g. Porphyra, Gracilaria, Gelidium, Polysiphonia, etc. Ectocarpus, Sargassum, Fucus, Laminaria, etc.

Question 9.
Write the characteristics of division that includes members like Chlamydomonas, Fucus, Gelidium, etc.
Answer:
Algae belongs to division Thallophyta.
Salient features of algae:

1. Habitat: Algae are mostly aquatic, few grow on other plants as epiphytes and some grow symbiotically. Some algae are epizoic i.e. growing or living non-parasitically on the exterior of living organisms.
   Aquatic algae grow in marine or fresh water. Most of them are free-living while some are symbiotic.
2. Structure: Plant body is thalloid i.e. undifferentiated into root, stem and leaves. They may be small, unicellular, microscopic like Cblorella (non-motile), Chlamydomonas (motile). They can be multicellular, unbranched, filamentous like Spirogyra or branched and filamentous like Chara. Sargassum is a huge macroscopic sea weed which measures more than 60 meters in length.
3. Cell wall: The algal cell wall contains either polysaccharides like cellulose / glucose or a variety of proteins or both. Reserve food material: Reserve food is in the form of starch and its other forms.
4. Photosynthetic pigments: Photosynthetic pigments like chlorophyll – a, chlorophyll – b, chlorophyll – c, chlorophyll – d, carotenes, xanthophylls, phycobilins are found in algae.
5. Reproduction: Reproduction takes place by vegetative, asexual and sexual method.
6. Life cycle: The life cycle shows phenomenon of alternation of generation, dominant haploid and reduced diploid phases.

Question 10.
Name the two algae from which agar is obtained.
Answer:
Red algae like Gelidium and Gracilaria are used to obtain agar-agar which is used as solidifying agent in tissue culture medium.

Question 11.
Identify the incorrectly labelled part in the figure of Funaria.
Answer:

Question 12.
Which are the first terrestrial plants to possess xylem and phloem?
Answer:
Pteridophytes are known as first vascular and true land plants.

Question 13.
Explain in detail three classes of algae.
Answer:

1. Chlorophyceae includes green algae.
2. These are mostly fresh water (few brackish water and marine).
3. Plant body is unicellular, colonial or filamentous.
4. Cell wall contains cellulose.
5. Chloroplasts are of various shapes like discoid, plate-like, reticulate, cup-shaped, ribbon-shaped or spiral with chlorophyll a and b.
6. Reserved food is in the form of starch.
7. Pyrenoids are located in the chloroplast.
8. Green algae like Chlorella are rich in protein, hence used as food even by space travelers, e.g. Chlamydomonas, Spirogyra, Chara, Volvox, Ulothrix, etc.

Characteristics of Phaeophyceae (Brown algae):

1. These algae are mostly marine, rarely fresh water.
2. Plant body is simple branched, filamentous (e.g. Ectocarpus) or profusely branched (e.g. Petalonia).
3. Cell wall has cellulose, fucans and algin.
4. Photosynthetic pigments like chlorophyll-a, chlorophyll-c and fucoxanthin are present.
5. Mannitol, laminarin are stored food materials. Body is usually differentiated into holdfast, stalk called stipe and leaf-like photosynthetic organ called frond.
6. Many species of marine algae are used as food. e.g. Laminaria, Sargassum.
7. Some species are used for the production of hydrocolloids (water holding substances), e.g. Ectocarpus, Fucus, etc.

Question 14.
Write ecological importance of Bryophytes.
Answer:
Economic importance of Bryophytes:
1. Some mosses provide food for herbivorous mammals, birds, etc.
2. Mosses along with lichens are the first living beings to grow on rocks. They decompose rocks to form soil and make them suitable for growth of higher plants.
3. Dense layers of mosses help in prevention of soil erosion, thus act as soil binders.

Question 15.
Mention one example each of aquatic and xerophytic pteridophytes.
Answer:
Habitat: Pteridophytes grow in moist and shady places, e.g. Ferns, Horsetail. Some are aquatic (Azolla, Marsilea), xerophytic (Equisetum) and epiphytic (Lycopodium).

Question 16.
State the uses of algae.
Answer:
(a) Many species of algae are used as food. For e.g. Chlorella (rich in cell proteins hence used as food supplement, even by space travelers), Sargassum, Laminaria, Porphyra, etc.
(b) Alginic acid is produced commercially from Kelps.
(c) Hydrocolloids like algin and carrageen are obtained from brown algae and red algae respectively.
(d) ‘Agar’ which is used as solidifying agent in tissue culture is obtained from red algae like Gelidium and Gracilaria.
(e) Brown algae like sea weeds are used a fodder for sheep, goat, etc.
[Students are expected to collect more information about the economic importance of algae.]
(f) Role of algae in environment.
Answer:
(a) Being photosynthetic, algae help in increasing the level of dissolved oxygen in their immediate environment.
(b) Algae are primary producers of energy rich compounds which forms the basis of food cycles in aquatic animals.
[Students are expected to find out more information about the role of algae in environment on internet.]

Question 17.
Mosses are used as packing material during transport of living material. Give reason.
Answer:
Mosses are also used as packing material for transport of living materials because they have significant water holding capacity.

Question 18.
Write the important characteristics of gymnosperms with respect to following points:
1. Vascular tissues
2. Roots
3. Spores
4. Leaves
Answer:
(b) Vascular tissues: They are vascular plants having xylem with tracheids and phloem with sieve cells.
(e) Roots: The root system is tap root type. In some gymnosperms, the roots form symbiotic association with other life forms. Coralloid roots of Cycas show association with blue green algae and roots of Pinus show association with endophytic fungi called mycorrhizae.
(g) Leaves: The leaves are dimorphic. The foliage leaves are green, simple needle like or pinnately compound, whereas scale leaves are small, membranous and brown.
(h) Spores: Spores are produced by microsporophyll (Male) and megasporophyll (Female).

Question 19.
What are the essential and accessory whorls in flower?
Answer:
Flower: Besides the essential whorls of microsporophylls (androecium) and megasporophylls (gynoecium), there are accessory whorls namely, calyx (sepals) and corolla (petals) arranged together to form flowers.

Question 20.
Write the characteristics of the class which includes Helianthus annuus.
Answer:
Habitat: Angiosperms is a group of highly evolved plants, primarily adapted to terrestrial habitat.

Question 21.
Secondary growth is absent in monocotyledonous plants. Justify.
Answer:
(a) In dicots, vascular bundles are conjoint, collateral and open type. Cambium is present between xylem and phloem for secondary growth.
(b) Whereas in monocots, vascular bundles are conjoint, collateral and closed type. Thus, due to absence of cambium, secondary growth does not occur in majority of monocots.

Question 22.
State characteristic of class monocotyledonae.
Answer:
b. Monocotyledonae:

1. These plants have single cotyledon in their embryo.
2. They have adventitious root system and stem is rarely branched.
3. Leaves generally have sheathing leaf base and parallel venation.
4. Flowers show trimerous symmetry.
5. The vascular bundles are conjoint, collateral and closed type.
6. Cambium is absent between xylem and phloem.
7. In Monocots, except few plants secondary growth is absent, e.g. Zea mays (Maize)

Question 23.
Draw a neat labelled diagram of:
1. Helianthus annuus (sunflower) plant.
2. Maize Plant.
Answer:
Two classes of Angiosperms are Dicotyledonae and Monocotyledonae.
а. Dicotyledonae:

1. These plants have two cotyledons in their embryo.
2. They have a tap root system and the stem is branched.
3. Leaves show reticulate venation.
4. Flowers show tetramerous or pentamerous symmetry.
5. Vascular bundles are conjoint, collateral and open type.
6. Cambium is present between xylem and phloem for secondary growth.
7. In dicots, secondary growth is commonly found.
   e. g. Helianthus annuus (Sunflower)

b. Monocotyledonae:

1. These plants have single cotyledon in their embryo.
2. They have adventitious root system and stem is rarely branched.
3. Leaves generally have sheathing leaf base and parallel venation.
4. Flowers show trimerous symmetry.
5. The vascular bundles are conjoint, collateral and closed type.
6. Cambium is absent between xylem and phloem.
7. In Monocots, except few plants secondary growth is absent, e.g. Zea mays (Maize)

Question 24.
Which is the diploid phase in life cycle of a plant?
Answer:
The life cycle of a plant includes two generations, sporophytic (diploid = 2n) and gametophytic (haploid = n)

Question 25.
Multiple Choice Questions:

Question 1.
Which of the following is not included in sub-kingdom Cryptogamae?
(A) Thallophyta
(B) Dicotyledonae
(C) Pteridophyta
(D) Bryophyta
Answer:
(B) Dicotyledonae

Question 2.
Unicellular, non-motile alga is
(A) Chara
(B) Chlorella
(C) Funaria
(D) Chlamydomonas
Answer:
(B) Chlorella

Question 3.
Which of the following is a brown algae?
(A) Laminaria
(B) Pteris
(C) Ulothrix
(D) Gelidium
Answer:
(A) Laminaria

Question 4.
Agar is obtained from group of algae.
(A) Rhodophyceae
(B) Chlorophyceae
(C) Phaeophyceae
(D) Both (A) and (C)
Answer:
(A) Rhodophyceae

Question 5.
In Chlamydomonas, pyrenoid is located in
(A) nucleus
(B) mitochondria
(C) chloroplast
(D) flagella
Answer:
(C) chloroplast

Question 6.
In bryophytes, represents sporophytic
generation.
(A) rhizoids
(B) thalloid
(C) capsule
(D) leafy plant body
Answer:
(C) capsule

Question 7.
Which of the following is an example of liverwort?
(A) Funaria
(B) Marchantia
(C) Polytrichum
(D) Sphagnum
Answer:
(B) Marchantia

Question 8.
The late Paleozoic era is regarded as the age of ______ .
(A) Thallophytes
(B) Gymnosperms
(C) Pteridophytes
(D) Angiosperms
Answer:
(C) Pteridophytes

Question 9.
Which of the following is an epiphytic pteridophyte?
(A) Azolla
(B) Equisetum
(C) Marsilea
(D) Lycopodium
Answer:
(D) Lycopodium

Question 10.
Complete the given analogy:
Lycopsida: _______:: Pteropsida: Pteris
(A) Adiantum
(B) Selaginella
(C) Equisetum
(D) Psilotum
Answer:
(B) Selaginella

Question 11.
Bryophytes differ from Pteridophytes in being
(A) vascular
(B) seeded
(C) non-vascular
(D) sporophytic
Answer:
(C) non-vascular

Question 12.
Endophytic fungi or mycorrhizae are found in the roots of
(A) Cycas
(B) Pinus
(C) Equisetum
(D) Hibiscus
Answer:
(B) Pinus

Question 13.
Gymnosperms are characterized by the absence of
(A) tracheids in xylem
(B) sieve cells in phloem
(C) heterosporous condition
(D) fruit formation
Answer:
(D) fruit formation

Question 14.
Complete the given analogy:
Tallest angiosperm : Eucalyptus :: Smallest angiosperm : _________ .
(A) Zanta pygmaea
(B) Sequoia sempervirens
(C) Taxodium mucronatum
(D) Wolffia
Answer:
(D) Wolffia

Question 15.
Select the INCORRECT statement with respect to angiosperms.
(A) Seeds are enclosed within a fruit.
(B) These plants show heteromorphic alternation of generation.
(C) Megaspores are borne on highly specialized microsporophyll.
(D) They are most advanced group of flowering plants.
Answer:
(C) Megaspores are borne on highly specialized microsporophyll.

Question 16.
Parallel venation is a characteristic feature of
(A) Monocotyledons
(B) Dicotyledons
(C) Pteridophytes
(D) Bryophytes
Answer:
(A) Monocotyledons

Question 17.
In gymnosperms and angiosperms _______ is much reduced.
(A) gametophyte
(B) root
(C) sporophyte
(D) vascular bundle
Answer:
(A) gametophyte

Question 18.
Presence of rhizoids in place of true roots is a characteristic of
(A) Gymnosperms
(B) Bryophyta
(C) Pteridophyta
(D) Angiosperms
Answer:
(B) Bryophyta

Question 19.
Competitive Corner:

Question 1.
Which one of the following statements is wrong?
(A) Laminaria and Sargassum are used as food.
(B) Algae increase the level of dissolved oxygen in the immediate environment.
(C) Algin is obtained from red algae, and carrageen from brown algae.
(D) Agar-agar is obtained from Gelidium and Gracilaria.
Hint: Algin is obtained from brown algae and carrageenan from red algae.
Answer:
(C) Algin is obtained from red algae, and carrageen from brown algae.

Question 2.
Select the CORRECT statement.
(A) Sequoia is one of the tallest trees.
(B) The leaves of gymnosperms are not well adapted to extremes of climate.
(C) Gymnosperms are both homosporous and heterosporous.
(D) Salvinia, Ginkgo and Pinus all are gymnosperms.
Hint: The leaves of gymnosperms are well adapted to withstand extremes of climate. Gymnosperms are heterosporous. Salvinia is a Pteridophyte.
Answer:
(A) Sequoia is one of the tallest trees.

Question 3.
In bryophytes and pteridophytes, transport of male gametes requires
(A) Birds
(B) Water
(C) Wind
(D) Insects
Answer:
(B) Water

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 1.
What is a carbonyl group?
Answer:
Carbonyl group : A functional group in which a carbon atom is attached to an oxygen atom by a double bond and remaining two valencies of carbon atom are free is called a carbonyl group and represented as . Carbonyl group is present in aldehydes and ketones.

Question 2.
What are carbonyl compounds?
Answer:
The organic compounds containing a carbonyl group are called carbonyl compounds. For example, acetaldehyde, , acetone, . As carbonyl group is common in aldehydes and ketones, their methods of preparation and properties show similarities.

Question 3.
What are carboxylic compounds?
Answer:
The compounds in which the functional group is – COOH are known as carboxylic compounds. Due to the – OH group bonded to group, carboxylic acids are distinct from aldehydes and ketones.

Question 4.
How are carbonyl compounds classified ?
OR
Name the compounds containing carbonyl group.
Answer:
The carbonyl compounds contain a group . They are classified as follows :

Question 5.
What are aliphatic aldehydes?
Answer:
The compounds in which the – CHO group (formyl group or aldehyde group) is attached directly to sp³ hybridized carbon atom that is saturated carbon atom are called aliphatic aldehydes. (Exception : Formaldehyde, H – CHO is also classified as aliphatic aldehyde though – CHO group is not attached to any carbon).

For example :

Question 6.
What are aromatic aldehydes ?
Answer:
The compounds in which – CHO group is attached directly to an aromatic ring are called aromatic aldehydes.

For example :

Question 7.
Explain the structure of carbonyl functional group.
Answer:

• In the carbonyl functional group, carbon atom is attached to an oxygen atom by a double bond and remaining two valencies of carbon atom are free, and it is represented as .
  
• The carbonyl carbon atom is sp²-hybridised forming coplanar three sigma (σ) bonds with the bond angle 120°.
• One sigma bond is formed with oxygen atom while other two sigma (σ) bonds are formed with hydrogen or carbon atoms.
• The remaining unhybridised 2p_z orbital of carbon atom overlaps with p orbital of oxygen atom colaterally forming a pi (π) bond. Hence, carbon atom is joined to oxygen atom by a double bond of which one is sigma and another is n.
• The oxygen atom in the carbonyl group has two lone pairs of electrons.
• The carbonyl bond is strong, short and polarized.
• The polarity of the carbonyl group is explained on the basis of resonance involving neutral and dipolar structures as shown below :
  

Question 8.
What are aliphatic ketones? How are they classified?
Answer:
Aliphatic ketones : The compounds in which group is attached to two alkyl groups are called aliphatic ketones.

Ketones are classified into two types :

1. Simple or symmetrical ketones and
2. mixed or unsymmetrical ketones.

1. Simple or symmetrical ketone : The ketone in which the carbonyl carbon is attached to two identical alkyl groups is called a simple or symmetrical ketone.

2. Mixed or unsymmetrical ketone : The ketone in which the carbonyl carbon is attached to two different alkyl groups is called a mixed or unsymmetrical ketone.

Question 9.
What are aliphatic carboxylic acids? Give their general formula.
Answer:
The organic compounds in which carboxyl (- COOH) group is bonded to an alkyl group are called aliphatic carboxylic acids or fatty acids. (Exception : Formic acid, H-COOH is also classified as aliphatic carboxylic acid though-COOH group is not attached to any carbon).

For example :

Question 10.
How are carboxylic acids classified ? Give examples. (3 marks)
Answer:
Carboxylic acids are classified according to the presence of number of carboxyl groups into mono-, di-, tri- and polycarboxylic acids.

• Monocarboxylic acids : These carboxylic acids contain one carboxyl group.
  Example :
• Dicarboxylic acids : These contain two carboxyl groups
  Example :
• Tricarboxylic acid : These contain three carboxyl groups
  

Question 11.
What are aromatic carboxylic acids ? Give examples.
Answer:
Aromatic carboxylic acids : These are the compounds in which one or more carboxyl groups (- COOH) are attached directly to the aromatic ring.

For example :

Question 12.
Give examples of common carboxylic acids which are used in daily life.
Answer:
Common carboxylic acids are widely distributed in nature, they are found in both the plants and animals.

• Acetic acid is main constituent of vinegar.
• Butyric acid of butter which is responsible for odour of rancid butter.
• L-lactic acid is present in curd.
• Citric acid is found in citrus fruits.
• Higher carboxylic acids such as palmitic acid, stearic acid and oleic acid are the components of animal fats and vegetable oils.

Nomenclature of Aldehydes :

(A) Common System :

• The names of aldehydes are derived from the common names of acids.
• The suffix ‘-ic acid’ of an acid is replaced by ‘aldehyde’.
• The positions of the substituents in the molecule are indicated by Greek letters α, β, γ, etc.
• starting from the carbon atom attached to the carbonyl group. E.g.
  

(B) I UP AC System :

• The longest carbon atoms chain containing aldehyde carbon atom is selected as a parent hydrocarbon.
• ‘e’ of the alkane is replaced by ‘al’. Alkane → Alkanal
• The position (locant) of aldehyde group need not be mentioned since it is always at the end position.
• The substituents in the alkyl group are prefixed in an alphabetical order by appropriate locants.
• When two – CHO groups are present at the two ends of the chain the ending ‘e’ of alkane is retained and the suffix  ‘-dial’ is added to the name of parent aldehyde.
• In IUPAC nomenclature an alicyclic compound -in which – CHO group is attached directly to the ring is named as a carbaldeliyde. The suffix ‘carbaldehyde’ is added after the full name of parent cycloalkane structure.

(C) Common or trivial names :

(1) The common name of a carboxylic acid is derived from the source from which it was first isolated.
The following table gives common names and the source or origin of name.

(2) In branched carboxylic acids, the position of substituents are indicated by Greek alphabet.

For example :

(D) IUPAC system of nomenclature :

• The longest continuous cha in of carbon atoms including the carbon atom of – COOH group ¡s selected.
• The carboxvlic acid is conside red (IS a derivative of the corresponding parent a/kane.
• The carbon atom of the – COOH group is always at terminal position, hence need not to be indicated while writing IUPAC name.
• The position of the other substitutents are indicated by the appropriate locants in alphabetical order.
• In case of dicarboxylic acids, ‘dioic acid’ is added to parent alkane.
• In an alicyclic compound having a carboxyl group directly attached to alicyclic ring is named as cycloalkane carboxylic acid.

Trivial and IUPAC names of carboxylic acids and aldhydes

Question 13.
Give the common names and IUPAC names of the Miowing aldehydes :
Answer:

Question 14.
Write the structures of the following compounds :
Answer:

Question 15.
What is the IUPAC name of the following compound?

Answer:
IUPAC name : 2-Amino butanoic acid

Question 16.
Write IUPAC name of

Answer:
IUPAC name : Ethanedioic acid

Question 17.
Write the structure and give IUPAC names of following carboxylic acid.

Answer:

Question 18.
Draw the structures of the following compounds :
Answer:

Question 19.
Give IUPAC names of the following carboxylic acids.
Answer:

Question 20.
Write the structures and IUPAC names of all isomers of carboxylic acids having molecular formula C₅H₁₀O₂. HOW many of them are chiral?
Answer:

Nomenclature of ketones :

(A) Common System :

• Ketones are named according to the alkyl groups attached to the carbonyl carbon atom followed by the word ketone.
• The substituents in the alkyl groups are indicated by Greek letters a, f, y, etc. starting from the carbon atom attached to the carbonyl group.

(B) IUPAC System :

• The longest continuous chain containing carbonyl carbon atom is selected as a parent hydrocarbon.
• ‘e’ of the alkane is replaced by ‘one’. Alkane → Alkanone
• The position of carbonyl group is represented by the lowest locant.
• The substituents in the alkyl groups are prefixed in the alphabetical order along with their positions by appropriate locants.
• When two C = O groups are present, then ending ‘e ’ of alkane is retained and the suffix – ‘dione ’ is added to the name of parent ketone indicating the locants of ketonic carbonyl groups.
• In case of polyfunctional ketones, higher priority group is given lower number.
• When ketonic carbonyl is a lower priority group it is named as ‘oxo’, preceded by the locant. In alicyclic ketones, carbonyl carbon is numbered as 1.

Question 21.
Give the common and IUPAC names of the following ketones :
Answer:

Question 22.
Give the structures of the following compounds :
Answer:

Question 23.
Give IUPAC names of the following compounds :
Answer:

Question 24.
Write the structures and give common names and IUPAC names of the carbonyl compounds represented by formula C₅H₁₀O.
Answer:

Question 25.
Write the structure and give IUPAC names of the following compounds :
Answer:

Question 26.
Write the structure of the following compounds :
Answer:

Question 27.
How is an aldehyde obtained from an alcohol ?
Answer:
When a primary alcohol is oxidized with potassium dichromate and dil. H₂SO₄ under controlled conditions, an aldehyde is obtained.

For example, when ethanol is oxidized with potassium dichromate and dil. H₂SO₄ under controlled conditions, acetaldehyde (ethanal) is obtained.

Question 28.
How is ketone obtained from an alcohol?
Answer:
When a secondary alcohol is oxidized with potassium dichromate and dii. H₂SO₄ under controlled conditions, a ketone is obtained.

For example. when 2-propanol is oxidized with potassium dichromate and dii, H₂SO₄ under controlled conditions, accIone (propanone) is obtained.

Question 29.
How are the following compounds obtained from alcohol:
(1) Methanal
(2) Propanal
(3) BuLanal
(4) 3-Methylpentanal?
Answer:

1. Mehanol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms methanal.
   
2. Propan-1-ol on controlled oxidation with K,Cr20 and dilute H₂SO₄ forms propanal.
   
3. Butan..l-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms butanal.
   
4. 3-Mcthylpcntan.1-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ gives 3-Methylpentanal.
   

Question 30.
How are the following compounds prepared from alcohol :
(1) Butanone
(2) Pentan-3-one
(3) 2,2-Dimethylpropanal?
Answer:

1. Butan-2-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms butanone.
   
2. Pentan-3-ol on oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms Pentan-3-one.
   
3. 2,2-Dimethylpropan- I -ol on oxidation with K₂Cr₂O₇ or KMnO₄ and dilute H₂SO₄ forms 2,2-Dimethyipropanal.
   

Question 31.
How is an aldehyde obtained from primary alcohol ?
Answer:
When vapours of primary alcohol is passed over heated copper at 573 K, dehydrogenation takes place, an aldehyde is formed.

For example : When vapours of isopropyl alcohol is passed over heated copper at 573 K, acetone (propanone) is obtained.

Question 32.
How is ketone obtained from secondary alcohol?
Answer:
When vapours of secondary alcohol is passed over heated copper at 573 K. dehydrogenation takes place, a ketone is obtained.

For example : When vapours of isopropyl alcohol is passed over heated copper at 573 K. acetone (propanone) is obtained.

Question 33.
How are the following compounds obtained from alkene :
(1) Formaldehyde
(2) Acetaldehyde and
(3) Acetone ?
Answer:
When a stream of ozonised oxygen is passed through a solution of an alkene, in organic solvent, an unstable addition cyclocompound, ozonide is formed which on reduction with zinc dust and water forms an aldehyde or a ketone or a mixture of both.

1. Formaldehyde : Under these conditions ethylene gives formaldehyde.
   
2. AcetuIdeh&: Symmetrically disubstituted alkene like but-2-ene gives acetaldehyde.
   
3. scctnne: Tetrasubshtwed alkene like 2,3-dimethyl but-2-ene gives acetone.
   

Question 34.
Write ozonolysis reaction for
(1) Propylene and
(2) Isobutylene.
Answer:
(1) Propylene on reaction with ozonised oxygen in the organic solvent forms propylene ozonide which on reduction with zinc dust and water forms acetaldehyde and formaldehyde.

Question 35.
How are the following compounds obtained from alkynes :
(1) Acetaldehyde
(2) Acetone?
Answer:

1. Acetaldehyde : On passing acetylene through warm 40% H₂SO₄ in the presence of 1 % H_gSO₄, vinyl alcohol is obtained which tautomerises and forms acetaldehyde. It is a hydration reaction.
   
2. Acetone : On passing propyne through warm 40 % H₂SO₄ in the presence of 1 % H_gSO₄, alkenol is obtained which on tautomerisation form acetone.
   

Question 36.
Predict the products when
(1) dimethyl acetylene
(2) ethyl acetylene and
(3) diethyl acetylene are treated with mercuric sulphate in dilute sulphuric acid.
Answer:
(1) Dimethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms ethyl methyl ketone by tautomerisation.

(2) Ethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms Butan-2-one by tautomerisation.

(3) Diethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms hexan-3-one.

Question 37.
Write the structures of aldehydes and ketones obtained by ozonolysis.

Answer:

Answer:

Question 38.
Predict the structures of ketones produced by hydration of but-l-yne and but-2-yne.
Answer:

Question 39.
How is acetaldehyde prepared from acetyl chloride?
Answer:
Acetyl chloride is reduced to acetaldehyde by hydrogen in presence of Pd catalyst poisoned with BaSO₄. This reaction is called Rosenmund reduction.

Question 40.
How is benzaldehyde obtained from benzoyl chloride?
OR
Write chemical equation for Rosenmund reduction.
Answer:
When benzoyl chloride is hydrogenated in the presence palladium on barium sulphate (Pd/BaSO₄), benzaldehyde is obtained. This reaction is called Rosenmund reduction.

Question 41.
How will you prepare acetophenone from benzene? (Friedel – Crafts acylation).
Answer:
When benzene is treated with acetyl chloride in the prcscncc of anhydrous aluminium chloride, acetophenonc is obtained. This reaction is known as Friedel – Crafts acylation.

Question 42.
How will you convert benzene into 1-phenylethanone?
Answer:

Question 43.
How will you obtain 4-Nitrobenzaldehyde from 4-Nitrotoluene ? (Friedel- Crafts reaction).
Answer:
When 4-nitrotoluene is treated with chromium oxide in acetic anhydride, a diacetate derivative is obtained which on acid hydrolysis produces 4-nitrobenzaldehyde.

Question 44.
How will you prepare Propanone (acetone) from Grignard reagent?
Answer:
Grignard reagent (methyl magnesium iodide) reacts with cadmium chloride to give dimethyl cadmium. When acetyl chloride reacts with dimethyl cadmium, propanone (acetone) is obtained.

Question 45.
How is acetophenone obtained from Grignard reagent ?
Answer:
Grignard reagent (methyl magnesium iodide) reacts with cadmium chloride to give dimethyl cadmium. When benzoyl chloride reacts with dimethyl cadmium, acetophenone is obtained.

Question 46.
How is benzyl methyl ketone obtained from Grignard reagent ?
OR
Convert: Acetyl chloride to benzyl methyl ketone.
Answer:
Grignard reagent (Benzyl magnesium chloride) reacts with cadmium chloride to give diphenyl cadmium. When acetyl chloride reacts with dibenzyl cadmium, benzyl methyl ketone is obtained.

Question 47.
How is an aldehyde obtained from alkyl nitrile ?
OR
What is Stephen reaction ?
OR
Write a note on Stephen reaction.
Answer:
(1) An ethereal solution of a nitrile is reduced to imine hydrochloride by SnCl₂ in the presence of HCl gas. Further, imine hydrochloride on acid hydrolysis gives aldehyde. This reaction is called Stephen reaction.


(2) Alternatively, nitriles are selectively reduced by diisobi.ityl aluminium hydride (DIBAI-H) lo imines which on acid hydrolysis to aldehydes.

Question 48.
How are following compounds prepared using Gngnard reagent
(1) Acetone
(2) Benzophenone?
Answer:
(1) Acetone: Acetoniti-ile (ethanenitrile) reacts with methyl magnesium iodide in presence of dry ether to give imine complex which on hydrolysis gives acctonc. During reaction acetonitrile and methyl magnesium iodide should be
taken in equimolecular proportion.

(2) Benzophenone: Benzonitrile reacts with phenyl magnesium bromide in presence of dry ether to give an imine complex which on acid hydrolysis gives a benzophenone. During reaction bcnzonitrile and phenyl magnesium bromide should be aken in equimolecular proportion.

Question 49.
Write the structures and IUPAC names of ketones produced by Friedel-Crafts acylation of benzene with
(i) C₂H₅COCl
(ii) C₆H₅COCl.
Answer:

Question 50.
Predict the products of the following conversions.
Answer:

Question 51.
How are the following preparations carried out ?

(1) Benzaldehyde from toluene. (Etard oxidation)
Answer:
When toluene is treated with solution of chromyl chloride (CrO₂Cl₂) in Cs₂, brown chromium complex is obtained, which on acid hydrolysis gives benzaldehyde. This reaction is called Etard reaction.

(2) Benzaldehyde from methyl arene.
Answer:
Methylarene is converted into a benzyllidene diacetate on treatment with chromium oxide in acetic anhydride at 273-278 K. The diacetate derivative on acid hydrolysis gives benzaldehyde.

(3) Benzaldehyde from toluene (commerical method).
Answer:
Side chain chlorination of toluene gives benzal chloride which on hydrolysis gives benzaldehyde.

(4) Benzaldehyde from benzene (Gattermann-Koch synthesis).
OR
Write chemical equation for Gatter- mann-Koch formylation.
Answer:
When benzene is treated with vapours of carbon monoxide and hydrogen chloride in the presence of a catalyst mixture of A1Cl₃ and CuCl under high pressure, benzaldehyde is obtained. This reaction is called Gattermann- Koch synthesis.

Preparation of aromatic ketones from hydrocarbons

Question 52.
Explain Friedel-Craft’s acylation reaction.
Answer:
The reaction in which hydrogen atom of benzene is replaced by an acyl group I in the presence of anhydrous AlCl₃ is called Friedel-Craft’s acylation. When benzene is heated with an acetyl chloride or acetic anhydride in the presence of anhydrous AlCl₃, forms acetophenone (1-Phenyl ethanone).

Electrophile : R – C + = O acylium ion Formation of the electrophile :

Question 53.
Give the preparation of acetophenone from benzene using
(i) acetyl chloride
(ii) acetic anhydride.
Answer:

The reaction in which hydrogen atom of benzene is replaced by an acyl group I in the presence of anhydrous AlCl₃ is called Friedel-Craft’s acylation. When benzene is heated with an acetyl chloride or acetic anhydride in the presence of anhydrous AlCl₃, forms acetophenone (1-Phenyl ethanone).

Electrophile : R – C + = O acylium ion Formation of the electrophile :

Question 54.
How will you prepare propionaldehyde from ethyl propionate?
Answer:
When ethyl propionate is reduced in presence of diisobutyl aluminium hydride (DIBAI-H), propionaldehyde is obtained.

Question 55.
Explain the structure of carboxyl group.
Answer:
In carboxyl group, the carboxyl carbon is sp²-hybridised and the bonds to the carboxyl carbon lie in one plane. The C-C = O and O = C-O bond angles are 120°. The carboxylic carbon is less electrophilic than carbonyl carbon because of the resonance structures shown below :

Question 56.
How is carboxylic acid obtained by the acid hydrolysis of an alkyl cyanide ?
Answer:
Alkyl cyanides or alkyl nitriles on acid or alkaline hydrolysis give corresponding carboxylic acids.

Acid Hydrolysis of Alkyl ankle: When alkyl cyanide is boiled wiLh dilute mineral acid, it gives corresponding carboxylic acid. In this, acid amide is obtained as the intermediate product.

Question 57.
How is ethanoic add obtained from methyl cyanide by acid hydrolysis?
Answer:
When methyl cyanide is heated with dilutc hydrochloric acid or dilute sulphuric acid. ethanoic acid is obtained.

Question 58.
How Is proplonlc acid obtained from an alkyl nitrile?
Answer:
When ethyl cyanide (propionitrile) is boiled with dilute HCI or dilute H₂SO₄, propionic acid is obtained.

Question 59.
How Is benzoic acid obtained from bcnzamide?
Answer:
When benzarnide is heated with dil. HCl. benzoic acid is obtained.

Question 60.
How is carboxylic acid obtained from acyl chlorides and acid anhydrides?
Answer:
When acyl chloride is hydrolysed with water, carboxylic acid is obtained. The reaction is carried out in presence of a base pyridine or NaOH to remove HCl generated.

Acetyl chloride reacts with water almost explosively while benzoyl chloride very slowly.

Acid anhydrides react with water to give carboxylic acid.

Question 61.
How is benzoic acid obtained from
(i) ethyl benzoate
(ii) styrene?
Answer:
(i) Benzoic acid from ethyl benzoate : When an ethyl benzoate is heated with dil. H₂SO₄, undergoes hydrolysis to form benzoic acid and ethyl alcohol.

(ii) Benzoic acid from styrene:

Question 62.
How is propanoic acid obtained from phenyl propanoate?
Answer:
When phenyl propanoate is heated with dil. NaOH, sodium salt is obtained, which on hydrolysis gives propanoic- acid.

Question 63.
How is propanoic acid obtained from methyl propanoate ?
OR
When methyl propanoate is heated with dil. NaOH, sodium, salt is obtained, which on hydrolysis gives propanoic acid.
Answer:

Question 64.
How is benzoic acid obtained from phenyl ethene?
Answer:
When phenyl ethene is heated with strong oxidising agents like acidic KMnO₄ or acidic K₂Cr₂O₇, benzoic acid is obtained.

Question 65.
How is adipic acid obtained from cyclohexene?
Answer:
When cyclohexene is heated with acidified KMnO₄, adipic acid (Hexane-1, 6-dioic acid) is obtained.

Question 66.
What is carbonation of Grignard reagent ? How is acetic acid prepared by this reaction ? How is ethanoic acid pepared from dry ice?
Answer:
Addition reaction of carbon dioxide (0 = C = 0) to Grignard reagent, forming a complex and further formation of carboxylic acid is called carbonation of Grignard reagent.

Example : When methyl magnesium iodide is added to solid carbon dioxide, a complex is formed which on acid hydrolysis forms acetic acid.

Question 67.
How is benzoic acid prepared from Grignard reagent?
OR
Write the preparation of benzoic acid from dry ice.
Answer:
When phenyl magnesium bromide is treated with dry ice (solid carbon dioxide) in the presence dry ether, complex is obtained which on acidification gives benzoic acid.

Question 68.
What are soaps ? How are they prepared ?
Answer:
The sodium or potassium salts of higher fatty acids are known as soaps. Soaps contain more than twelve carbon atoms.

When fat or oil is hydrolysed using sodium or potassium hydroxide solution, soap obtained remains in colloidal form. Soap and glycerol are separated by adding sodium chloride. Soap precipitates out due to common ion effect, and glycerol remains in the solution can be recovered by fractional distillation.

Question 69.
How is benzoic acid prepared from alkyl benzenes ?
OR
How will you convert the following :
(1) n-butyl benzene to benzoic acid.
(2) Toluene to benzoic acid.
(3) Cumene to benzoic acid ?
Answer:
When an alkyl benzene is heated with strong oxidizing agents like acidic or alkaline KMnO₄ or acidified K₂Cr₂O₇ etc. gives aromatic carboxylic acid. The alkyl side chain gets oxidised to -COOH group irrespective of the size of the chain.

Question 70.
Lower aldehydes and ketones are water soluble whereas higher homologues are insoluble. Explain, why.
Answer:
(1) The oxygen atom of shows hydrogen bonding with water molecule.

(2) As a result of this, the lower aldehydes and ketones are water soluble (example – acetaldehyde, acetone). As the molecular mass increases, the proportion of hydrocarbon part of the molecule increases which cannot form hydrogen bonding with water and the solubility of higher homologues in water decreases.

Question 71.
Carboxylic acids have higher boiling points than those of alcohols, aldehydes, ketones, ethers, hydrocarbons of comparable molecular masses. Explain, why.
Answer:
(1) Carboxylic group (-COOH) in acids is highly polar. in liquid state, pair of carboxylic acid molecules is held together by two intermolecular hydrogen bonds, have higher aggregations and in the vapour.

(2) Intermolecular hydrogen bonding in carboxylic acids state most of the carboxylic acid.s exist as dimmers in which two molecules are held by two hydrogen R – R bonds. Acidic hydrogen of one molecule forms hydrogen bond with carbonyl oxygen of the other molecule. This doubles the size of the molecule resulting in increase in o – intermolecular van der Waals forces, which in turn results in high boiling points. Therefore, carboxylic acids possess higher boiling points than those of alcohols, aldehydes, ketones. ether, hydrocarbons of comparable molecular masses.

Question 72.
Lower aliphatic carboxylic acids are miscible with water while higher carboxylic acids are immiscible.
Answer:
(1) Lower aliphatic carboxylic acids are miscible with water due to the formation of hydrogen bonding with water molecules.

(2) Hydrogen bonding between acid and water. As the molecular mass increases (he solubility of carboxylic acids in water decreases. The insolubility of carboxylic acids is due to increased hydrophobic interaction of hydrocarbon parts with water.

Question 73.
Explain why carboxylic acids are much weaker acids than mineral acids.
Answer:
Carboxylic acids are the organic compounds which are acidic in nature. However, compared to mineral acids like HCI or H₂SO₄. the carboxylic acids are weaker acids.

The strength of acidity depends upon their ability to release H⁺ ions. Greater the ease with which they release H⁺ ions, stronger is the acid.

Carboxylic acids when dissolved in water, pcoduce H⁺ due to its dissociation. (it does not dissociate completely.)

The mineral acid-s like HCI, H₂SO₄ release H⁺ ion to a larger extent as they dissociatc almost complctcly in aqueous solution for e.g. HCl → H⁺ Cl^– thus carboxylic acids are weaker than mineral acids. The equilibrium exists in aqueous solution of carboxylic acid as

Since concentration of water practically remains constant

where K_a is acidity constant
Larger the value K_a greater is the extent of ionization and stronger is the acid. But strength of acids is expressed in ternis of their pK_a values. Smaller the value of pK_a. the stronger is the carboxylic acid. Here pK_a value of carboxylic acids is higher than mineral acids. Hence, carboxylic acids arc weaker than mineral acids.

Question 74.
Carboxylic acids are more acidic than phenols and alcohols. Explain. Why?
Answer:
(1) Carboxylic acid loses a proton as compared to phenol. Consider the ionization of carboxylic acid and phenol

Due to delocalization the negative charge over the ortho and para positions of aromatic ring, phenoxide anion is more stable than phenol. Thus phenol easily undergoes ionization

However, alcohol and alkoxide ion are single structures. In an alkoxide anion the negative charge is localized on a single oxygen atom. Hence, phenols are more acidic than alcohols.

(2) Carboxylic acid has two resonance hybride non equivalent structures (I & II) while carboxylate anion has two resonance hybrid equivalent structures (III & IV). The carboxylate ion is more stable than carboxylic acid and equilibrium is shifted towards the direction of increased ionization.

(3) Carboxylate ion has two equivalent resonance structures with nejative charge is delocalized over two electro negative oxygen atoms. Phenoxide anion has non-equivalent resonance structures in which negative charge is
delocalized over one oxygen atom and less electronegative carbon atom. As a result carboxylate anion is more stable than phenoxide ion. Hence carboxylic acids ionize to the greater extent than phenol furnishing higher concentration of H⁺ ions. Therefore, carboxylic acids are more acidic than phenols and alcohols.

Question 75.
Trichloro acetic acid is a stronger acid than acetic acid. ExplaIn.
Answer:
(1) The acidic nature of carboxylic acid is due to the ability to release H ions. Greater the ease with which they release H⁺ ions, stronger is the acid. Any factor that stabilizes the carboxylate ion would help the release of H⁺
ions and thus increase the strength of the acid. The electron-withdrawing group attached to -carbon atom increases the strength of the acid. In trichloroacetic acid, three chloro substituents on s-carbon atom of acetic acid makes the electrons withdrawing effect more pronounced and the negative charge of carboxylate ion formed gets dispersed.

Thus, increases the stability of carboxylate ion.

(2) The acetate ion formed gets destabilised due to the electron releasing effect of a methyl group (+ I effect). As a result, acetic acid dissociates to a lesser extent.

(3) The trichloro acetate ion formed gels stabilised due to electron-withdrawing effect of three 3Cl atoms (- I effect). As a result. trichloro acetic acid dissociates to a greater extent. Trichioro acetic acid having lower pK_a value than acetic acid. Hence. trichloro acetic acid is a stronger acid than acetic acid.

Question 76.
Which is the stronger acid in each of the following pairs ?
(1) CH₃-COOH and CH₂ = CH-COOH
Answer:
CH₂ = CH-COOH is the stronger acid than CH₃-COOH

(2) C₆H₅-COOH and C₆H₅-CH₂-COOH
Answer:
C₆H₅-COOH is the stronger acid than C₆H₅-CH₂-COOH

Answer:

(4) CH₃-CH₂-COOH and NC-CH₂-COOH
Answer:
NC-CH₂-COOH is the stronger acid than CH₃-CH₂-COOH

(5) (CH₂)₂CH-CH₂-COOH and (CH₃)₂NH-CH₂-COOH
Answer:
(CH₃)₂NH-CH₂-COOH is the stronger acid than (CH₃)₂CH-CH₂-COOH

(6) O₂N-CH₂-C00H and Cl-CH₂-COOH
Answer:
NO₂-CH₂-COOH is the stronger acid than Cl-CH₂-COOH.

Question 77.
Arrange the following acids in their increasing order of acidic strength.

(1) Acetic acid, chloroacetic acid, propionic acid, formic acid.
Answer:
Propionic acid < acetic acid < formic acid < chloroacetic acid

(2) Bromoacetic acid, chloroacetic acid, fluoroacetic acid, iodoacetic acid.
Answer:
Iodoacetic acid < bromoacetic acid < chloroacetic acid < fluoroacetic acid.

(3) Acetic acid, chloroacetic acid, dichloroacetic acid, trichloroacetic acid.
Answer:
Acetic acid < chloroacetic acid < dichloroacetic acid < trichloroacetic acid.

(4) n-butyric acid, 3-chlorobutyric acid, 2-chlorobutyric acid, 3-chlorobutyric acid.
Answer:
n-Butyric acid < 3-chlorobutyric acid < 2-chlorobutyric acid < 1-chlorobutyric acid.

(5) Acetic acid, benzoic acid, p-methoxy benzoic acid, p-nitrobenzoic acid.
Answer:
Acetic acid < benzoic acid < p-methoxy benzoic acid < p-nitrobenzoic acid.

(6) Acetic acid, phenyl acetic acid, p-nitro phenyl acetic acid.
Answer:
Acetic acid < phenyl acetic acid < p-nitro phenyl acetic acid.

(7) Benzoic acid, p-toluic acid, p-chlorobenzoic acid.
Answer:
p-toluic acid < benzoic acid < p-chlorobenzoic acid.

Question 78.
Arrange the following carboxylic acids in order of increasing acidity : m-Nitrobenzoic acid, Trichloroacetic acid, benzoic acid, a-Chlorobutyric acid.
Answer:
Acidity in the increasing order : Benzoic acid, m-nitrobenzoic acid, a-chlorobutyric acid, trichloroacetic acid.

Question 79.
Arrange the following carboxylic acids with increasing order of their acidic strength and justify your answer.

Question 80.
Explain polarity of carbonyl group.
Answer:
The polarity of a carbonyl group is duc to higher electronegativity of oxygen compared to carbon. It is explained on the basis of resonance involving neutral and dipolar structures.

The carbonyl carbon has positive polarity (see structures (A) and (D)). Therefore, it is electron deficient. As a result, this carbon atom is electrophilic (electron loving) and is susceptible to attack by a nucleophile (Nu : ^–).

Question 81.
Explain SchifPs reagent test.
OR
What is a SchifPs reagent ? How is it used to detect aldehydes ?
Answer:

• Schiff’s reagent is prepared by dissolving pink p-rosaniline hydrochloride (dye Fuchsin) in water and passing SO₂ gas till the pink solution is decolourised.
• Schiff s reagent is an oxidising agent.
• When an aldehyde is added to Schiff s reagent, the colourless solution turns pink or in magenta colour and aldehyde is oxidised to a carboxylic acid.
• This test is not given by ketones, hence, used to distinguish between aldehyde and ketone.

Question 82.
Which colour is obtained when SchifFs reagent is treated with acetaldehyde?
Answer:
When Schiff s reagent is treated with acetaldehyde, pink colour is obtained.

Question 83.
What is Tollen’s reagent?
Answer:

• Tollen’s reagent is an ammoniacal silver nitrate, [Ag(NH₃)₂]⁺ OH^–.
• It is prepared by adding NH₄OH solution to silver nitrate solution.
  
• It is a stronger oxidising agent than Fehling solution. Aldehyde when heated with Tollen’s reagent, silver mirror is deposited.

Question 84.
Explain Tollen’s reagent test.
OR
Explain silver mirror test.
Answer:

• Tollen’s reagent is an (ammoniacal silver nitrate) [Ag(NH₃)₂]⁺ OH^–.
• When an aldehyde, like acetaldehyde is heated with Tollen’s reagent, it is oxidised to acetic acid and silver ions Ag⁺ in Tollen’s reagent complex are reduced to silver Ag giving greyish black precipitate or deposition of silver on inner surface of the test tube which shines like a mirror. Hence this test is also called silver mirror test.
  
• This test is not given by ketones.
• Hence Tollen’s reagent is used to distinguish between aldehydes and ketones.

Question 85.
What is Fehling’s solution? How is it prepared?
Answer:

• Fehling’s solution is a complex of cupric ions with tartaric acid.
• It is a mild oxidising agent.
• It is prepared by mixing equal amount of Fehling’s solution ‘A’ containing CuSO₄ solution and Fehling’s solution ‘B’ containing sodium potassium tartrate (Rochelle salt) in caustic soda (NaOH) solution.
• It is used to detect aldehydes that decolourise deep blue colour of the solution and give red precipitate of Cu₂O.

Question 86.
Explain Fehling’s solution test.
Answer:

• Fehling’s solution is a mixture of Fehling’s solution ‘A’ containing CuSO₄ solution and Fehling’s solution ‘B’ containing sodium potassium tartrate (Rochella salt) in caustic soda (NaOH) solution.
• When an aldehyde is heated with Fehling’s solution, the deep blue colour of the solution disappears and Cu⁺² (cupric ion) is reduced to Cu⁺ ion a red precipitate of cuprous oxide, Cu₂O is obtained while aldehyde is oxidised to a carboxylate ion.
• For example,
  
• This test is not given by ketones, since they cannot be oxidised by Fehling solution.
• Aromatic aldehydes are not oxidised by Fehling solution.
• Hence this test is used to distinguish between aldehydes and ketones.

Question 87.
What is the action of the following reagents on ethanal :
(1) Fehling’s solution,
(2) Tollen’s reagent or Ammonical silver nitrate ?
Answer:

1. When ethanal is heated with Fehling’s solution, the deep blue colour of the solution disappears and a red precipitate of Cu₂O is obtained.
   
2. When ethanal is heated with Tollen’s reagent a greyish black precipitate or deposition of silver is obtained.
   

Question 88.
Why is benzaldehyde NOT oxidized by Fehling solution ?
Answer:
When benzaldehyde is treated with Fehling solution, it does not reduce cupric ion (Cu⁺²). Fehling solution does not oxidise benzaldehyde. Thus, Fehling test cannot be used for aromatic aldehyde.

Question 89.
Explain laboratory test for ketonic group or sodium nitroprusside test.
Answer:
Laboratory test for ketonic group : Sodium nitroprusside test : When a freshly prepared sodium nitroprusside solution is added to a ketone, mixture is shaken well and basified by adding sodium hydroxide solution drop by drop, red colour appears in the solution, which indicates the presence of ketonic (> C = O) group.

The anion of ketone formed by alkali reacts with nitroprusside ion to form a red coloured complex which indicates the presence of the ketonic group.

Question 90.
What is the action of hydrogen cyanide on the following :
(1) Acetaldehyde
(2) Acetone
(3) Benzaldehyde?
Answer:
(1) Action of HCN on acetaldehyde : When acetaldehyde is treated with hydrogen cyanide, acetaldehyde cyanohydrin is formed.

(2) Action of HCN on acetone : When acetone is treated with hydrogen cyanide, acetone cyanohydrin is formed.

(3) Action of HCN on benzaldehyde : When benzaldehyde is treated with hydrogen cyanide, benzaldehyde cyanohydrin is formed.

Question 91.
What is the action of hydrogen cyanide in basic medium on (1) butanone (2) 2,4-dichlorobenzaldehyde?
Answer:
(1) Action of hydrogen cyanide on butanone : When butanone is treated with hydrogen cyanide, butanone cyanohydrin is obtained.

(2) Action of hydrogen cyanide on 2,4-dichlorobenzaldehyde : When 2, 4-dichloro benzaldehyde is treated with hydrogen cyanide, cyanohydrin of 2,4-dichloro benzaldehyde is obtained.

Question 92.
What is the action of sodium bisulphite on :
(1) Acetaldehyde
(2) Acetone (propanone)?
Answer:
(1) Acetaldehyde reacts with saturated aqueous solution of sodium bisulphite (NaHSO₃) and forms crystalline acetaldehyde sodium bisulphite. It is water soluble crystalline solid.

(2) Acetone reacts with saturated aqueous solution of sodium bisuiphite (NaHSO₃) and forms crystalline acetone sodium bisuiphite. It is water soluble crystalline solid.

Question 93.
A carbonyl compound ‘A’ having molecular formula C₅H₁₀O forms crystalline precipitate with sodium bisulphite and gives positive iodoform test but does not reduce Fehling solution. Write the structure of carbonyl compound.
Answer:
A carbonyl compound C₅H₁₀O has two structures.

Pentan-2-one forms crystalline precipitate with sodium bisulphite and gives positive iodoform test. But does not reduce Fehling solution.

Pentan-3-one does not react with iodine and NaOH because it does not contain group.

Question 94.
How does alcohols react with aldehydes and ketones?
Answer:
Aldehyde reacts with one molecule of anhydrous monohydric alcohol in presence of dry hydrogen chloride to give alkoxyalcohol known as hemiacetal, which further reacts with one more molecule of anhydrous monohydric alcohol to give a geminaldialkoxy compound known as acetal as shown in the reaction.

Ketones react with 1, 2 – or 1, 3 – diols in presence of dry hydrogen chloride to give five or six-membered cyclic ketals.

Question 95.
What is the action of ethanol on acetaldehyde ? What is the action of ethylene glycol on acetone ?
Answer:
Acetaldehyde reacts with one equivalent of monohydric alcohol in the presence of dry hydrogen chloride to form an intermediate known as hemiacetal, which further adds another molecule of alcohol to form a gem-dialkoxy compound known as acetal. Acetone reacts with ethylene glycol under similar conditions to form cyclic products known as ethylene glycol ketals.

Question 96.
Write the structure of product in the following reactions:

Answer:

Answer:

Question 97.
How does Grignard reagent react with the carbonyl compounds (or aldehydes and ketones)?
Answer:
The carbonyl compounds like aldehydes and ketones react with Grignard reagent (R – Mg – X) in dry ether and form a complex which on further hydrolysis with acid forms the corresponding alcohol.

Question 98.
What is the action of Grignard reagent, CH₃ – Mg – I on : (1) formaldehyde (2) acetone?
Answer:
(1) Grignard reagent with formaldehyde gives a primary alcohol.
Formaldehyde on reaction with Grignard reagent, CH₃ – Mg – I in dry ether forms a complex which on hydrolysis with dilute HCl forms ethyl alcohol.

(2) Acetone on reaction with Grignard reagent, CH₃ – Mg – I in dry ether forms a complex which on hydrolysis with dilute HC1 forms tert-butyl alcohol.

Question 99.
Explain the mechanism of addition reactions of ammonia derivatives H₂N-Z with carbonyl compounds (aldehydes or ketones).
Answer:
Derivatives of ammonia H₂N-Z reacts with carbonyl compounds (aldehydes or ketones) in weakly acidic medium to give addition products, which loses a water molecule to give a final product imine derivatives. A substituted imine is called a Schiff base. Schiff bases are solids and have sharp melting points.

General reaction :

Question 100.
What Is the action of ethylamine on :
(1) acetaldehyde
(2) acetone ?
Answer:
(1) Acetaldehyde on reaction with ethyl amine forms imine (Schiff base).

(2) Acetone on reaction with ethyl amine forms imine (Schiff base).

Question 101.
What are oximes? Which functional group do they contain?
Answer:
Oximes : These are the compounds obtained by the reactions of carbonyl compounds namely aldehydes and ketones with hydroxyl amine NH₂OH.

Question 102.
What is the action of hydroxyl amine (NH₂OH) on (1) acetaldehyde (2) acetone?
Answer:
(1) Acetaldehyde on reaction with hydroxyl amine (in weakly acidic medium) forms crystalline acetaldoxime.

(2) Acetone on reaction with hydroxyl amine (in weakly acidic medium) forms crystalline acetoxime.

Question 103.
What are hydrazones?
Answer:
Carbonyl compounds like aldehydes and ketones react with hydrazine forming compounds like hydrazones. For example, acetaldehyde on reaction with hydrazine gives acetaldehyde hydrazone.

Question 104.
Which compound can convert
Answer:
The compound which can convert.

Question 105.
What is the action of hydrazine on (1) formaldehyde (2) acetone ?
Answer:
(1) Formaldehyde on reaction with hydrazine forms formaldehyde hydrazone.

(2) Acetone with hydrazine forms acetone hydrazone.

Question 106.
What are phenylhydrazones?
Answer:
The carbonyl compounds like aldehydes and ketones on reaction with phenylhydrazine form hydrazones. For example, acetaldehyde on reaction with phenylhydrazine forms acetaldehydephenylhydrazone.

Question 107.
What is the action of phenylhydrazine on (1) formaldehyde (2) acetone (propanone) ?
Answer:
(1) Formaldehyde on reaction with phenylhydrazine forms formaldehydephenylhydrazone.

(2) Acetone on reaction with phenylhydrazine forms acetone phenylhydrazone.

Question 108.
What is the action of 2,4-Dinitrophenylhydrazine on
(1) Acetaldehyde
(2) Acetone
(3) Butanone
(4) Benzaldehyde ?
OR
Complete and rewrite the balanced chemical equation : Butanone + 2, 4-Dinitrophenylhydrazine.
Answer:
(1) Acetaldehyde on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

(2) Acetone on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

(3) Butanone on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

Question 109.
What is the action of semi carbazide on (1) Acetaldehyde (2) Acetone?
Ans.
(1) Acetaldehyde on reaction with semicarbazide forms scrnicarbazone derivative.

(2) Acetone on reaction with sernicarbazide forms semicarbazone derivative.

Question 120.
Write the structures of carbonyl compounds and ammonia derivatives that combine to give following imines.
Answer:

Question 121.
Write the structure of the products obtained from the following ketones by action of hydrazine in presence of (1) slightly acidic medium (2) strong base KOH.

Answer:
(1) In slightly acidic medium

(2) In the presence of a strong base KOH

Question 122.
Explain haloform reaction.
Answer:
A ketone containing -COCH₃ group is oxidised by sodium hypohalite a mixture of (sodium hydroxide and halogen) results in the formation of sodium salt of carboxylic acid having one carbon atom less than that of ketone and methyl group is converted to haloform.

Acetaldehyde is the only aldehyde which gives haloform reaction. In this reaction, R may be hydrogen, methyl group or aryl group and X may be Cl, Br or I. The reaction is given by all methyl ketones (CH₃ – CO – R) and all alcohols containing CH₃ – (CHOH) group.

When a methyl ketone is warmed with iodine and sodium hydroxide, a yellow precipitate of iodoform is obtained. The iodoform reaction is used as a qualitative test for detection of CH₃CO-group in a organic compound.

Question 128.
Identify the compounds, amongst the following, that give positive iodoform test.
(CH₃)₂CHOH, (CH₃)₃COH, CH₃COCH₂CH₂CH₃, CH₃CH₂CHO, CH₃CH₂CH(OH)CH₂CH₃, CH₃CH₂OH, C₆H₅COCH₂CH₃, CH₃CHO, C₆H₅CH₂CH₂OH and CH₃CH(OH)CH₂CH₂CH₃.
Answer:
For an iodoform test, the carbonyl compound must have  group.
The compounds that give positive iodoform test :

Question 129.
Explain cross aldol condensation.
Answer:
(1) An aldol condensation between two different carbonyl compounds (aldehydes and or ketones) takes place even though one of the two carbonyl compounds molecules does not contain a-hydrogen atom e.g. HCHO and C₆H₅CHO.

(2) If both aldehydes or ketones contain two a-hydrogen atoms each, then a mixture of four products, is formed. For example, a mixture of ethanal and propanal on reaction with dilute alkali followed by heating gives a mixture of four products.

Self aldol condensation:

Cross aldol condensation:

Question 130.
Write the structure of the major product of the following crossed aldol condensation.
Answer:
(1) Formaldehyde and propionaldehyde :

(2) Benzaldehyde with acetone:

Question 131.
Explain aldol condensation reaction of propionaldehyde.
Answer:
Since propionaldehyde has an a-hydrogen atom it undergoes aldol condensation with alkali Ba(OH)₂, forming 3-Hydroxy-2-methylpentanal.

3-Hydroxy-2-methylpentanal on heating undergoes dehydration and forms 2-Methylpent-2-enal.

Question 132.
If a mixture of formaldehyde and acetaldehyde is subjected to aldol condensation, predict the products formed and draw their structures.
Answer:
Since formaldehyde, does not have α-hydrogen atom it will not undergo self aldol condensation. Since acetaldehydehas a-hydrogen atom, it will undergo self aldol condensation.

Formaldehyde and acetaldehyde undergo cross aldol condensation.

Hence two aldol condensation products will be obtained.

Question 133.
Indicate by equations, what happens when a mixture of acetaldehyde and acetone are treated with alkali.
Answer:
When a mixture of acetaldehyde and acetone is treated with alkali, Ba(OH)₂, they undergo self aldol condensation and cross aldol condensation.
(1) Self aldol condensation acetaldehyde :

(2) Self aldol condensation of acetone:

(3) Cro5s aldol condensation:

Question 134.
Explain Cannizzaro reaction.
OR
Write a note on Cannizzaro reaction.
OR
Write a note on self oxidation-reduction reaction of aldehyde with suitable example.
Answer:

• Aldehydes which do not have a-hydrogen atom, on heating with concentrated alkali (50% aqueous or ethanolic solution of NaOH or KOH) undergo self oxidation and reduction reaction or redox reaction.
• This self redox reaction or disproportionation reaction is called Cannizzaro reaction.
• In this reaction one molecule of the aldehyde is oxidised to carboxylic acid while the second molecule of the aldehyde is reduced to alcohol (carboxylic acid formed, reacts with alkali, NaOH and forms a salt R – COONa).
• When formaldehyde (methanal) is heated with 50% NaOH solution, methanol (reduction product) and sodium formate (oxidation product) are formed.
  
• Ketones and aldehydes like acetaldehyde, propionaldehyde, etc. having a – H atom do not give Cannizzaro reaction.

Question 135.
Explain cross Cannizzaro reaction with example.
OR
Write the reaction for the action of 50 % NaOH on a mixture of formaldehyde and benzaldehyde.
Answer:
The reaction between two different aldehydes, not having a-hydrogen atoms is called cross Cannizzaro reaction. These two aldehydes undergo disproportionation in presence of concentrated alkali to give four products. However, if one of the aldehydes is formaldehyde, the reaction yields exclusively formate and alcohol to corresponding aldehyde.

Formaldehyde and benzaldehyde since do not have a-hydrogen atom, will undergo Cannizzaro (redox) reactions.

(1) Self Cannizzaro (redox) reaction :

(2) Cmss Cannizzaro (redox) reaction:

Question 136.
What is the action of cone, potassium hydroxide on benzaldehyde?
Answer:
When benzaldehyde is heated with concentrated potassium hydroxide in presence of methanol, a mixture of potassium benzoate and phenyl methanol (benzyl alcohol) is obtained.

Question 137.
Differentiate between Cannizzaro reaction and Aldol reaction.
Answer:

Cannizzaro reaction	Aldol reaction
1. It is given by aldehydes not having alpha hydrogen atom. 2.  In this reaction an aldehyde is converted to the corres­ponding acid and an alcohol. 3.  It is a disproportionate ion reaction. 4.  It requires concentrated alkali as a catalyst.	1. It is given by aldehydes and ketones possessing alpha hydrogen atom. 2.  In this reaction aldehydes and ketones are converted into aldol and ketols, respectively. 3.  It is an addition reaction. 4.  It requires dilute alkali as a catalyst.

Question 138.
Write the chemical equations for aldol condensation or Cannizzaro reaction that the following compounds undergo :
(1) Propanal
(2) 2-Methyl propanal (isobutyraldehyde)
(3) Pentanal
(4) 3-Methylbutanal
(5) Acetophenone
(6) p-Methoxybenzaldehyde
(7) 2-Methyl cyclohexanone
(8) Chloral
(9) Cyclopentanone
(10) Phenyl acetaldehyde
(11) 1-Phenyl propan-l-one.
Answer:
(1) Propanal (Aldol condensation) : Propanal contains α-H atom. Two molecules of propanal undergo self condensation in presence of dil. alkali to form 3-Hydroxy-2-methyl pentanal.

(2) 2-NIethvl propanal (Canniuaro reaction) : Two molecules of them undergo cannizzaro reaction in the presence of 50% alkali to form sodium isobutyrate and isohutyl alcohol.

(3) Pentanal (Aldol condensation) : Pentanal contains a-H atom. Two molecules of them undergo self condensation in the presence of dil. alkali to form 3-Hydroxy-2-propyl heptanal.

(4) 3-Methyl butanal (Aldol condensation) : 3-Methyl butanal contains a-H atom. Two molecules of them undergo self condensation in the presence of dil. alkali to form 3-Hydroxy-2-isopropyl-5-methyl hexanol.

(5) Acetophenone (Aldol condensation) : Acetophenone contains a-H atom. Two molecules of them undergo self condensation in the presence of base to form 3-Hydroxy-1, 3-diphenyl but-l-one.

(6) p-Methoxybenzaldehyde (Cannizzaro reaction) :

(7) 2-Methyl cyclohexanone

(8) Chloral (Cannizzaro’s reaction) : There is no α-H atom CCl₃CHO, therefore it undergoes Cannizzaro’s reaction.

(9) Cyclopentanone (Aldol condensation) : Cyclopentanone contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 2-(l-Hydroxy-1 cyclopentyl) cyclo pentane-l-one.

(10) Phenyl acetaldehyde (Aldol condensation) : Phenyl acetaldehyde contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 3-Hydroxy-2, 4-diphenylbutanal.

(11) 1-phenyl propan-l-one (Aldol condensation) : 1-phenyl propan-l-one contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 3-Hydroxy-2-methyl-l, 3-diphenyl pentan-l-one
3-Hydroy-2-methyl-1, 3-diphenyl pentan-l-one

Question 139.
Write a note on Clemmensen reduction.
OR
Explain Clemmeusen’s reduction.
OR
Explain the reduction of carbonyl group into methylene group.
Answer:

• The carbonyl groupon reduciion with zinc amalgam (Zn – Hg) in concentrated hydrochloric acid is converted into methylene group ( – CH₂ -).
  
• Aldehydes and kctoncs on reaction with Zn – Hg in concentrated HCl forms corresponding alkanes. ibis reduction is called Clemmensen reduction.
  
• Acetaldehyde on reduction with Zn – Hg in concentrated HCl forms ethane.
  
• Acetone on reduction with Zn – Hg in concentrated HCl forms propane.
  

Question 140.
Explain Wolff-Kishner reduction.
Answer:
Hydrazine (NH₂-NH₂) reduces carbonyl groupof aldehydes or ketones to metylene group  When aldehyde or ketone is heated with hydrazine in the presence of base such as potassium hydroxide and ethylene glycol, an alkane is obtained due to reduction of carbonyl compound.

Question 141.
Compound A (C₅H₁₀O) form a phenyl hydrazone and gives a negative Tollen’s reagent test and iodoform test. On reduction with Zn-Hg/HCl, compound A gives n-pentane. Write the structure of ‘A’.
Answer:
Since A (C₅H₁₀O) forms a phenyl hydrozone, it is a carbonyl compound. Since it gives negative Tollen’s reagent test, it is not an aldehyde but it must be a ketone.

Since it doesn’t give iodoform test, it doesn’t have group.
Hence the structure of ‘A’ will be,

Question 142.
Identify A and B in the following reactions :

Answer:

 

Answer:

 

Answer:

 

Answer:

Question 143.
What is the action of concentrated nitric acid on (1) Benzaldehyde (2) Benzophenone?
Answer:

1. Benzaldehyde on reaction with concentrated nitric acid in presence of cone. H₂SO₄ forms m-nitrobenzaldehyde
   
2. Benzophenone on reaction with concentrated nitric acid in presence of cone. H₂SO₄ forms m-nitrobenzophenone
   

Question 144.
Explain laboratory tests for carboxyl (- COOH) group.
Answer:
The presence of – COOH group in carboxylic acids is identified by the following tests :

(1) Litmus test : (valid for water soluble substances)
Aqueous solution of Organic compound containing – COOH group turns blue litmus red which indicates the presence of acidic functional group.

(2) Sodium bicarbonate test : When sodium bicarbonate is added to an organic compound containing – COOH group, a brisk effervescence of carbon dioxide gas is evolved. Water insoluble acid goes in solution and gives precipitate an acidification with cone. HCl. This indicates the presence of -COOH group.

(3) Ester test : One drop of concentrated sulfuric acid is added to a mixture of given organic compound containing – COOH group and one mL of ethanol, the reaction mixture is heated for 5 minutes in hot water bath. After this, hot solution is poured in a beaker containing water, fruity smell of ester confirms the presence of carboxylic acid.

Question 145.
How is acid chloride obtained from carboxylic acid?
Answer:
Carboxylic acid on heating with thionyl chloride (SOCl₂), phosphorus trichloride (PCl₃) or phosphorus pentachloride (PCl₅) give corresponding acid chlorides. In this reaction – OH of carboxyl group is replaced by -Cl.

The reactions are :
(1) Action on SOCl₂ on carboxylic acid :

Example : Acetic acid reacts with thionyl chloride to give acetyl chloride.

(2) Action of PCl₃ on carboxylic acid (ethanoic acid) :

Example : Action of phosphorus trichloride on acetic acid gives acetyl chloride.

(3) Action of PCI₅ on carboxIic acid :

Question 146.
How will you convert 3,5-Dinitrobenzoic acid to 3,5-Dinitrobenzoyl chloride?
Answer:
When 3,5-Dinitrobenzoic acid is heated with phosphorus pentachloride, 3,5-Dinitrobenzoyl chloride is obtained.

Question 147.
How is acid amide obtained from carboxylic acid?
Answer:
Carboxylic acid or acid chloride with ammonia salts, which on further strong heating at high temperature decompose to give amides.

When acetic acid is treated with ammonia, ammonium acetate is obtained. Ammonium acetate on strong heating decomposes to form acetamide.

When acetyl chloride is treated with ammonia, acetamide is obtained

Question 148.
How is acid anhydride obtained from carboxylic acid?
Answer:
When carboxylic acid is heated with strong dehydrating agent like phosphorus pentoxide or concentrated sulphuric acid, an acid anhydride is obtained,

The reaction is reversible and anhydride is hydrolysed back to acid.

Alternatively, when sodium acetate is heated with acetyl chloride, acetic anhydride is obtained. This reaction is irreversible.

Question 149.
What is decarboxylation of an acid ? How is it done?
OR
What happens when sodium acetate is heated with soda lime ?
Answer:
Removal of a carboxylic group from acid is called decarboxylation. Decarboxylation of an acid is carried out by heating anhydrous sodium salts of carboxylic acids with soda lime (NaOH + CaO). The product hydrocarbons obtained contain one carbon atom less than the carboxylic acid.

When sodium acetate is heated with soda lime, methane is obtained.

Question 150.
How is alcohol obtained from carboxylic acid ?
OR
What is the action of LiAlH₄/H₃O⁺on ethanoic acid? Write balanced equation for the conversion :
Cyclopropane carboxylic acid to Cyclopropylmethanol.
Answer:
Carboxylic acids are reduced to primary alcohols using powerful reducing agent lithium aluminium hydride.

(i) When ethanoic acid is reduced in the presence of LiAlH4 in dry ether, forms ethanol.

(ii) When cyclopropane carboxylic acid is reduced in the presence of lithium aluminium hydride in dry ether, forms cyclopropyl methanol.

Question 151.
What is the action 9f following compounds on cyclohexanone in presence of dry hydrogen chloride?
(1) Ethyl alcohol
(2) Ethylene glycol
Answer:
(1) With Ethyl alcohol : Cyclohcxanonc reacts with one equivalent of monohydric ethyl alcohol lo form hemi ketal, which further adds another molecule of alcohol to form a gem-dialkox compound known as ketal.

(2) With Ethylene glycol : cyclohexanone reaCts with ethylene glycol to form cyclic ketal.

Question 152.
Answer the following in one sentence.

1. Name the compound which reacts with formaldehyde to produce ethyl alcohol.
Answer:
The compound which reacts with formaldehyde to produce ethyl alcohol is methyl magnesium iodide.

2. What are imines ?
Answer:
These are the compounds obtained by the reactions of carbonyl compounds namely aldehydes and ketones with primary amine.

3. Why does skin have burning sensation, when an ant bites ?
Answer:
When an ant bites, formic acid is released from an ant which gives burning sensation as the acid comes in contact with the skin.

4. What is the percentage of acetic acid in vinegar?
Answer:
The percentage of acetic acid in vinegar is 6 to 8%.

5. Which reagent is used to distinguish formic acid and acetic acid?
Answer:
The reagent used to distinguish formic acid and acetic acid is ammoniacal silver nitrate.

6. What happens when acetyl chloride is treated with dibenzyl cadmium ? Give reaction.
Answer:

7. Complete the following reaction :

Answer:

8. Give reason : Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.
Answer:
Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the catalyst aluminium chloride (Lewis acid) gets bonded to carboxyl group.

9. Write the name of two compounds which do not contain carbonyl group but show iodoform test.
Answer:
The name of two compounds which do not contain carbonyl group but show iodoform test are ethanol

10. Give reason : In semicarbazide, – NH₂ group bonded to carbonyl group is not involved in the formation of semicarbazone.
Answer:
NH₂ group attached to – NH group in semicarbonide is more active than NH2 group attached to carbonyl group due to electron density difference.

11. Fehling solution does not oxidise benzaldehyde but Tollen’s reagent oxidises benzaldehyde. Give reason.
Answer:
When benzaldehyde is heated with Fehling solution, there is no change in colour of the solution, Cu²⁺ ion is not reduced, hence Fehling solution does not oxidise benzaldehyde. However, Tollen’s reagent oxidises benzaldehyde to give silver mirror test.

12. Give reason : Direct attachment of vinyl group to carboxylic group increases the acidity of corresponding acids.
Answer:
Direct attachment of vinyl group to carboxylic group increases the acidity of corresponding acids due to greater electronegativity of sp²-hybridised carbon to which carboxyl carbon is attached.

Multiple Choice Questions

Question 153.
Select and write the most appropriate answer from the given alternatives for each sub-question :

1. IUPAC name of

(a) l-Phenylhexan-2-one
(b) 6-Phenylhexan-5-one
(c) l-Benzylhexan-2-one
(d) Dodecan-5-one
Answer:
(a) l-Phenylhexan-2-one

2. The general formula of carbonyl compounds is
(a) C_nH_2n+1OH
(b) C_nH_2nO
(c) C_nH_2nO₂
(d) C_nH_2n+1O
Answer:
(b) C_nH_2nO

3. Aldehydes and ketones are
(a) chain isomers
(b) functional isomers
(c) geometrical isomers
(d) position isomers
Answer:
(b) functional isomers

4. Identify ‘C’ in the following reaction,
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \stackrel{\mathrm{KCN}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{H}^{+} \mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{C}\)
(a) Propan – 1 – ol
(b) Propanone
(c) 2-Ethyl-3-pentanone
(d) Propanal
Answer:
(d) Propanal

5. Grignard reagent when reacted with alkyl cyanide followed by hydrolysis gives
(a) an aldehyde
(b) a ketone
(c) a primary alcohol
(d) a secondary alcohol
Answer:
(b) a ketone

6. Identify ‘ B ’ in the following reaction :
\(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{N} \stackrel{\mathrm{CH}_{3} \mathrm{MgI}}{\longrightarrow} \mathrm{A} \underset{\mathrm{HCl}}{\stackrel{2 \mathrm{HOH}}{\longrightarrow}} \mathrm{B}+\mathrm{NH}_{3}+\mathrm{MgIOH}\)
(a) Magnesium intermediate
(b) Ethanol
(c) Propanal
(d) Propanone
Answer:
(d) Propanone

7. \(\mathrm{A}+\mathrm{B} \stackrel{\text { dry ether }}{\longrightarrow} \mathrm{Col}\) Complex \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+}\) Ethyl Methyl Ketone. In the above reaction, A and B are
(a) Formonitrile, Propyl magnesium bromide
(b) Ethyl cyanide, Ethyl magnesium bromide
(c) Hydrogen cyanide, Ethyl magnesium bromide
(d) Acetonitrile, Ethyl magnesium bromide
Answer:
(d) Acetonitrile, Ethyl magnesium bromide

8. A dilute solution of p-rosaniline hydrochloride in water whose pink colour has been discharged by passing sulphur dioxide, does not restore its colour by
(a) HCHO
(b) CH₂CHO
(c) (CH₃)₂COCH₃
(d) CCl₃CHO
Answer:
(c) (CH₃)₂COCH₃

9. The reagent with which both acetaldehyde and acetone reacts easily is –
(a) Fehling’s solution
(b) Tollen’s reagent
(c) Grignard reagent
(d) Schiff s reagent
Answer:
(c) Grignard reagent

10. Isopropyl methyl ketone when treated with Zn-Hg and concentrated hydrochloric acid give
(a) iso-butane
(b) iso-pentane
(c) n-pentane
(d) neo-pentane
Answer:
(b) iso-pentane

11. The formation of acetone cyanohydrin from acetone is an example of
(a) Nucleophilic addition
(b) Nucleophilic substitution
(c) Electrophilic addition
(d) Electrophilic substitution
Answer:
(a) Nucleophilic addition

12. Which of the following is Fehling solution ‘A’?
(a) CuSO₄ solution
(b) CaSO₄ solution
(c) NaOH solution
(d) Sodium potassium tartarate solution
Answer:
(a) CuSO₄ solution

13. The compound ‘X’ upon alkaline hydrolysis gives a product which reacts with phenylhydrazine but does not reduce ammoniacal silver nitrate solution. A possible structure for ‘X’ is
(a) CH₃CHCl CH₂Cl
(b) CH₃CCl₂CH₃
(c) CH₃CH₂CH₂Cl
(d) CH₃CH₂CHCl₂
Answer:
(b) CH₃CCl₂CH₃

14. Which of the following is the correct statement with respect to aldehyde and ketones ?
(a) Ketones are reducing agents
(b) Aldehydes are good reducing agents
(c) Cannizzaro reaction is an addition reaction
(d) Ketones do not react with Grignard reagent
Answer:
(b) Aldehydes are good reducing agents

15. Acetaldehyde acts as
(a) a catalyst
(b) a reducing agent
(c) an oxidizing agent
(d) a mordant
Answer:
(b) a reducing agent

16. An organic compound (A) C₃H₈0 on oxidation gives (B) C₃H₆O₂. The compound A may be
(a) an ester
(b) an alcohol
(c) an aldehyde
(d) a ketone
Answer:
(b) an alcohol

17. Identify ‘B’ from the following reaction :
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{NH}_{2} \mathrm{OH} \rightarrow \mathrm{A} \stackrel{\mathrm{Na} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\Delta} \mathrm{B}\)
(a) Propan-1-amine
(b) Propan-2-amine
(c) Isopropylamine
(d) Dimethylamine
Answer:
(a) Propan-1-amine

18. Sodium borohydride does not reduce
(a) – COOH group
(b) – NO₂ group
(c) – X atom
(d) – CHO group
Answer:
(a) – COOH group

19. An aldehyde when warmed with Zn/Hg and cone. HCl gives
(a) alcohol
(b) hydrocarbon
(c) carboxylic acid
(d) ketone
Answer:
(b) hydrocarbon

20. Acetaldol is
(a) 3-hydroxy butanol
(b) 3-hydroxy butanal
(c) 2-hydroxy propanal
(d) 3-hydroxy pentanal
Answer:
(b) 3-hydroxy butanal

21. Acetone can be reduced to propane, the reduction is called
(a) Clemmensen’s reduction
(b) catalytic reduction
(c) Rosenmund’s reduction
(d) partial reduction
Answer:
(a) Clemmensen’s reduction

22. Which of the following reagents can react with acetaldehyde to give water soluble white crystal-line solid?
(a) NaHSO₄
(b) NaHSO₃
(c) Na₂SO₃
(d) Na₂SO₄
Answer:
(b) NaHSO₃

23. Which of the following compounds does NOT undergo aldol condensation?

Answer:
(a)

24. Formalin is 40% aqueous solution of :
(a) Methanal
(b) Methanoic acid
(c) Methanol
(d) Methanamine
Answer:
(a) Methanal

25. Which of the following compounds do not produce pink colour with Schiff s reagent?
(a) Formaldehyde
(b) 2-propanone
(c) 3-pentanone
(d) Both (b) and (c)
Answer:
(d) Both (b) and (c)

26. Both aldehydes and ketones can react with
(a) Tollen’s reagent
(b) the Grignard reagent
(c) Fehling’s solution
(d) Schiffs reagent
Answer:
(b) the Grignard reagent

27. Aldol reaction is.
(a) an addition reaction
(b) an elimination reaction
(c) a self-reduction reaction
(d) a disproportionate ion reaction
Answer:
(a) an addition reaction

28. The reaction in which two molecules combine to form a new molecule with the elimination of a small molecule like water is called
(a) an oxidation reaction
(b) a condensation reaction
(c) a hydrolysis reaction
(d) a redox reaction
Answer:
(b) a condensation reaction

29. Benzaldehyde undergoes Cannizzaro’s reaction to give
(a) sodium benzoate and methyl alcohol
(b) sodium benzoate and benzyl alcohol
(c) benzyllic acid and benzyl alcohol
(d) phenol and benzoic acid
Answer:
(b) sodium benzoate and benzyl alcohol

30. Identify ‘X’ in the following reaction :
CH₃-CHO + X → CH₃-CH = N-NH-C₆H₅ + H₂O
(a) C₆H₅-NH₂
(b) C₆H₅-NH-NH₂
(c) C₆H₅-N = NH
(d) C₆H₅-NH-NH-CH₃
Answer:
(b) C₆H₅-NH-NH₂

31. What happens when propanal is treated with zinc amalgam and conc.HCl ?
(a) Propan-l-ol
(b) Propan-2-ol
(c) Propane
(d) Propanone
Answer:
(c) Propane

32. Identify ‘ B ’ in the following reaction :
\(2 \mathrm{CH}_{3}-\mathrm{CHO} \frac{\text { dil. base or acid }}{300 \mathrm{~K}} \mathrm{~A} \underset{\text { dehydration }}{\text { dehy }} \mathrm{B}+\mathrm{H}_{2} \mathrm{O}\)
(a) CH₃-CH(OH)-CH₂-CHO
(b) CH₃-CH₂-CH(OH)-CHO
(c) CH₃-CH = CH-CHO
(d) CH₃-CO-CH₃
Answer:
(c) CH₃-CH = CH-CHO

33. The blue colour of Fehling’s solution is due to
(a) Cu₂O
(b) CuCO₃
(c) CuO
(d) Cu⁺⁺ ions
Answer:
(d) Cu⁺⁺ ions

34. How is Schiff s reagent prepared?
(a) By passing CO₂ through p-rosaniline solution
(b) By passing NO₂ through p-rosaniline solution
(c) By passing SO₂ through p-rosaniline solution
(d) By passing NH₃ through silver nitrate solution
Answer:
(c) By passing SO₂ through p-rosaniline solution

35. Benzaldehyde when treated with cone. HNO₃ gives
(a) o-nitrobenzaldehyde
(b) p-nitrobenzaldehyde
(c) m-nitrobenzaldehyde
(d) a mixture of -o and -p-nitrobenzaldehyde
Answer:
(c) m-nitrobenzaldehyde

36. Which of the following carbonyl compounds undergoes aldol condensation ?
(a) Benzaldehyde
(b) Benzophenone
(c) Acetophenone
(d) tert-Butyl phenyl ketone
Answer:
(c) Acetophenone

37. Which of the following carbonyl compounds undergoes self redox reaction in presence of concentrated base ?
(a) 3-Methylpentanal
(b) 2-Chlorobutanal
(c) 2,2-Dimethylpropanal
(d) tert-Butyl methyl ketone
Answer:
(c) 2,2-Dimethylpropanal

38. The smell of bitter almond is given by the compound.
(a) Benzoic acid
(b) Benzaldehyde
(c) Vanillin
(d) Cinnamaldehyde
Answer:
(b) Benzaldehyde

39. Which of the following will not give yellow precipitate when treated with NaOH and H?
(a) 3-Methylbutan-2-one
(b) 2-methylpentan-3-one
(c) Propanone
(d) Hexan-2-one
Answer:
(b) 2-methylpentan-3-one

40. A β-hydroxyl carbonyl compound is obtained by the action of NaOH on
(a) HCHO
(b) C6H₅CHO
(c) CR₃CHO
(d) CH₃CHO
Answer:
(d) CH₃CHO

41. Decarboxylation of sodium propionate gives
(a) methane
(b) ethane
(c) propane
(d) ethene
Answer:
(b) ethane

42. Ester on hydrolysis with dil HCl gives
(a) RCOOH + ROH
(b) RCOR + ROH
(c) ROH + ROH
(d) RCOR + RCOOH
Answer:
(a) RCOOH + ROH

43. \(\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{3} \stackrel{\mathrm{CrO}_{3}}{\longrightarrow} \mathrm{A}\) The compound A is
(a) acetic acid
(b) propionic acid
(c) formic acid
(d) benzoic acid
Answer:
(a) acetic acid

44. The reaction of C₆H₅CH = CHCHO with LiAlH₄ gives
(a) C₆H₅CH₂CH₂CH₂OH
(b) C₆H₅CH₂CH₂CHO
(c) C₆H₅CH = CHCH₂OH
(d) C₆H₅CH₂CHOHCH₃
Answer:
(a) C₆H₅CH₂CH₂CH₂OH

45. A mixture of sodium benzoate and sodalime on heating yields
(a) methane
(b) benzene
(c) sodium benzoate
(d) calcium benzoate
Answer:
(b) benzene

46. Which is the strongest acid?
(a) CH₃COOH
(b) CH₃CH₂COOH
(c) (CH₃)₃CCOOH
(d) CICH₂COOH
Answer:
(d) CICH₂COOH

47. Benzaldehyde when treated with alkaline KMnO₄ yields
(a) Benzyl alcohol
(b) Benzoic acid
(c) CO₂ and H₂O
(d) Salicylic acid
Answer:
(a) Benzyl alcohol

48. Acetonitrile on acidic hydrolysis gives
(a) HCOOH
(b) CH₃NC
(c) CH₃COONa
(d) CH₃COOH
Answer:
(d) CH₃COOH

49. The organic compounds A and B reacts with sodium metal and liberates hydrogen gas. A and B reacts together to give ethyl acetate. The A and B are
(a) CH₃COOH & C₂H₅OH
(b) HCOOH & C₂H₅OH
(c) CH₃COOH & HCOOH
(d) CH₃COOH & CH₃OH
Answer:
(a) CH₃COOH & C₂H₅OH

50. Which one of the following undergoes reaction with 50% NaOH to give to corresponding alcohol and acid?
(a) Phenol
(b) Benzoic acid
(c) Benzaldehyde
(d) Butanal
Answer:
(c) Benzaldehyde

51. Identify the reactant in the following reaction.

Answer:
(c) CO

52. The strongest acid is

Answer:
(c)

53. Predict the product in the following reaction

The compound A is
(a) butane
(b) propane
(c) ethane
(d) propene
Answer:
(b) propane

54. Ethyl benzoate when heated with dil H₂SO₄ gives
(a) acetic acid
(b) benzoic acid
(c) ethanoic acid
(d) phenyl methanol
Answer:
(b) benzoic acid

55. Margarine contains
(a) acetaldehyde
(b) propionaldehyde
(c) butyraldehyde
(d) formaldehyde
Answer:
(c) butyraldehyde

56. Monocarboxylic acids have the general formula
(a) C_nH_2n+1O₂
(b) C_nH_2nO₂
(c) C_nH_2nO
(d) C_nH_2n-1O₂
Answer:
(b) C_nH_2nO₂

57. Formic acid is obtained from
(a) vinegar
(b) red ants
(c) butter
(d) oil
Answer:
(b) red ants

58. Butter contains
(a) lactic acid
(b) butyric acid
(c) citric acid
(d) acetic acid
Answer:
(b) butyric acid

59. Glacial acetic acid is
(a) HCOOH
(b) CH₃COOH
(c) CH₃CH₂COOH
(d) C₃H₇COOH
Answer:
(b) CH₃COOH

60. Which of the following acids is optically active?
(a) Oxalic acid
(b) Salicylic acid
(c) Acetic acid
(d) Lactic acid
Answer:
(d) Lactic acid

61. Lactic acid is
(a) propionic acid
(b) α-hydroxy propionic acid
(c) p-hydroxy benzoic acid
(d) butyric acid
Answer:
(b) α-hydroxy propionic acid

62. The carbon atom of the carboxylic group is
(a) sp³-hybridized
(b) sp²-hybridized
(c) sp-hybridized
(d) unhybridized
Answer:
(b) sp²-hybridized

63. The common name of carboxylic fatty acids is derived from
(a) the name of parent alkanes
(b) the name of corresponding aldehydes
(c) from their original sources
(d) the name of alkyl group present in them
Answer:
(c) from their original sources

64. The IUPAC name of a-methylpropionic acid is
(a) Propanoic acid
(b) Butanoic acid
(c) 2-Methylpropanoic acid
(d) 2-Methylbutanoic acid
Answer:
(c) 2-Methylpropanoic acid

65. For the nomenclature of carboxylic acids, the suffix used is
(a) -ane
(b) -oic
(c) -al
(d) -ol
Answer:
(b) -oic

66. Propionic acid can be prepared by the
(a) action of propyl magnesium chloride on dry ice
(b) alkaline hydrolysis of propyl cyanide
(c) acid hydrolysis of ethyl cyanide
(d) oxidation of Propanone
Answer:
(c) acid hydrolysis of ethyl cyanide

67. The intermediate compound formed during hy-drolysis of acetonitrile to acetic acid is
(a) acetone
(b) acetamide
(c) ammonium acetate
(d) ethyl ammonium chloride
Answer:
(b) acetamide

68. Carbonation of CH₃MgI gives organic compound. The same compound can also be obtained by
(a) oxidation of Methanol
(b) oxidation of Methanal
(c) acid hydrolysis of acetonitrile
(d) alkaline hydrolysis of ethyl cyanide
Answer:
(c) acid hydrolysis of acetonitrile

69. The acid that cannot be prepared by the action of Grignard reagent on dry ice is
(a) methanoic acid
(b) ethanoic acid
(c) propanoic acid
(d) butanoic acid
Answer:
(a) methanoic acid

70. The compound which on acid hydrolysis followed by oxidation gives acetic acid is
(a) CH₃I
(b) CH₂Cl₂
(c) ClCH₂CH₂C1
(d) CH₃CHCl₂
Answer:
(d) CH₃CHCl₂

71. The hydrolysis product of alkyl cyanide is
(a) primary amine
(b) amides
(c) aldehyde
(d) carboxylic acid
Answer:
(d) carboxylic acid

72. To prepare acetic acid,
(a) methyl alcohol is oxidized with KMnO₄
(b) calcium acetate is distilled with calcium for-mate under dry conditions
(c) acetaldehyde is oxidized in the presence of K₂Cr₂O₇ and dil. H₂SO₄
(d) glycerol is heated with H₂SO₄
Answer:
(c) acetaldehyde is oxidized in the presence of K₂Cr₂O₇ and dil. H₂SO₄

73. Solid carbon dioxide when treated with etheral solution of C₂H₅MgBr followed by acid hydrolyzis gives
(a) propanoic acid
(b) ethanoic acid
(c) propionic acid
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

74. Which of the following compound does not give acetic acid on oxidation ?
(a) Ethanol
(b) Propan-l-ol
(c) Propan-2-ol
(d) 2-Methyl propan-2-ol
Answer:
(b) Propan-l-ol

75. A carboxylic acid resembles an alcohol with respect to its reaction with
(a) acidified K₂Cr₂O₇
(b) washing soda
(c) caustic soda
(d) sodium metal
Answer:
(d) sodium metal

76. Acetic acid can be converted into acetic anhydride on heating with
(a) POCl₃
(b) PCl₃
(c) PCI₅
(d) P₂O₅
Answer:
(d) P₂O₅

77. Acetyl chloride reacts with ammonia to give
(a) ammonium acetate
(b) ethylammonium chloride
(c) ethylamine
(d) acetamide
Answer:
(d) acetamide

78. The reagent that reacts with acetic acid to give sodium acetate with liberation of carbon dioxide gas is
(a) sodium metal
(b) caustic soda
(c) caustic potash
(d) baking soda
Answer:
(d) baking soda

79. An alkene on hydration gives a compound, which reacts with propionic acid to produce isopropyl propionate. The alkene is
(a) CH₂ = CH₂
(b) CH₃-CH = CH₂
(c) CH₃ – CH₂ – CH = CH₂
(d) CH₃ – CH = CH – CH₃
Answer:
(b) CH₃-CH = CH₂

80. Both the compounds ‘A’ and ‘B’ react with sodium metal to liberate hydrogen gas and react with each other to give Methylethanoate. The compounds ‘A’ and ‘B’ are
(a) C₂H₅ – COOH and CH₃ – OH
(b) C₂H₅ – COOH and C₂H₅ – OH
(c) CH₃ – COOH and C₂H₅ – OH
(d) CH₃ – COOH and CH₃ – OH
Answer:
(d) CH₃ – COOH and CH₃ – OH

81. Identify the product ‘D’ in the following series of reactions.

(a) CH₃COOCH₃
(b) CH₃COOC₂H₅
(c) C₂H₅COOCH₃
(d) C₂H₅COOC₂H₅
Answer:
(b) CH₃COOC₂H₅

82. Acetyl chloride on heating with sodium acetate gives
(a) ethyl acetate
(b) acetamide
(c) acetic anhydride
(d) acetaldehyde
Answer:
(c) acetic anhydride

83. Carboxylic acids are acidic in nature because
(a) it dissociates to give H⁺ ions
(b) it donates proton
(c) it reacts with active metal and liberates hydro-gen gas
(d) all of these
Answer:
(d) all of these

84. Carboxylic acids on heating with P₂O₅ give
(a) acid chlorides
(b) alkyl halides
(c) acid amides
(d) acid anhydrides
Answer:
(d) acid anhydrides

85.

Answer:
(c)

86. The compound having general formula is called

(a) diester
(b) acid anhydride
(c) hemiacetal
(d) acetal
Answer:
(d) acetal

87. Identify the strongest acid amongst the following :
(a) Chloroacetic acid
(b) Acetic acid
(c) Trichloroacetic acid
(d) Dichloroacetic acid
Answer:
(c) Trichloroacetic acid

88. Acetaldehyde, when treated with which among the following reagents does ‘not’ undergo addition reaction?
(a) ammonia
(b) hydroxyl amine
(c) ammoniacal silver nitrate
(d) semicarbazide
Answer:
(c) ammoniacal silver nitrate

89. Popcorn has butter flavour which contains
(a) butan-l-one
(b) butane-2, 3-dione
(c) butan-2-one
(d) butyric acid
Answer:
(b) butane-2, 3-dione

90. On acid hydrolysis, propane nitrile gives
(a) propanal
(b) acetic acid
(c) propionamide
(d) propanoic acid
Answer:
(d) propanoic acid

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 9 Business Communication Skills of a Secretary Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 9 Business Communication Skills of a Secretary

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Lay out of a letter means ____________
(a) physical appearance
(b) arrangement of its parts
(c) body
Answer:
(b) arrangement of its parts

Question 2.
Written communication is called ____________
(a) compliments
(b) consideration
(c) correspondence
Answer:
(c) correspondence

Question 3.
Communication that takes place internally between various departments of an organization ____________
(a) external communication
(b) internal communication
(c) verbal communication
Answer:
(b) internal communication

Question 4.
Communication that takes place between business organization and outsiders like bank, suppliers, creditors, etc ____________
(a) internal communication
(b) external communication
(c) verbal communication
Answer:
(b) external communication

Question 5.
Communication conveying message in spoken form is called ____________
(a) Non-verbal communication
(b) Verbal communication
(c) Internal communication
Answer:
(b) Verbal communication

Question 6.
Facial expression is a form of ____________ communication.
(a) Verbal communication
(b) Non-verbal communication
(c) Internal communication
Answer:
(b) Non-verbal communication

Question 7.
Information or ideas in written form is ____________
(a) oral communication
(b) written communication
(c) body language
Answer:
(b) written communication

Question 8.
A website for publishing informal online articles is called ____________
(a) blog
(b) cellular phone
(c) video conferencing
Answer:
(a) blog

Question 9.
Two-way personalised webscam based communication is called as ____________
(a) website
(b) blog
(c) video conferencing
Answer:
(c) video conferencing

Question 10.
____________ is a skill of being able to understand the feelings of another person.
(a) Empathy
(b) Sympathy
(c) Body language
Answer:
(a) Empathy

Question 11.
A ____________ introduces the firm.
(a) inside
(b) address, heading
(c) reference number
Answer:
(b) address, heading

Question 12.
Language of business letter should be ____________
(a) simple
(b) hard
(c) flowery
Answer:
(a) simple

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) Correspondence	(1) Greeting to the recipient of the letter
(b) Salutation	(2) Message not included in the body of the letter
(c) Postscript	(3) Non-verbal communication
(d) Hand gestures	(4) Introduction of the sender
(e) Conciseness	(5) Written communication
	(6) Inside address
	(7) Brief matter
	(8) Other documents attached
	(9) Verbal communication
	(10) Heading

Answer:

Group ‘A’	Group ‘B’
(a) Correspondence	(5) Written communication
(b) Salutation	(1) Greeting to the recipient of the letter
(c) Postscript	(2) Message not included in the body of the letter
(d) Hand gestures	(3) Non-verbal communication
(e) Conciseness	(7) Brief matter

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
Branch of general communication concerned with business activities.
Answer:
Business communication

Question 2.
Communication that takes place internally between departments.
Answer:
Internal communication

Question 3.
Communication that takes place between organization and outsider.
Answer:
External communication

Question 4.
An electronic mail sending messages using electronic devices through the internet.
Answer:
E-mail

Question 5.
A website for publishing informal articles.
Answer:
Blog

Question 6.
Online interactive groups using advanced mobile and web-based technologies.
Answer:
Social media network

Question 7.
The gesture of a dancer.
Answer:
Non-verbal communication

Question 8.
Skill to understand and share the feelings of another person.
Answer:
Empathy

Question 9.
The name and address of the letter writer.
Answer:
Heading

Question 10.
A matter wrote after completing the letter.
Answer:
Postscript

Question 11.
Copies were sent along with the letter.
Answer:
Enclosure

Question 12.
The final part of the letter.
Answer:
Signature

1D. State whether the following statements are True or False.

Question 1.
A report is a systematic presentation of facts, figures, and conclusions about a topic.
Answer:
True

Question 2.
A notice is a written summary of transactions conducted at the meeting.
Answer:
False

Question 3.
Minutes give precise information regarding an important event that is about to take place.
Answer:
False

Question 4.
Letterhead includes the name and address of the sender.
Answer:
True

Question 5.
A letter written in a logical sequence is coherence.
Answer:
True

Question 6.
“You’ attitude means writing the word you many times in a letter.
Answer:
False

Question 7.
Harsh and rude words should be avoided in a letter.
Answer:
True

Question 8.
Minimum words should be used in a letter.
Answer:
True

Question 9.
The additional information written after the letter is completed is Postscript.
Answer:
True

Question 10.
The letter sent to other people at the same time is ‘Carbon Copy Notation’.
Answer:
True

Question 11.
Salutation includes documents, cheques, etc. attached with the letter.
Answer:
False

Question 12.
A complimentary close is written below the body of the letter in a polite manner.
Answer:
True

1E. Find the odd one.

Question 1.
Twitter, Youtube, Facebook, Written communication
Answer:
Written communication

Question 2.
E-mail, website, blog, completeness
Answer:
Completeness

Question 3.
Notice, Reports, Minutes, Non-verbal communication
Answer:
Non-verbal communication

Question 4.
Coherence, Consideration, Cheerfulness, Margin
Answer:
Margin

1F. Complete the sentences.

Question 1.
The process of communication conveying a message in spoken form is called as ____________
Answer:
verbal communication

Question 2.
Communication which is neither written or spoken is known as ____________
Answer:
Non-verbal communication

Question 3.
The participants are able to see and hear each other and also display visual data models etc. under ____________
Answer:
video conferencing

Question 4.
Communication that involves hearing and understanding what a person says to you is ____________
Answer:
active listening

Question 5.
Proper arrangement of various parts of the letter is called as ____________
Answer:
layout

Question 6.
The number written on the left-hand side below the heading to give a quick reference to the matter concerned is called ____________
Answer:
Reference Number

Question 7.
The reader gets the idea of the matter of the letter without reading the letter completely by reading ____________
Answer:
Subject

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) Written summary	……………………….
(2) Spirit of hopeness	………………………..
(3) Logical sequence	……………………….
(4) ……………………….	Your’s faithfully

(Coherence, Minutes, Cheerfulness, Complimentary close)
Answer:

Group ‘A’	Group ‘B’
(1) Written summary	Minutes
(2) Spirit of hopeness	Cheerfulness
(3) Logical sequence	Coherence
(4) Your’s faithfully	Your’s faithfully

1H. Answer in one sentence.

Question 1.
Name the Electronic device in which mails can be sent.
Answer:
Mails can be through E-mail (Internet).

Question 2.
Name the device that provides a short message service.
Answer:
Mobile/Cellular phone provides short message service.

Question 3.
What is the final part of the letter called?
Answer:
The final part of the letter is called ‘Signature’.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
Written communication includes body language, facial expression, etc.
Answer:
Non Verbal communication includes body language, facial expression, etc.

Question 2.
Internal communication takes place between business organizations and outsiders.
Answer:
External communication takes place between business organizations and outsiders.

Question 3.
Coherence is also called the use of ‘You attitude’.
Answer:
Consideration is also called the use of ‘You attitude’.

Question 4.
A written summary of a business transacted at the meeting is called Notice.
Answer:
A written summary of a business transacted at the meeting is called Minutes.

Question 5.
Oral Communication is a permanent record.
Answer:
Written Communication is a permanent record.

Question 6.
The skill of understanding the feelings of another person is called Sympathy.
Answer:
The skill of understanding the feelings of another person is called Empathy.

Question 7.
A letter without clarity is invalid.
Answer:
A letter without a signature is invalid.

2. Explain the following terms/concepts.

Question 1.
Non-Verbal Communication
Answer:

• Communication that does not involve written or spoken words is called non-verbal communication.
• It takes place by using body language, facial expression, eye contact, silence, symbols, signs, gestures, etc.

Question 2.
Notice
Answer:

• It is an intimation by the company to the concerned persons or members or directors about the day, date, time, place, and business to a transacted at the meeting.
• It is to be sent to all the concerned persons through registered post.
• It should also include the agenda which is going to be discussed at the meeting.

3. Answer in brief.

Question 1.
Explain any four modes of Electronic devices.
Answer:
Modes of electronic device:
(a) E-mail:
It means sending messages through the internet. An E-mail has made businesses come closer to their customers. It also saves time for writing letters, posting a letter, etc.

(b) Websites:
It is a set of interconnected web pages which are located on a single web. It contains the information provided by the owner of the website. It can be accessed through the internet or through a private local area network. Each website has a unique internet address called “Uniform Resource Locator”(URL).

(c) Social media network:
It is a very popular online interactive group created by using mobile and web-based technologies. It helps the business organization to interact with the consumers and communicate about their products and services.

(d) Video-conferencing:
It can be done through computers by providing a video link between two or more people. The participants can see and hear each other while communicating and can also display visual data also.

4. Justify the following statements.

Question 1.
The letterhead introduces the firm.
Answer:

• The letterhead (or the head address) consists of the name of the company sending out the letter, its registered address, its emblem or logo, telephone number, fax number, telegraphic address, telex number, website, and e-mail address, and same times a brief line describing the nature of business of the company.
• The letterhead is usually printed at the top center of the page.
• When the addressee of the letter receives the letter, his attention first goes to the letterhead.
• On seeing the letterhead, he knows about the sender’s firm.
• The layout of the letterhead, it’s printing, and its style help to create an image of the company.
• An attractive letterhead creates a good impression.
• The letterhead speaks a lot of the sender’s organization.
• Thus, the letterhead introduces the firm.

Question 2.
Postscript should not be used in business letters.
Answer:

• Postscript means ‘written afterward’. It is like a footnote added to the letter.
• It means additional matter not written in the main body of the letter but added after completing the letter.
• It is necessary if some information is received late and is required to be conveyed along with the original letter.
• To have a postscript in the letter is not a good practice and hence should be avoided.
• It indicates the possible carelessness of the writer.
• It should be used only for communicating an urgent message.
• Thus, postscript should not be used in business letters.

Question 3.
Courtesy is one of the requisites of a good business letter.
Answer:

• Every business letter must be courteously worded.
• The language used should be polite, pleasing, and convincing.
• Courtesy and polite tone are possible by using terms like ‘please’, ‘thank you, ‘very kind of you’ etc.
• It is specifically necessary for letters sent to the employees.
• It includes polite manners, kindness, and consideration for others.
• A courteous letter builds goodwill for the organization.
• Lack of courtesy may prove to be costly and troublesome to a business unit.
• A business letter and serves as a ‘silent salesman’ or ‘silent ambassador’ only when it is courteous in meaning and language.
• Courtesy facilitates the expansion of business activities.
• Thus, courtesy is one of the requisites of a good business letter.

Question 4.
A business letter is a silent salesman.
Answer:

• In modern times, the scale of the business has increased to such an extent that direct and personal dealing with the customers is difficult except in case of retail trade.
• Correspondence is, thus, necessary to widen the boundaries of markets and expand the scale of business.
• Effectively drafted and neatly typed letters play the role of a salesman in creating and expanding the market for goods and services.
• When letters are courteous it creates a good impression.
• Letters convince the prospects of the novelty, specialty, and utility of the products and induce them to place orders.
• Thus, a business letter is a silent salesman.

Question 5.
A letter without a date is incomplete.
Answer:

• The date is an important part of a business letter.
• The date includes the day, month, and year on which the letter is sent to the addressee.
• The reference of data is useful for the fulfillment of business transactions on a particular date.
• A letter with a date can serve as evidence in a court of law.
• The date on the letter facilitates quick reference to old letters and filing of letters in chronological order.
• A letter without a date is incomplete and legally invalid. It is like a body without ahead.
• Thus, a letter without a date is incomplete.

Question 6.
The margin on all sides is a waste of paper.
Answer:

• In order to make a letter appear systematic and attractive, it is better to maintain margins on both sides of the letter.
• Proper margin makes the letter attractive.
• Proper margin must be kept on the left-hand side and right-hand side.
• Proper space must be left at the bottom of the letter.
• Thus, the margin on all sides is a waste of paper.

Question 7.
The signature provides legal value to the letter.
Answer:

• It usually consists of the name of the writer, his official designation, if any, and the department he is concerned with.
• It is placed directly below the complimentary close.
• The signature must be handwritten in ink and placed two or three line spaces below the complimentary close.
• Business letters, today, carry the typed name and also the designation of the signatory below the signature.
• This helps the reader to identify the person by his name and his official position in the company.
• A letter without a date is invalid, likewise, a letter without a signature is also invalid.
• The unsigned letter has no significance.
• Thus, a signature provides legal value to the letter.

5. Answer the following questions.

Question 1.
Explain the physical appearance of a letter.
Answer:
‘Physical appearance’ simply means ‘The look’ of the letter. How it should appear or present itself to the receiver of the letter, what impression should it create? All these depend on the ‘Look’ or ‘Outer appearance’ of the letter. Therefore to make the letter look attractive, attention should be given to the following points.

(a) Paper:
A paper of superior quality should be used. The high cost of such paper is a sort of investment and not an expenditure. A paper of good quality always creates a good impression. Proper colour of the paper must be selected (generally white colour). A quality paper is useful to keep records for long period.

(b) Typing:
The typed letters are more attractive and legible than handwritten letters. However, the typing must be neat, accurate, and clean. Spelling must be checked for its correctness. Electronic typewriters or computers with printers must be used.

(c) Margin:
The proper margin on both sides gives the letter a presentable appearance. The size of the margin depends upon the size of the paper. It is also necessary to leave space at the bottom of the letter.

(d) Spacing:
Proper and enough space between the lines should be left. Spacing should be uniform. Proper spacing between the paragraphs increases the attractiveness of letter writing.

(e) Letterhead:
The design of the letterhead should be simple but attractive and impressive. It should include all necessary details like address, telephone number, E-mail, Fax No. etc. The letterhead has advertising value. It introduces the sender to the receiver of the letter.

(f) Folding:
Though the folding of a letter is a simple matter but not an unimportant one. Folds should be minimum. The number of folds depends on the size of paper used for writing a letter and the size of the envelope used for sending it.

(g) Envelope:
The envelope used for sending a letter should be of the proper size, color, and quality. A right-sized envelope should be used. In simple words, the envelope should neither be too small nor too big. If a window envelope is used, to save time and labour then the letter must be folded in such a way that the inside address shall appear exactly on the window (of the envelope).
It Must Be Noted That – “There Are No Exact Rules for Physical Appearance of A Business Letter. It Depends on Circumstances.”

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 10 Correspondence with Directors Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 10 Correspondence with Directors

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Secretary is ____________ to Directors.
(a) owner
(b) servant
(c) member
Answer:
(b) servant

Question 2.
Directors are the ____________
(a) owners
(b) representative of shareholders
(c) creditors of the company
Answer:
(b) representative of shareholders

Question 3.
The provision regarding qualification shares of a director is contained in the ____________
(a) Articles of Association
(b) Memorandum of Association
(c) Prospectus
Answer:
(a) Articles of Association

1B. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
A person is elected by the shareholders of a company.
Answer:
Director

Question 2.
Maximum information in minimum words.
Answer:
Brevity

1C. Complete the sentences.

Question 1.
The Directors are ____________ of shareholders.
Answer:
representative

Question 2.
The Directors are responsible for making ____________
Answer:
decision, plans and policies

Question 3.
The report prepared by the directors at the end of every Financial Year is called ____________
Answer:
Directors report

Question 4.
The gap between two consecutive Board meetings should not be more than ____________
Answer:
120 days

1D. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) ………………………	The link between members and the directors
(2) Brevity	……………………………..

(Secretary, Concise and compact information)
Answer:

Group ‘A’	Group ‘B’
(1) Secretary	The link between members and the directors
(2) Brevity	Concise and compact information

1E. Correct the underlined word and rewrite the following sentences.

Question 1.
Brevity means not using harsh words while writing a letter.
Answer:
Politeness means not using harsh words while writing a letter.

Question 2.
The First Board meeting should be held with 60 days from the date of its incorporation.
Answer:
The First Board meeting should be held with 30 days from the date of its incorporation.

Question 3.
Accuracy means avoiding unnecessary details and irrelevant information in a letter.
Answer:
Brevity means avoiding unnecessary details and irrelevant information in a letter.

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 2 Solutions Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 2 Solutions

Question 1.
What is solution?
Answer:
The solution is a homogeneous mixture of two or more components or pure substances. When the size of particles of the components is of the order of 10^-10 m, then the solution is called a true solution. E.g. An aqueous solution of sugar.

Question 2.
Explain : (1) Homogeneous solution
(2) Heterogeneous solution.
Answer:
(1) Homogeneous solution : A solution in which solute and solvent form uniform homogeneous one phase due to attraction between their molecules/particles is called homogeneous solution.
E.g. A solution of NaCl or sugar.

(2) Heterogeneous solution : A solution consisting of two or more phases is called a heterogeneous solution. E.g. A colloidal solution of starch.

Question 3.
Define : (1) Solvent (2) Solute.
Answer:
(1) Solvent : The component in which solution formation takes place and which constitutes larger proportion of a solution is called solvent. For example, in an aqueous solution of sugar, water is the solvent.

(2) Solute : In a solution the component which dissolves and constitutes smaller proportion of a solution is called a solute. For example, in a sugar solution, sugar is the solute.

Question 4.
What are the different types of solutions?
Answer:
A solution consists of a solvent and a solute. Since the physical states of a solvent and a solute may be gaseous, liquid or a solid, there are nine types of solutions.

Question 5.
Mention the solvent and solute in the following :
(1) Smoke
(2) Moisture
(3) Alloy
(4) Soda water.
Answer:

Substance	Solvent	Solute
1. Smoke	Gas	Solid
2. Moisture	Gas	Liquid
3. Alloy	Solid	Solid
4. Soda water	Liquid	Gas

Question 6.
What is a saturated solution?
Answer:
A solution which contains the maximum amount of dissolved solute and further the solute can’t be dissolved is called a saturated solution.

There exists a dynamic equilibrium which is represented as,

Question 7.
What is a supersaturated solution?
Answer:
A solution containing a solute more than that required to form a saturated solution at equilibrium is called a supersaturated solution.

When a tiny crystal of a solute is added to supersaturated solution, the excess solute separates out and forms saturated solution.

Question 8.
What is solubility of a solute ?
Answer:
Solubility : It is defined as amount of a solute present per unit volume in its saturated solution at a specific temperature.
It is expressed in mol L^-1 or mol dm^-3.

Question 9.
Explain the factors on which the solubility of a substance (solute) depends.
Answer:
The extent of dissolution of a substance (solute) depends upon the following factors :
(1) Nature of a solute : A solute may be crystalline, amorphous, ionic or covalent. Hence accordingly its tendency to dissolve changes. The substances having similar intermolecular forces tend to dissolve in each other.

(2) Nature of solvents : Solvents are classified as polar and nonpolar. Polar solutes dissolve in polar solvents. For example, ionic compounds dissolve in polar solvent like water. A solvent other than water is called a nonaqueous solvent. For example, C₆H₆, CCl₄, etc. Solutions in these solvents are called nonaqueous solutions.
The solvent may be a gas, a liquid or a solid.

(3) Amount of a solvent : More amount of a solvent, will dissolve more quantity of the solute.

(4) Temperature : Depending on the nature of a solvent and a solute the solubility changes with termperature. The effect depends on the heat of solution, hydration energy, etc. Generally as the temperature increases, solubility of solid increases and that of gases decreases.

(5) Pressure :

• Pressure has no effect on the solubilities of solids and liquids since they are incompressible.
• The effect of pressure is important only for solutions which involve gases as solutes. With the increase in pressure and decrease in temperature, the solubility of gases increases.

Question 10.
Explain with the help of Le Chatelier’s principle the effect of temperature on solubility.
Answer:

1. The effect of temperature on solubility depends on enthalpy of solution.
2. For example, dissolution of KCl in water is an endothermic process since heat is absorbed during dissolution. In according to Le Chatelier’s principle by increasing temperature the solubility of KCl increases.
3. Dissolution of CaCl₂, Li₂SO₄, H₂O in water is an exothermic process since heat is evolved during dissolution. In this, according to Le Chatelier’s principle by increasing the temperature the solubility decreases.

Question 11.
Explain the solubility of gases in liquids.
Answer:

• Gases are soluble in water and other liquids and their solubility depends upon the nature of the gas.
• Non-polar gases like O₂, have less solubility in polar solvents.
• Polar gases like CO₂, NH₃, HCl, etc. are more soluble in polar solvent like water. CO₂ forms H₂CO₃, while NH₃ forms NH₄OH in aqueous solutions.
• The solubility of gases in liquids increases with the increase in pressure and the decrease in temperature.

Question 12.
Why does the solubility of a gas decrease with increase in temperature ?
OR
How does solubility of a gas in water varies with temperature ?
Answer:

• The gases are soluble in water and other liquids.
• According to Charles’ law, the volume of a given mass of a gas increases with the increase in temperature at constant pressure.
• Hence, the volume of the dissolved gas increases with the increase in temperature.
• This enormous increase in volume of the gases cannot be accommodated by the solvent molecules, hence excess of the gases escape out in the form of bubbles.

Therefore, the solubility of gases in liquids decreases with temperature.

Question 13.
Explain the effect of pressure on the solubility of the gases.
OR
State and explain Henry’s law.
Answer:
(1) Since the gases are compressible, their solubility in the liquids is influenced by external pressure of the gas. The solubility of gases increases with the increase in pressure.

(2) Henry’s law : It states that the solubility of a gas. in a liquid at constant temperature is proportional to the pressure of the gas above the solution.
(i) If S is the solubility of a gas in mol dm^-3 at a pressure P and constant temperature then by Henry’s law,
S ∝ P or S = K_H × P
where K_H is called Henry’s law constant.
(ii) If P = 1 atm, then S = K_H.
(iii) If several gases are present, then the solubility of any gas in the mixture is proportional to its partial pressure at given temperature.

(3) Illustration of Henry’s law : In case of aerated or carbonated drink beverage, the bottle is filled by dissolving CO₂ gas at high pressure and then sealed.
Above the liquid surface there is air and undissolved CO₂. Due to high pressure, the amount of dissolved CO₂ is large.

When the cap of the aerated bottle is removed, the pressure on the solution is lowered, hence excess of CO₂ and air escape out in the form of effervescence. Thus by decreasing the pressure, solubility of CO₂ is decreased.

Question 14.
What are the exceptions to Henry’s law? Why ?
Answer:
(1) The gases like NH₃ and CO₂ do not obey Henry’s law.
(2) This is because, these gases react with water,
\(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \longrightarrow \mathrm{NH}_{4 \text { (aq) }}^{+}+\mathrm{OH}_{(\text {aq })}^{-}\)
CO_2(g) + H₂O(l) → H₂CO_3(aq)
(3) Due to reactions of the gases like NH₃, CO_2(g), they have higher solubilities than expected by Henry’s law.

Question 15.
Oxygen gas is slightly soluble in water but it is highly soluble in blood. Explain.
Answer:
The vital constituent of blood, namely haemoglobin reacts with oxygen increasing the solubility of oxygen.
Haemoglobin + 4O_2(g) → Haemoglobin O₈
This oxygenated blood is circulated to the various parts of body, for the supply of oxygen.

Question 16.
Obtain the units of Henry’s law constant.
Answer:
By Henry’s law, S = K_H × P, where S is solubility of the gas in mol dm^-3, P is the pressure of the gas in atmosphere (or in bar) and K_H is Henry’s law constant.
∴ K_H = \(\frac{S^{\left(\mathrm{mol} \mathrm{dm}^{-3}\right)}}{P_{(\mathrm{atm})}}\) mol dm^-3 atm^-1 or mol dm^-3 bar^-1
Hence the units of Henry’s law constant K_H are mol dm^-3atm^-1.

Solved Examples 2.4

Question 17.
Solve the following :
(1) For a gas, the Henry’s law constant is 1.25 × 10^-3 mol dm^-3 atm^-1 at 25 °C. Calculate the solubility of the given gas at 2.5 atm and 25 °C.
Solution :
Given : Henry’s law constant = K_H
= 1.25 × 10^-3 mol dm^-3 atm^-1
Pressure of the gas = P = 2.5 atm
Solubility of the gas = S = ?
By Henry’s law
S = K_H × P
= 1.25 × 10^-3 mol dm^-3 atm^-1 × 2.5 atm
= 3.125 × 10^-3 mol dm^-3
Ans. Solubility of gas = 3.125 × 10^-3 mol dm^-3

(2) The solubility of dissolved oxygen to 27 °C is 2.6 × 10^-3 mol dm^-3 at 2 atm. Find its solubility at 8.4 atm and 27 °C.
Solution :
Given : Solubility of O₂
= S₁ = 2.6 × 10^-3 mol dm^-3
Initial pressure of O₂ = P₁ = 2 atm
Final pressure of O₂ = P₂ = 8.4 atm
Solubility of O₂ = S₂ = ?
(i) By Henry’s law,
S₁ = K_H × P₁
∴ Henry’s law constant K_H is,
K_H = \(\frac{S_{1}}{P_{1}}=\frac{2.6 \times 10^{-3}}{2}\)
= 1.3 × 10^-3 mol dm^-3 atm^-1
(ii) Now, S₂ = K_H × P₂ = 1.3 × 10^-3 × 8.4
= 10.92 × 10^-3
= 1.092 × 10^-2 mol dm^-3
Ans. Solubility of O₂ = 1.092 × 10^-2 mol dm^-3

(3) Henry’s law constant for the solubility of methane in benzene is 4.27 × 10^-5 mm^-1 Hg mold dm^-3 at constant temperature. Calculate the solubility of methane at 760 mm Hg pressure at same temperature.
Solution :
Given : Henry’s law constant = K_H
= 4.27 × 10^-5mm^-1 Hg mol dm^-3
Pressure of the gas = P = 760 mm Hg
K_H = 4.27 × 10^-5 mm^-1 mol dm^-3
= 4.27 × 10^-5 × 760 atm^-1 mol dm^-3
= 3245 × 10^-5 atm^-1 mol dm^-3
= 3.245 × 10^-2 atm^-1 mol dm^-3
P = 760 mm = \(\frac{760}{760}\) atm = 1 atm
By Henry’s law,
S = K_H × P = 3.245 × 10^-2 atm^-1 mol dm^-3 × 1 atm
= 3.245 × 10^-2 mol dm^-3
Ans. Solubility of methane = 3.245 × 10^-2 mol dm^-3

(4) The solubility of ethane at 25 °C is 0.92 × 10^-3 g dm^-3 at 1000 mm Hg pressure. Calculate Henry’s law constant.
Solution :
Given : Solubility of ethane
= S = 0.92 × 10^-5 g dm^-3
Pressure of ethane = P = 1000 mm Hg
Molar mass of ethane (C₂H₆) = 30 g mol^-1
Henry’s law constant = K_H = ?
S = 0.92 × 10^-3 g dm^-3
= \(\frac{0.92}{30}\) × 10^-3
= 3.067 × 10^-5 mol dm^-3
P = 1000 mm = \(\frac{1000}{760}\)atm = 1.316 atm
By Henry’s law,
S = K_H × P
∴ K_H = \(\frac{S}{P}=\frac{3.067 \times 10^{-5}}{1.316}\)
= 2.33 × 10^-5 mol dm^-3 atm^-1
Ans. Henry’s law constant = K_H
= 2.33 × 10^-5 mol dm^-3 atm^-1

(5) The solubility of nitrogen at 30 °C is 2.5 × 10^-3 g dm^-3 at 760 mm pressure. What will be its solubility in mol dm^-3 at 20,000 mm and same temperature?
Solution :
Given : Initial solubility of N₂ = S₁ =2.5 × 10^-3 g dm^-3
Initial pressure = P₁ = 760 mm
Final pressure = P₂ = 20,000 mm
Final solubility = S₂ in mol dm^-3 = ?
Molar mass of N₂ gas = 28 g mol^-1
S₁ = 2.33 × 10^-5 mol dm^-3
= \(\frac{2.5 \times 10^{-3}}{28}\) mol dm^-3
= 8.93 × 10^-5 mol dm^-3
P₁ = \(\frac{760}{760}\) = 1 atm
P₂ = \(\frac{20,000}{760}\) = 26.32 atm
By Henry’s law,
S₁ = K_H × P₁
∴ K_H = \(\frac{S_{1}}{P_{1}}=\frac{8.93 \times 10^{-5}}{1}\)
= 8.93 × 10^-5 mol dm^-3 atm^-1
S₂ = K_H × P₂ = 8.93 × 10^-5 × 26.32
= 2.35 × 10^-3 mol dm^-3
Ans. Solubility of N₂ gas
= 2.35 × 10^-3 mol dm^-3.

[Alternative method:
S₁ = K_HP₁ and S₂ = K_HP₂
∴ \(\frac{S_{2}}{S_{1}}=\frac{K_{\mathrm{H}} P_{2}}{K_{\mathrm{H}} P_{1}}=\frac{P_{2}}{P_{1}}\)
∴ S₂ = S₁ × \(\frac{P_{2}}{P_{1}}\) = 8.93 × 10^-5 × \(\frac{20,000}{760}\)
= 2.35 × 10^-3 mol dm^-3]

Question 18.
Marine life like fish prefers to stay at lower level in water. Explain.
OR
Explain, why do aquatic animals prefer to stay at lower level of water during summer?
Answer:

1. The solubility of oxygen gas decreases with the increase in temperature.
2. In sea or lake water, the temperature of upper level is higher than the lower level.
3. Therefore the dissolved oxygen content in water is more at lower level than at higher level required for the marine life.

Hence marine life like fish prefers to stay at lower level than upper level of water.

Question 19.
State and explain Raoult’s law.
Answer:
Statement of Raoult’s law : The law states that, at constant temperature, the partial vapour pressure of any volatile component of a solution is equal to the product of vapour pressure of the pure component and the mole fraction of that component in the solution.

Let P₀ and P be the respective vapour pressures of a pure volatile component and a solution. If x₁ is the mole fraction of a solvent then by Raoult’s law,
P = x₁ × P₀.

Explanation : Consider a solution containing two volatile components A and B having mole fractions x₁ and x₂ respectively.
Let \(P_{1}^{0}\) and \(P_{2}^{0}\) be the vapour pressures of pure components (or liquids) A and B respectively.
Then by Raoult’s law, vapour pressure of component A = P₁ = X₁ × \(P_{1}^{0}\), vapour pressure of component B = P₂ = x₂ × \(P_{2}^{0}\).
Here P₁ and P₂ represent partial vapour pressures of the two liquid components in the solution.
Hence the total vapour pressure, P_T of the solution will be,
P_T = P₁ + P₂
∴ P_T = x₁\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
∵ x₁ + x₂ = 1
∴ x₁ = 1 – x₂
∴ P_T = (1 – x₂)\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
= \(P_{1}^{0}\) – x₂\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
= (\(P_{2}^{0}\) – \(P_{1}^{0}\))x₂ + \(P_{1}^{0}\)
With the help of above equation, the vapour pressures of solutions having different concentrations (or mole fractions) can be calculated.

Question 20.
Explain the variation of vapour pressure with mole fraction of a solute in a liquid mixture.
Answer:
Consider a liquid mixture of two liquid components A and B having vapour pressures \(P_{1}^{0}\) and \(P_{2}^{0}\) and mole fractions x₁ and x₂ respectively.
By Raoult’s law, the vapour pressures P₁ and P₂ are,
P₁ = x₁\(P_{1}^{0}\) and P₂ = x₂\(P_{2}^{0}\)
The vapour pressure of the solution is,
P_T = P₁ +P₂ = x₁\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)

∴ P_T = (\(P_{2}^{0}\) – \(P_{1}^{0}\))x₂ + \(P_{1}^{0}\)
The plot of P_T versus x₂ is a straight line. The plots of P₁ versus x₁, and P₂ versus x₂ are straight lines passing through the origin.
When x₁ = 1, x₂ = 0, P_T = \(P_{1}^{0}\) and when x₁ = 0, x₂ = 1, P_T = \(P_{2}^{0}\) as shown by lines I and II in Fig. 2.2

Question 21.
Explain the composition of vapour phase above a liquid mixture.
Answer:
Consider a liquid mixture of two liquids A and B having vapour pressures \(P_{\mathrm{A}}^{0}\) and \(P_{\mathrm{B}}^{0}\) and mole fractions x₁ and x₂ respectively in the liquid phase.
By Raoult’s law, the vapour pressures of two liquids will be,
P_A = x₁\(P_{\mathrm{A}}^{0}\) and P_B = x₂\(P_{\mathrm{B}}^{0}\)
The total vapour pressure of this liquid mixture is,
P_T = P_A + P_B
P_T = x₁\(P_{\mathrm{A}}^{0}\) + x₂\(P_{\mathrm{B}}^{0}\)
The vapour above liquid surface contains A and B. If y₁ and y₂ are the mole fractions of A and B components respectively in the vapour phase, then by Dalton’s law of partial pressures,
P_A = y₁P_T and P_B = y₂P_T
and total vapour pressure is,
P_T = y₁P_T + y₂P_T.

Question 22.
What are ideal and nonideal solutions ?
Answer:

• Ideal solutions : These are solutions which obey Raoult’s law over an entire range of concentrations at constant temperature.
• Nonideal solutions : These are solutions which do not obey Raoult’s law over the entire range of concentrations.

Question 23.
What are the characteristics of ideal solutions ?
Answer:

• The ideal solutions obey Raoult’s law over entire range of concentrations at constant temperature.
• In the formation of an ideal solution, heat is neither evolved nor absorbed and enthalpy change for mixing is zero, i.e. Δ_mix H = 0.
• In the formation of an ideal solution, there is no volume change on mixing two liquid components and the volume of solution is equal to the sum of volumes of two liquid components. Δ_mix V = 0
• In the ideal solution, solvent-solvent, solute-solute and solvent-solute interactions are comparable.
• The vapour pressure of an ideal solution lies between vapour pressures of two pure components.

Question 24.
Give an example of an ideal solution.
Answer:
A liquid mixture of benzene and toluene which have nearly identical physical properties and inter molecular forces forms an ideal solution.

Question 25.
What are the characteristics of nonideal solutions.
Answer:

• Nonideal solutions do not obey Raoult’s law over the entire range of concentrations.
• The vapour pressures of these solutions may be higher or lower than ideal solutions.
• These solutions exhibit two types of deviations from Raoult’s law namely (a) positive deviation and (b) negative deviation.
• These solutions give azeotropic mixtures.

Question 26.
Explain solutions with positive deviations from Raoult’s law.
Answer:
(i) A solution or a liquid mixture which has higher vapour pressure than theoretically calculated by Raoult’s law or higher than those of pure components is called a nonideal solution with positive deviation.

(ii) In these solutions, solute-solvent intermolecular attractions are weaker than those between solvent-solvent and solute-solute interactions.
(iii) For example, solutions of acetone and ethanol, carbon disulphide and acetone, etc.

Question 27.
Explain solutions with negative deviations from Raoult’s law.
Answer:
(1) A solution or a liquid mixture which has lower vapour pressure than theoretically calculated by Raoult’s law or lower than those of pure components is called a nonideal solution with negative deviation.

(2) In these solutions, the intermolecular interactions between solvent and solute molecules are stronger than solvent-solvent or solute-solute interactions.
(3) For example, solutions of phenol and aniline, chloroform and acetone, etc.

Solved Examples 2.5

Question 28.
Solve the following :

(1) The vapour pressures of two liquids A and B are 400 mm Hg and 600 mm Hg respectively at 47 °C. A solution is prepared by dissolving 10 g of A of molar mass 60 g mol^-1 in 80 g of B of molar mass 40 g mol^-1. Find the vapour pressure of the solution.
Solution :
Given : \(P_{\mathrm{A}}^{0}\) = 400 mm Hg; \(P_{\mathrm{B}}^{0}\) = 650 mm Hg.
W_A = 10 g and W_B = 80 g.
M_A = 60 g mol^-1; M_B = 40 g mol^-1, P_soln = ?
n_A = \(\frac{W_{\mathrm{A}}}{M_{\mathrm{A}}}=\frac{10}{60}\) = 0.1667 mol
n_B = \(\frac{W_{\mathrm{B}}}{M_{\mathrm{B}}}=\frac{80}{40}\) = 2 mol
Total number of moles = n = n_A + n_B
= 0.1667 + 2
= 2.1667 mol

= 0.07693 × 400 + 0.9230 × 600
= 30.772 + 553.8
= 584.572 mm Hg
Ans. Vapour pressure of solution
= P_soln = 584.572 mm Hg

(2) 2.5 mol of a liquid A is mixed with 4.5 mol of liquid B at 25 °C. If the vapour pressures of A and B at 25 °C are 160 mm Hg and 230 mm Hg respectively, calculate the vapour pressure of the liquid mixture.
Solution :
Given : n_A = 2.5 mol; n_B = 4.5 mol,
\(P_{\mathrm{A}}^{0}\) = 160 mm Hg; \(P_{\mathrm{B}}^{0}\) = 230 mm Hg, P_soln = ?
Total number of moles = n = n_A + n_B
= 2.5 + 4.5
= 7.0 mol
Mole fraction of A = x_A = \(\frac{n_{\mathrm{A}}}{n}=\frac{2.5}{7}\) = 0.3571
Mole fraction of B = 1 – x_A = 1 – 0.3571
= 0.6429
\(P_{\text {soln }}=x_{\mathrm{A}} P_{\mathrm{A}}^{0}+x_{\mathrm{B}} P_{\mathrm{B}}^{0}\)
= 0.3571 × 160 + 0.6429 × 230
= 57.136 + 147.9
= 205 mm Hg
Ans. P_soln = 205 mm Hg

(3) The vapour pressures of pure liquids A and B are 400 mm Hg and 650 mm Hg respectively at 330 K. Find the composition of liquid and vapour if total vapour pressure of solution is 600 mm Hg.
Solution :
Given : \(P_{\mathrm{A}}^{0}\) = 400 mm Hg; \(P_{\mathrm{B}}^{0}\) = 650 mm Hg,
P_T = 600 mm Hg, T = 330 K; x_A = ? x_B = ?, y₁ = ? y₂ = ?
(x is mole fraction in liquid phase while y is mole fraction in vapour phase.)
P_T = (\(P_{\mathrm{A}}^{0}\) – \(P_{\mathrm{B}}^{0}\))x_B + \(P_{\mathrm{A}}^{0}\)
600 = (650 – 400)x_B + 400
= 250x_B + 400
∴ x_B = \(\frac{600-400}{250}\) = 0.8
∵ x_A + x_B = 1
∴ x_A = 1 – x_B = 1 – 0.8 = 0.2
The composition of A and B in liquid mixture is, x_A = 0.2 and x_B = 0.8.
If P₁ and P₂ are vapour pressures (or partial pressures) of A and B in vapour phase then by Raoult’s law,
P₁ = x_A × \(P_{\mathrm{A}}^{0}\) = 0.2 × 400 = 80 mm Hg
P₂ = x_B × \(P_{\mathrm{B}}^{0}\) = 0.8 × 650 = 520 mm Hg
If y₁ and y₂ are mole fractions of A and B respectively in vapour phase then, by Dalton’s law,
P₁ = y₁P_T

(or y₂ = 1 – y₁ = 1 – 0.1333 = 0.8667)
Ans. Composition of liquid : x_A = 0.2 and x_B = 0.8
Composition of vapour: y_A = 0.1333 and y_B = 0.8667

(4) A mixture of two liquids A and B have vapour pressures 3.4 × 10⁴ Nm^-2 and 5.2 × 10 Nm^-2. If the mole fractions of A is 0.85, find the vapour pressure of the solution.
Solution:
Given : Vapour pressure of pure liquid A
= \(P_{\mathrm{A}}^{0}\) = 3.4 × 10⁴ Nm^-2
Vapour pressure of pure liquid B
= \(P_{\mathrm{B}}^{0}\) = 5.2 × 10⁴ Nm^-2
Mole fraction of A = x_A = 0.85
Mole fraction of B = x_B = 1 – x_A
= 1 – 0.85
= 0.15
The vapour solution is given by
\(P_{\text {soln }}=X_{A} P_{A}^{0}+X_{B} P_{B}^{0}\)
= 0.85 × 3.4 × 10 + 0.15 × 5.2 × 10⁴
=2.89 × 10⁴ + 0.78 × 10⁴
=(2.89 + 0.78) × 10⁴
P_soln = 3.67 × 10⁴ Nm^-2
Ans. Vapour pressure of a solution = 3.67 × 10⁴ Nm^-2

Question 29.
Define the term colligative property. Give examples.
Answer:
(1) Colligative Property : The property of a solution which depends on the total number of particles of the solute (molecules, ions) present in the solution and does not depend on the nature or chemical composition of solute particles is called colligative property of the solution.

(2) Examples of colligative properties : (a) lowering or relative lowering of vapour pressure of a solution (b) elevation in the boiling point (c) depression in the freezing point (d) osmotic pressure.

Question 30.
Explain and define the term vapour pressure of a liquid.
OR
What is vapour pressure of a liquid?
Answer:

(1) If a volatile liquid is placed in an open vessel, the liquid molecules have a tendency to escape in a gaseous state forming vapour and diffuse into surroundings. Hence evaporation takes place continuously, and no equilibrium is attained.

(2) If a liquid is placed in a closed vessel, the vapour molecules get accumulated on its surface. These vapour molecules are in continuous random motion. They collide with each other, with walls of a container and surface of the liquid, and return to the liquid state. This reverse phenomenon is called condensation.

(3) After some time, rates of evaporation and condensation become equal and an equilibrium is established between liquid and vapour phases. At this stage the vapour exerts a constant pressure called vapour pressure on liquid surface at constant temperature.

(4) Vapour pressure : The pressure exerted by the vapour of a liquid (or solid) when it is in equilibrium with the liquid (or solid) phase at a constant temperature is called the vapour pressure of the liquid (or solid).

(5) The vapour pressure of a liquid increases with the increase in temperature.

Question 31.
Explain the following terms :
(1) Relative vapour pressure of a solution
(2) Lowering of vapour pressure of a solution
(3) Relative lowering of vapour pressure.
Answer:
(1) Relative vapour pressure of a solution : If Po is the vapour pressure of a pure liquid (solvent) and P is the vapour pressure of a solution after adding a nonvolatile solute, then, relative vapour pressure = \(\frac{P}{P_{0}}\).

(2) Lowering of vapour pressure of a solution :
When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure.
If P₀ is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P₀, then, (P₀ – P) is the lowering of the vapour pressure.

(3) Relative lowering of vapour pressure : If P₀ and P are the respective vapour pressures of a pure liquid (solvent) and the solution containing a non-volatile solute then P < P₀. Hence, P₀ – P represents the lowering of the vapour pressure due to addition of a nonvolatile solute.
∴ Relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}=\frac{\Delta P}{P_{0}}\)

Question 32.
State and explain Raoult’s law for solutions of nonvolatile solutes.
Answer:
(a) Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature.

(b) Explanation : Let P0 and P be the vapour pressures of a pure solvent and a solution respectively. If x₁ is the mole fraction of the solvent then Raoult’s law can be represented as,
P = x₁P₀
For a binary solution containing one solute, if x₁ and x₂ are mole fractions of a solvent and a solute respectively then,
x₁ + x₂ = 1
∴ x₁ = 1 – x₂
∴ P = x₁P₀
= (1 – x₂) P₀
= P₀ – x₂P₀
= P₀ – P = x₂P₀
∴ P₀ – P = x₂P₀
∴ x₂ = \(\frac{P_{0}-P}{P_{0}}\)
P₀ – P = ΔP is the lowering of vapour pressure
∴ x₂ = \(\frac{\Delta P}{P_{0}}\)
In the equation, P₀ – P/P₀ is called relative lowering of vapour pressure.
Hence Raoult’s law can also be stated as the relative lowering of vapour pressure is equal to mole fraction of the solute.

Question 33.
Show that relative lowering of vapour pressure is a colligative property.
Answer:
Consider a binary solution containing a nonvolatile solute. If P₀ and P are vapour pressures of a pure solvent and the solution respectively then, Lowering of vapour pressure = ΔP = P₀ – P
Relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}\)
By Raoult’s law,
\(\frac{P_{0}-P}{P_{0}}=x_{2}\)
where x₂ is mole fraction of the solute. Therefore the relative lowering of vapour pressure is a colligative property.

Question 34.
Explain the variation of vapour pressure with mole fraction of a solvent in solution.
OR
Explain the variation of vapour pressure with the concentration of a solution.
Answer:
The vapour pressure of a pure solvent decreases when a nonvolatile solute is dissolved in it. Consider a pure solvent with vapour pressure P₀ and mole fraction x₁. For a solution containing a non-volatile solute, if x₁ and x₂ are the mole fractions of a solvent and a solute respectively, then x₁ + x₂ = 1 and x₁ < 1.

By Raoult’s law, P = x₁ × P₀. As the mole fraction of a solvent in the solution increases the vapour pressure increases as shown in the above figure 2.5. When x₁ becomes equal to 1, the vapour pressure becomes P₀, i.e. the vapour pressure of a pure solvent.

If at any mole fraction of a solvent, the vapour of the solution is P, then the lowering of vapour pressure will be, ΔP = P₀ – P.

Question 35.
Derive a relation between relative lowering of vapour pressure and molar mass of non-volatile solute.
Answer:
Consider a solution in which W₁ gram of a solvent of molar mass (or molecular weight) M₂ contains W₂ gram of a solute of molar mass M₂. Then
number of moles of a solvent = n₁ = \(\frac{W_{1}}{M_{1}}\)
Number of moles of a solute = n₂ = \(\frac{W_{2}}{M_{2}}\)
∴ Total number of moles = n = n₁ + n₂
Mole fraction of the solvent = x₁ = \(\frac{n_{1}}{n}\)
Mole fraction of the solute = x₂ = \(\frac{n_{2}}{n}\)

In the case of an ideal solution which is a dilute solution, the concentration and the number of moles of the solute are very low, i.e. n₂ << n₁

If P₀ and P are the vapour pressures of a pure solvent and a solution respectively, then relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}\)
By Raoult’s law

Hence by measuring the vapour pressure of a pure solvent and a solution, the molar mass of the dissolved nonvolatile substances can be determined.

Solved Examples 2.6 – 2.7

Question 36.
Solve the following :

(1) The vapour pressure of a pure liquid at 298 K is 4 × 10⁴ Nm^-2. When a nonvolatile solute is dissolved the vapour pressure becomes 3.65 × 10⁴ Nm^-2. Calculate (A) relative vapour pressure, (B) lowering of vapour pressure and (C) relative lowering of vapour pressure.
Solution :
Given : P₀ = 4 × 10⁴ Nm^-2
P = 3.65 × 10⁴ Nm^-2
(A) Relative vapour pressure = \(\frac{P}{P_{0}}\)
= \(\frac{3.65 \times 10^{4}}{4 \times 10^{4}}\)
= 0.9125

(B) Lowering of vapour pressure = ΔP = P₀ – P
= 4 × 10⁴ – 3.65 × 10⁴
= (4 – 3.65) × 10⁴
= 0.35 × 10⁴ Nm^-2 = 3.5 × 10³ Nm^-2

(C) Relative lowering of vapour pressure is given by,

Ans. (A) 0.9125 (B) 3.5 × 10³ Nm^-2 (C) 0.0875

(2) A pure liquid has vapour pressure 5.2 × 10⁴ Pa at 298 K. When a solute is dissolved, the mole fraction of it is 0.02 in the solution. Find the vapour pressure of the solution.
Solution :
Given : P₀ = 5.2 × 10⁴ Pa
x₂ = 0.02
P = ?

Ans. Vapour pressure of the solution
= 5.096 × 10⁴ Pa

(3) The vapour pressure of pure benzene is 640 mm of Hg. 2.175 × 10^-3 kg of nonvolatile solute is added to 39 gram of benzene, the vapour pressure of solution is 600 mm of Hg. Calculate molar mass of solute (C = 12, H = 1).
Solution :
Given : P₀ = 640 mm Hg
W₁ = 39 g benzene = 39 × 10^-3 kg
W₂ = 2.175 × 10^-3 kg
P = 600 mm Hg
M₁ = 78 × 10^-3 kg, M₂ = ?

Ans. Molar mass of a solute = 69.6 × 10^-3 kg mol^-1

(4) In an experiment, 18.04 g of mannitol were dissolved in 100 g of water. The vapour pressure of water was lowered by 0.309 mm Hg from 17.535 mm Hg. Calculate the molar mass of mannitol.
Solution :
Given : Mass of a solute (mannitol)
= W₂ = 18.04 g
Mass of a solvent (water) = W₁ = 100 g
Vapour pressure of a solvent (water)
= P₀
= 17.535 mm Hg
Lowering of vapour pressure = ΔP = 0.309 mm Hg
Molar mass of H₂O = M₁ = 18 g mol^-1
Molar mass of solute (mannitol) = M₂ = ?

Ans. Molar mass of mannitol = 184.3 g mol^-1

(5) The vapour pressure of 2.1% solution of a non-electrolyte in water at 100 °C is 755 mm Hg. Calculate the molar mass of the solute.
Solution :
Given : At 100 °C, vapour pressure of water = P₀ = 760 mm Hg
Vapour pressure of the solution = P = 755 mm Hg
Since the solution is 2.1 % by mass,
Mass of solute (nonelectrolyte) = W₂ = 2.1 g
Mass of water = W₁ = 100 – 2.1 = 97.9 g
Molar mass of water = M₁ = 18 g mol^-1
Molar mass of nonelectrolyte = M₂ = ?

Ans. Molar mass of nonelectrolyte = 58.69 g mol^-1

(6) Calculate the mass of a nonvolatile solute (molar mass 40 × 10^-3 kg/mol) which is dissolved in 114 × 10^-3 kg octane to reduce its vapour pressure to 80%.
Solution :
Given : Molar mass of solute = M₂ =40 × 10^-3 kg mol^-1
Mass of solvent (octane) = W₁ = 114 × 10^-3 kg
If vapour pressure of octane = P₀ = 100
Vapour pressure of solution = P = 80
Molar mass of octane (C₈H₁₈) = 114 × 10^-3 kg mol^-1
Mass of solute = W₂ = ?
In this solution, since the vapour pressure is decreased by greater extent, 100 – 80 = 20% the solution must be concentrated and number of moles of solute must be large so that n₁ + n₂ ≠ n₁
Hence Raoult’s law must be represented as,

∴ Mass of solute dissolved = W
= moles × molar mass = 0.25 × 40 × 10^-3
= 0.01 kg
Ans. Mass of solute dissolved = 0.01 kg

(7) The vapour pressure of water is 16.8 mm Hg at a certain temperature. If the vapour pressure of the solution is 16.78 mm, find the molality of the solution.
Solution :
Given : P₀ = 16.8 mm Hg; P = 16.78 mm Hg
Molar mass of water = M₁ = 18 g mol^-1
Molality of the solution = m = ?

Ans. Molality of the solution = 0.66 m

Question 37.
Explain the effect of temperature on the vapour pressure of a liquid.
Answer:
The vapour pressure of a liquid is the pressure of the vapour in equilibrium with the liquid at a given temperature. The evaporation of a liquid requires thermal energy. Hence, as temperature rises, the vapour pressure rises until it becomes equal to the external pressure, generally the atmospheric pressure, 101.3 kNm^-2 (1 atm). This temperature is called the normal boiling point of the liquid.

Question 38.
What is boiling point of liquid?
OR
Define boiling point.
Answer:
The boiling point of a liquid is defined as the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, i.e., the atmospheric pressure (1 atm), e.g., the boiling point of water at 1 atm is 373 K. m

Question 39.
Explain the elevation in the boiling point of a solution.
Answer:
The elevation in the boiling point of a solution is defined as the difference between the boiling points of the solution and the pure solvents at a given pressure, e.g. If T₀ and T are the boiling points of a pure solvent and a solution, then the elevation in boiling point, ΔT_b = T – T₀. It is a colligative property.

Question 40.
What are the units of molal elevation constant?
Answer:
Boiling point elevation, ΔT_b is given by,
ΔT_b = K_b × m
where m is molality in mol kg^-1 and K_b is molal elevation constant or abullioscopic constant.

∴ K_b has units K kg mol^-1 (or °C kg mol^-1)
Therefore, molal elevation constant is the elevation in boiling point produced by 1 molal solution of a nonvolatile solute.

Question 41.
Derive the relation between molar mass of the solute and boiling point elevation.
Answer:
The boiling point elevation, ΔT_b of a solution is directly proportional to molality (m) of the solution.
∴ ΔT_b ∝ m
ΔT_b = K_b m
where K_b is a proportionality constant
If m = 1 molal, then
ΔT_b = K_b
where K_b is called molal elevation constant.
The molality of the solution is given by,
Number of moles of the solute, m = \(\frac{\text { Number of moles of the solute }}{\text { Weight of the solvent in } \mathrm{kg}}\)
Let W₁ = Weight (in gram) of a solvent,
W₂ = Weight (in gram) of a solute
M₂ = Molecular weight of the solute
Then the molality (m) of the solution is given by

If the weights and molecular weight are expressed in kg, then,
\(\Delta T_{\mathrm{b}}=K_{\mathrm{b}} \times \frac{W_{2}}{W_{1} M_{2}}\)

Question 42.
Define molal elevation constant (Ebullioscopic constant)? Does it depend on the nature of a solute ? What are its units ?
Answer:
(1) Molal elevation constant (Ebullioscopic constant) : It is defined as the elevation in boiling point, produced by dissolving one mole of a solute in 1 kg (or 1000 gram) of a solvent (i.e. 1 molal solution).
The elevation in the boiling point,
ΔT_b is given by ΔT_b = K_b × m
where Kb is molal elevation constant and m is molality of the solution.
∴ When m = 1, ΔT_b = K_b
(2) K_b depends only on the nature of the solvent.
(3) K_b does not depend on the nature of the solute.
(4) It does not depend on concentration of the solution.
(5) The units of molal elevation constant are :
(A) K kg mol^-1 and (B) Km^-1.

Solved Examples 2.8

Question 43.
Solve the following :

(1) Calculate the (i) elevation in the boiling point and (ii) the boiling point of 0.05 m aqueous solution of glucose. (K_b = 0.52 Km^-1)
Solution :
Given : Concentration of the solution = m = 0.05 m
Molal elevation constant = K_b = 1.86 K kg mol^-1
Boiling point of pure water = T₀ = 373 K
Elevation in the boiling point = ΔT_b = ?
Boiling point of solution = T_b = ?

(i) ΔT_b = K_b × m
= 0.52 × 0.05
= 0.026K

(ii) The elevation in the boiling point is given by,
ΔT_b = T_b – T₀
∴ Boiling point of a solution,
T_b = T₀ + ΔT_b
= 373 + 0.026
= 373.026 K
Ans. ΔT_b = 0.026 K, T_b = 373.026 K

(2) 0.18 molal aqueous solution of a substance boils at 373.25 K. Calculate the molal elevation constant of water. (Boiling point of water is 373.15 K)
Solution :
Given : Concentration of solution = m = 0.18 m
Boiling point of water = T₀ = 373.15 K
Boiling of the solution = T_b = 373.25
Molal elevation constant = K_b = ?
Elevation in boiling point = ΔT_b
= T_b – T₀
= 373.25 – 373.15
= 0.1 K
ΔT_b = K_b × m
∴ K_b = \(\frac{\Delta T_{\mathrm{b}}}{m}=\frac{0.1}{0.18}\) = 0.5556 K kg mol^-1
Ans. Molal elevation constant of water
= 0.5556 K kg mol^-1

(3) The boiling point of benzene is 353.23 K. When 1.80 gram of non-volatile solute was dissolved in 90 gram of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute. [K_b for benzene = 2.53 K kg mol^-1]
Solution :
Given : \(T_{b}^{0}\) = 353.23 K; T_b = 354.11 K
W₁ = 90 g; W2 = 1.8 g; K_b = 2.53 K kg mol^-1
Molar mass = M₂ = ?
ΔT_b = T_b – \(T_{b}^{0}\) = 354.11 – 353.23 = 0.88 K

Ans. Molar mass = M₂ = 57.5 g mol^-1

(4) Boiling point of a solvent is 80.2°C. When 0.419 g of the solute of molar mass 252.4 g mol^-1 is dissolved in 75 g of the above solvent, the boiling point of the solution is found to be 80.256 °C. Find the molal elevation constant.
Solution :
Given : T₀ = (273 + 80.2) K
T_b = 273 + 80.256 (K)
W₁ = 75g
W₂ = 0.419 g
M₂ = 252.4 g mol^-1
K_b = ?
ΔT_b = T_b – T₀
= (273 + 80.256) – (273 + 80.2)
= 0.056 K

Ans. K_b = 2.53 K kg mol^-1

(5) 3.795 g of sulphur is dissolved in 100 g of CS₂. This solution boils at 319.81 K. What is the molecular formula of sulphur in solution? Theboiling point of CS₂ is 319.45 K.
(Given that K_b for CS₂ =2.42 K kg mol^-1 and atomic mass of S = 32.)
Solution :
Given : W₂ = 3.795 g; W₁ = 100 g CS₂
T_b = 319.81 K; T₀ = 319.45 K
K_b = 2.42 K kg mol^-1 M₂ = ?
ΔT_b = T_b – T₀ = 319.81 – 319.45 = 0.36 K

Number of S atoms in molecule = \(\frac{255.1}{32}\)
= 7.972
≅ 8
Ans. Formula of sulphur in CS₂ solution = S₈

(6) Calculate the mass in grams of an impurity of molar mass 100 g mol^-1 which would be required to raise the boiling point of 50 g of chloroform by 0.30°C. (K_b for chloroform = 3.63 K kg mol^-1)
Solution :
Given : M₂ = 100 g mol^-1
W₁ = 50 g chloroform
W₂ = ?
ΔT_b = 0.30°C.
K_b = 3.63 K kg mol^-1

Ans. Weight of impurity = 0.4132 g

(7) A solution containing 0.5126 g of naphthalene (molar mass = 128.17 g mol^-1) in 50.0 g of CCl₄ gives a boiling point elevation of 0.402 °C. While a solution of 0.6216 g of unknown solute in the same mass of the solvent gives a boiling point elevation of 0.647 °C. Find the molar mass of the unknown solute. (K_b for CCl₄ = 5.03 K kg mol^-1 of solvent)
Solution:
Given : For naphthalene solution :
W₂ = 0.5126 g
W₁ = 50 g
M₂ = 128.17 g mol^-1
K_b = 5.03 K kg mol^-1
ΔT_b = 0.402°C
For unknown solute :
W’₁ = 50g
W’₂ = 0.6216 g
ΔT’_b = 0.647°C
M’₂ = ?
For the solution of unknown substance,

Ans. Molecular weight of unknown substance = 96.65 g mol^-1

(8) A solution containing 0.73 g of camphor (molar mass 152 g mol^-1) in 36.8 g of acetone (boiling point 56.3°C) boils at 56.55°C. A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46°C. Calculate the molar mass of the unknown compound.
Solution:
Given : Mass of solute = W₂ = 0.73 g camphor
Mass of solvent = W₁ = 36.8 g
Molar Mass of solute = M₂ = 152 g mol^-1
Boiling point of acetone = T₀ = (273 + 56.3) K
Mass of unknown solute = W’₂ = 0.564 g
Mass of solvent = W’₁ = 36.8 g
Boiling point of solution of camphor = T_b = (273 + 56.55) K
Boiling point solution of unknown compound = T’_b = (273 + 56.46) K
Molar mass of unknown compound = M’₂ = ?

For camphor solution,
∴ ΔT_b = T_b – T₀
= (273 + 56.55) – (273 + 56.3)
= 0.25 K

For a solution of unknown compound
ΔT’_b = ΔT’_b – T₀
= (273 + 56.46) – (273 + 56.3)
= 0.16 K

Ans. Molar mass of the compound = 183.5 g mol^-1

(9) 0.12 molal solution of a substance boils at 373.21°K. Calculate molal elevation constant of the solvent.
Solution :
Given : Concentration of solution = m = 0.12
Boiling point of the solution = T_b = 373.21 K
For solvent (water) T₀ = 373.15 K
Molal elevation constant, K_b = ?
ΔT_b = T_b – T₀
= 373.21 – 373.15
= 0.06 K
ΔT_b = K_b × m
∴ K_b = \(\frac{\Delta T_{\mathrm{b}}}{m}\)
= \(\frac{0.06}{0.12}\)
=0.5 K kg mol^-1
Ans. Molal elevation constant = K_b = 0.5 K kg mol^-1

(10) Boiling point of water at 750 mm of Hg is 99.63 °C. How much sucrose must be added to 500 g of water so that it boils at 100 °C? (K_b = 5.02 K kg mol^-1)
Solution :
Given : Boiling point of water = T₀ = 273 + 99.63
= 372.63 K
Boiling point of a solution = T_b = 273 + 100 = 373 K
Mass of a solvent (water) = W₁ = 500 g
Molar mass of sucrose (C₁₂H₂₂O₁₁) = M2
= 342 g mol^-1
K_b = 5.02 K kg mol^-1
Mass of solute (sucrose) = W₂ = ?
ΔT_b = T_b – T₀
= 373 – 372.63
= 0.37 K

Ans. Mass of sucrose required to be added = 12.60 g

(11) A solution of phosphorus prepared by dissolving 0.0175 kg of phosphorus in 0.08 kg of CS₂ has a boiling point 319.87 K. If K_b for CS₂ is 2.4 K kg mol-1 and atomic mass of phosphorus is 31 × 10^-3 kg mol^-1, find the formula of phos-phorus. (Boiling point of CS₂ = 319.45 K)
Solution :
Given : Mass of solvent (CS₂) = W₁ = 0.08 kg
Mass of phosphorus = W₂ = 0.0175 kg
Boiling point of CS₂ = T₀ = 319.45 K
Boiling point of solution = T_b = 319.87 K
Molal elevation constant = K_b = 2.4 K kg mol^-1
Atomic mass of phosphorus = 31 × 10^-3 kg mol^-1
Molecular formula of phosphorus = ?
ΔT_b = T_b – T₀ = 319.87 – 319.45 = 0.42 K

∴ Number of atoms in a molecule of phosphorus = \(\frac{\text { molar mass of phosphorus }}{\text { atomic mass of phosphorus }}\)
= \(\frac {125}{31}\)
= 4.032
≅ 4
Hence the molecular formula of phosphorus is P₄ in CS₂.
Ans. Molecular formula of phosphorus = P₄

(12) Boiling point of water at 750 mm of Hg is 99.63 °C. How much sucrose must be added to 500 g of water so that it boils at 100 °C? (K_b = 0.52 K kg mol^-1)
Solution :
Given : Pressure = P = 750 mm Hg
T₀ = 273 + 99.63 = 372.63 K
T_b = 273 + 100 = 373 K
K_b = 0.52 K kg mol^-1
Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342 × 10^-3 kg mol^-1
W₁ = 500 g = 0.5 kg
Mass of sucrose to be added = W₂ = ?
ΔT_b = T_b – T₀ = 373 – 372.63 = 0.37 K

Ans. Mass of sucrose to be added = 121.7 × 10^-3 kg
= 121.7 g

(13) 35 % (W/W) solution of ethylene glycol in water, an anti-freezer used in automobiles in radiators as a coolant. It lowers freezing point of water to -17.6 °C. Calculate the mole fraction of the components.
Solution :
35% (W/W) means 100 g solution contains 35 g ethylene glycol (CH₂OH-CH₂OH) and 65 g H₂O.
Molar mass of water = 18 g mol^-1
Molar mass of ethylene glycol (CH₂OH-CH₂OH) = 62 g mol^-1
Number of moles of water = n₁ = \(\frac{65}{18}\) = 3.611 mol
Number of moles of ethylene glycol = n₂ = \(\frac{35}{62}\)
= 0.5645 mol
Total moles = n = n₁ + n₂ = 3.611 + 0.5645
= 4.1755 mol
Mole fraction of ethylene glycol = x₂
\(=\frac{n_{2}}{n_{1}}=\frac{0.5645}{4.1755}\) = 0.1352
∴ Mole fraction of water = 1 – x₂ = 1 – 0.1352
= 0.8648
Ans. Mole fraction of water = 0.8648
Mole fraction of ethylene glycol = 0.1352

Question 44.
Define freezing point of a liquid.
Answer:
The freezing point of a liquid is defined as the temperature at which the solid coexists in the equilibrium with the liquid and the vapour pressure of the liquid and the solid are equal.

Question 45.
Explain the depression in the freezing point of a solution.
Answer:
The depression in the freezing point of a solution is defined as the difference between the freezing points of a pure solvent and that of the solution.
If T₀ and T are the respective freezing points of a pure solvent and a solution, then the depression in the freezing point ΔT_f is given by,
ΔT_f = T₀ – T (T < T₀)
The depression in the freezing point (ΔT_f) is a colligative property.

Question 46.
What causes depression in freezing point ?
OR
Explain, freezing point depression as a consequence of vapour pressure lowering.
Answer:
The freezing point of a liquid is the temperature at which the liquid and the solid have the same vapour pressure.

Addition of a nonvolatile solute to a liquid decreases the freezing point, i.e., the freezing point of the solution is less than that of the pure solvent. This is due to the lowering of the vapour pressure of the solvent by the addition of the nonvolatile solute.

When a liquid is cooled from the point A, its vapour pressure decreases and at the point B, it freezes (solidifies).

In case of a solution, since the vapour pressure is lowered, the freezing point decreases. Hence, if the solution is cooled from the point A’, it freezes at lower temperature B’, than the pure liquid. This is also due to separation of solvent molecules due to solute molecules decreasing their intermolecular attraction.

If T₀ and T are the freezing points of a pure solvent and the solution, then, the depression in the freezing point is given by,
ΔT_f = T₀ – T.
This depression ΔT_f depends on the lowering of the vapour pressure (P₀ – P). ΔT_f ∝ (P₀ – P), where P₀ and P are the vapour pressures of the pure liquid and the solution.

Question 47.
What is a relationship between freezing point depression and concentration of solute (or solution)?
Answer:
It is observed experimentally that as the concentration of a solution increases, the freezing point of the solution decreases and hence the depression in the freezing point (ΔT_f) increases.

The depression in the freezing point of a solution is directly proportional to the molal concentration (expressed in mol kg^-1) of the solution.
Thus,
ΔT_f ∝ m
where m is the molality of the solution.
∴ ΔT_f = Kfm, where Kf is a constant of proportionality. If m = 1 molal,
ΔT_f = K_f. Hence K_f is called the cryoscopic constant or molal depression constant. K_f is characteristic of the solvent.

Question 48.
What are the units of molal depression constant or cryoscopic constant?
Answer:
The freezing point depression, ΔT_f is given by,
ΔT_f = K_f × m
where m is molality in mol kg^-1 and K_f is molal depression constant or cryoscopic constant.

∴ K_f has unit K kg mol^-1 (or °C kg mol^-1)
Therefore cryoscopic constant is the depression in freezing point produced by 1 molal solution of a nonvolatile solute.

Question 49.
Write the formula to determine molar mass of a solute using freezing point depression method.
Answer:
M₂ = \(\frac{K_{\mathrm{f}} \times W_{2} \times 1000}{W_{1} \times \Delta T_{\mathrm{f}}}\)
where
K_f = Molal depression constant
ΔT_f = Depression in freezing point
W₁ = Mass of a solvent
W₂ = Mass of a solute.
M₂ = Molar mass of solute

Question 50.
Define cryoscopic constant (or molal depression constant).
Answer:
Molal depression constant : It is defined as the depression in freezing point, produced by dissolving one mole of a solute in 1 kg (or 1000 g) of a solvent (i.e. 1 molal solution).

Solved Examples 2.9

Question 51.
Solve the following :

(1) 1.35 g of a substance when dissolved in 55 g acetic acid produced a depression of 0.618°C in a freezing point. Calculate the molar mass of the dissolved substance. (K_f = 3.865 K kg mol^-1)
Solution :
Given : Mass of a solvent = W₁ = 55 g
Mass of a solute = W₂ = 1.35 g
Depression in freezing point = ΔT_f = 0.618 °C
Molal depression constant = K_f = 3.865 K kg mol^-1
Molar mass of a solute = M₂ = ?

Ans. Molar mass of the substance = 153.5 g mol^-1

(2) When certain amount of sucrose is dissolved in 1 kg of water, the freezing point of the solution is found to be 272.8 K. If the molecular mass of sucrose is 342 g mol^-1 and K_f for water is 1.86 K kg mol^-1, calculate the amount of sucrose present in the solution.
Solution :
Given : W₁ = Mass of the solvent = 1 kg
T₀ = Freezing point of pure water = 273 K
T_f = Freezing point of the solution = 272.8 K
ΔT_f = T₀ – T_f = Depression in freezing point
= 273 K – 272.8 K
= 0.2 K
M₂ = Molecular mass of the solute
= 342 × 10^-3 kg mol^-1
K_f = Molal freezing point depression constant for water
= 1.86 K kg mol^-1
W₂ = Mass of the solute = ?
ΔT_f = T₀ – T_f

Ans. Mass of sucrose = 36.78 × 10^-3 kg

(3) The freezing point of a pure solvent is 315 K. On addition of 0.5 mole of urea in 1 kg of the solvent, the freezing point decreases by 3 K. Calculate the molal depression constant for the solvent.
Solution :
Given :
m = Molality of urea = 0.5 m
ΔT_f = Depression in the freezing point = 3 K
K_f = Molal depression constant for the solvent = ?
ΔT_f = K_f × m
∴ K_f = \(\frac{\Delta T_{\mathrm{f}}}{m}\)
= \(\frac{3}{0.5}\)
= 6 K kg mol^-1
Ans. Molal depression constant = 6 K kg mol^-1

(4) The freezing point of pure benzene is 278.4 K. Calculate the freezing point of the solution when 2g of a solute having molecular weight 100 is added to 100 g of benzene. (K_f for benzene = 5.12 K kg mol^-1)
Solution :
Given : K_f for benzene = 5.12 K kg mol^-1
T₀ = Freezing point of the solvent = 278.4 K
W₁ = Mass of the solvent = 100 g = 0.1 kg
W₂ = Mass of the solute = 2g = 2 × 10^-3 kg
M₂ = Molecular mass of the solute = 100 g mol^-1
= 100 × 10^-3 kg mol^-1
T = Freezing point of the solution = ?

Ans. Freezing point of the solution = 277.376 K

(5) 0.635 × 10^-3 kg of a substance of molar mass 190 × 10^-3 kg mol^-1 was dissolved in 30.5 × 10^-3 kg of a solvent. If the depression in the freezing point is 0.62 °C, find the molal depression constant of the solvent.
Solution :
Given : Mass of solvent = W₁ = 30.5 × 10^-3 kg
Mass of solute = W₂ = 0.635 × 10^-3 kg
Depression in freezing point = ΔT_f = 0.62 °C (or K)
Molar mass of the substance = M₂ = 190 × 10^-3 kg mol^-1
Molal depression constant = K_f = ?

Ans. Molal depression constant = K_f
= 5.66 K kg mol^-1

(6) The boiling point of an aqueous solution is 100.18 °C. Find the freezing point of the solution. (Given : K_b = 0.52 K kg mol^-1, K_f = 1.86 K kg mol^-1)
Solution :
Given : T_b = 100.18 °C + 273 = 373.18 K
K_b = 0.52 K kg mol^-1; K_f = 1.86 K kg mol^-1
Boiling point of water = T₀ = 373 K
T_f = ?
ΔT_b = T_b – T₀ = 373.18 – 373 = 0.18 K
If m is the molality of the solution, then
ΔT_b = K_b × m and ΔT_f = K_f × m

Freezing point of water = T₀ = 273 K
ΔT_f = T₀ – T_f
Hence the freezing point of the solution is,
T_f = T₀ – ΔT_f = 273 – 0.6438 = 272.3562 K
OR Freezing point of solution is -0.6438 °C.
Ans. Freezing point of the solution
= 272.3562 K
= -0.6438 °C

(7) 1.0 × 10^-3 kg of urea when dissolved in 0.0985 kg of a solvent, decreases freezing point of the solvent by 0.211 K. 1.6 × 10^-3 kg of another nonelectrolyte solute when dissolved in 0.086 kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of the another solute. (Given molar mass of urea = 60)
Solution :
Given : Mass of Urea = W₂ = 1 × 10^-3 kg
Mass of solvent = W₁ = 0.0985 kg
Depression in freezing point = ΔT_f = 0.211 K
Molar mass of urea = M₂ = 60 g mol^-1
= 60 × 10^-3 kg mol^-1
Mass of another solute = W’₂ = 1.6 × 10^-3 kg
Mass of solvent = W’₁ = 0.086 kg
Depression in freezing point = ΔT’_f = 0.34 K
Molar mass of another solute = 60 × 10^-3 kg mol^-1

From urea solution:


For solution of another solute :

Ans. Molar mass of solute = 68.23 g mol^-1.

Question 52.
What do you understand by the terms :
(1) permeable membrane
(2) semipermeable membrane?
Answer:
(1) Permeable membrane : A membrane which allows free transfer of the solute molecules from a solution of a higher concentration to a solution of a lower concentration through it is called a permeable membrane and the transfer is called diffusion, e.g., a membrane of a paper.

(2) Semipermeable membrane : A membrane which allows free passage of only the solvent molecules but not the large solute molecules or ions of large molecular mass from a solution of a lower concentration (or a pure solvent) to a solution of higher concentration through it, is called a semi-permeable membrane, e.g., parchment paper, complex like Cu₂[Fe(CN)₆], etc.

Question 53.
Define and explain osmosis.
Answer:
(1) Definition : It is defined as a spontaneous uni-directional flow of the solvent molecules from a pure solvent or a dilute solution to the more concentrated solution through a semipermeable membrane.

Example : A flow of water molecules from a dilute solution into a concentrated glucose solution through a parchment paper.

(2) Explanation : Consider a vessel divided into two compartments by a semipermeable membrane. When one compartment is filled with a pure solvent or a dilute solution and another by concentrated solution, there is a spontaneous of solvent molecules to the concentrated solution. This arises due to higher vapour pressure of a pure solvent or dilute solution than concentrated solution.

Question 54.
Explain the relation between osmotic pressure and concentration of solution.
Ans.
(1) Consider V dm³ of a solution in which n₁ moles of a solvent contains n₂ moles of a nonvolatile solute at absolute temperature T.

(2) The osmotic pressure, n of a solution is given by,
π = \(\frac{n R T}{V}\)
R is gas constant having value 0.08206 dm³ atm K^-1 mol^-1 (OR L atm K^-1 mol^-1). Since concentration, C of a solution is in mol dm^-3 or molarity is,
C = \(\frac{n}{V}\) mol dm^-3 or M
∴ π = CRT
(If concentration C is expressed in mol m^-3 and R = 8.314 J K^-1mol^-1, then π will be in SI units, pascals or Nm^-2.)

Question 55.
Explain the terms :
(1) Isotonic solutions
(2) Hypotonic solutions
(3) Hypertonic solutions.
Answer:
(1) Isotonic solutions : The solutions having the same osmotic pressure at a given temperature are called isotonic solutions.

Explanation : If two solutions of substances A and B contain n_A and n_B moles dissolved in volume V (in dm³) of the solutions, then their concentrations are,
C_A = \(\frac{n_{\mathrm{A}}}{V}\) (in mol dm^-3) and
C_B = \(\frac{n_{\mathrm{B}}}{V}\) (in mol dm^-3)
If the absolute temperature of both the solutions is T, then by the van’t Hoff equation,
π_A = C_ARTand π_B = C_BRT where π_A and π_B are their osmotic pressures.
For the isotonic solutions,
π_A = π_B
∴ C_A = C_B
∴ \(\frac{n_{\mathrm{A}}}{V}=\frac{n_{\mathrm{B}}}{V}\)
∴ n_A = n_B
Hence, equal volumes of the isotonic solutions at the same temperature will contain equal number of moles (hence, equal number of molecules) of the substances.

(2) Hypotonic solutions : When two solutions have different osmotic pressures, then the solution having lower osmotic pressure is said to be a hypotonic solution with respect to the other solution.
Explanation : Consider two solutions of the substances A and B having osmotic pressures π_A and π_B. If π_B is less than π_A, then the solution B is a hypotonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then, C_B < C_A. Hence, for equal volumes of the solutions, n_A < n_A.

(3) Hypertonic solutions : When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.
Explanation : Consider two solutions of substances A and B having osmotic pressures π_A and π_B. If π_B is greater than π_A, then the solution B is a hypertonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then C_B > C_A. Hence, for equal volume of the solutions, n_B > n_A.

Question 56.
Explain colligative properties of electrolytes.
Answer:

• The electrolytic solutions do not exhibit colligative properties similar to nonelectrolytes.
• The colligative properties of electrolytes are higher than those shown by equimolar solutions of nonelectrolytes.
• The molar masses of electrolytes determined by colligative properties are found to be considerably lower than their actual molar masses.

Question 57.
Why are the colligative properties of electrolytic solutions greater than those for non-electrolytic solutions with same concentration ?
Answer:

• The electrolytes in polar solvents or aqueous solutions dissociate into two or more ions whereas nonelectrolytes do not dissociate.
• Consequently the number of particles in electrolytic solutions are considerably higher than equimolar nonelectrolytic solutions.
• Therefore the colligative properties of electrolytes are greater than nonelectrolytes with same concentration in solution.

Question 58.
What is abnormal colligative property? Explain the reasons.
Answer:
Abnormal colligative property : When the experimentally measured colligative property of a solution is different from that calculated theoretically by the van’t Hoff equation or by the laws of osmosis, then the solution is said to have abnormal colligative property.

Explanation : The colligative property depends on the number of solute particles in the solution but it is independent of their nature. Abnormal values of them arise when the dissolved solute undergoes a molecular change like dissociation or association in the solution.
The observed colligative property (or abnormal colligative property ) may be higher or lower than the theoretical value.

(i) Dissociation of the solute molecules : When a solute like an electrolyte is dissolved in a polar solvent like water, it undergoes dissociation, which results in the increase in the number of particles in the solution.

Hence, the observed value of the colligative property becomes higher than the theoretical value, e.g., when one mole of KCl is dissolved in the solution then due to dissociation, KCl → K⁺ + Cl^–, the number of particles increases, hence, the colligative properties like osmotic pressure elevation in the boiling point, etc. increase.

(ii) Association of the solute molecules : When a solute like a nonelectrolyte is dissolved in a nonpolar solvent like benzene, it undergoes association forming molecules of higher molecular mass. Hence, the number of the particles in the solution decreases. Therefore the colligative properties like osmotic pressure, elevation in the boiling point, etc., are lower than the theoretical value, e.g., nA → An.
2CH₃COOH → (CH₃COOH)₂
2C₆H₅COOH → (C₆H₅COOH)₂

Question 59.
Explain abnormal osmotic pressure.
Answer:

• When the experimentally observed osmotic pressure is different than theoretically calculated value by van’t Hoffs equation then it is called abnormal osmotic pressure.
• This arises when the dissolved solute undergoes a molecular change like association or dissociation.

Question 60.
Explain abnormal molecular masses.
Answer:
When the observed molecular masses obtained from their colligative properties of the substances are different (higher or lower) than the theoretical or normal values calculated from their molecular formulae, then they are called abnormal molecular masses.

Question 61.
How is van’t Hoff factor related to molecular mass of the substance ?
Answer:
The van’t Hoff factor is also defined as,
actual moles of particles in solution after


In case of nonelectrolytes, i < 1.
In case of electrolytes, i > 1. For example for KNO₃ and NaCl, i = 2, for Na₂SO₄, CaCl₂, i = 3, etc.

Question 62.
Write modification of expressions of colligative properties with the help of van’t Hoff factor.
Answer:
The modified expressions of colligative properties with the help of van’t Hoff factor i are as follows :
(1) By Raoult’s law :

Question 63.
Obtain a relation between degree of dissociation and molar mass for an electrolyte.
Answer:
Consider an electrolyte AxBy.

If i is van’t Hoff factor and n is total number of ions produced from dissociation of one mole of electrolyte then, the degree of dissociation α is given by,
α = \(\frac{i-1}{n-1}\)
Now if M_th and M_ob are theoretical arid observed molecular (or molar) masses respectively, then,
i = \(\frac{M_{\mathrm{th}}}{M_{\mathrm{ob}}}\)
∵ α(n – 1) = i – 1

This is a relation between degree of dissociation and molecular masses of the dissolved electrolyte AxBy.

Question 64.
Solve the following :

(1) 300 mL solution at 27 °C contains 0.2 mol of a nonvolatile solute. Calculate osmotic pressure of the solution.
Solution :
Given : V= 300 ml = 0.3 dm³ ;
T= 273 + 27 = 300 K
π = 0.2 mol, π = ?
π = \(\frac{n R T}{V}\)
= \(\frac{0.2 \times 0.08206 \times 300}{0.3}\)
= 16.41 atm
Ans. Osmotic pressure = π = 16.41 atm.

(2) 200 mL solution at 27 °C contains 10 g of a nonvolatile solute of molar mass 65 g mol^-1. What is osmotic pressure of the solution ?
Solution :
Given : V = 200 mL = 0.2 dm³; W = 10 g
M = 65 g mol^-1; T= 273 + 27 = 300 K; π = ?

Ans. Osmotic pressure = π = 18.93 atm.

(3) Calculate the osmotic pressure of 0.2 M glucose solution at 300 K. (R = 8.314 J mol^-1 K^-1)
Solution :
Given : Concentration of the solution
= C = 0.2 M
= 0.2 mol dm^-3
= 0.2 × 10³ mol m^-3
Temperature = T= 300 K
= 8.314 J mol^-1 K^-1
The osmotic pressure, π is given by,
π = CRT
= 0.2 × 10³ × 8.314 × 300
= 4.988 × 10⁵ Nm^-2 (or Pa)
Ans. Osmotic Pressure = 4.988 × 10⁵ Nm^-2

(4) A solution of cane sugar containing 18 g L^-1 has an osmotic pressure 1.25 atm. Calculate the temperature of the solution. (Molar mass of cane sugar = 342, R = 0.082 lit atm mol^-1 K^-1)
Solution:
Given : Amount of cane sugar = W = 18 g L^-1
Osmotic pressure = π = 1.25 atm
Molar mass of cane sugar = M = 342 g mol^-1
Temperature = T = ?
Number of moles of cane sugar
= \(\frac{W}{M}\)
= \(\frac{18}{342}\)
= 0.05263 mol
∴ Concentration of solution = C
= \(\frac{n}{V}\)
= \(\frac{0.05263}{1}\)
= 0.05263 mol lit^-1
π = CRT
∴ T = \(\frac{\pi}{\mathcal{C R}}\)
= \(\frac{1.25}{0.05263 \times 0.08206}\)
= 289.4 K
Ans. Temperature of solution = 289.4 K

(5) Equal volumes of two solutions, one containing glucose and another urea have osmotic pressures 12.6 atm and 23.8 atm at 25 °C. Calculate the ratio of number of moles of urea to glucose.
Solution :
Given : Osmotic pressure of glucose solution = π₁
= 12.6 atm
Osmotic pressure of urea solution = π₂
= 23.8 atm
Number of moles of glucose = n₁
Number of moles of urea = n₂

(6) One litre of a solution containing 25 g glucose has osmotic pressure 3.4 atm at 300 K. If by diluting the solution the osmotic pressure becomes 2.0 atm at the same temperature, what would be its concentration ? (Molar mass of glucose = 180)
Solution :
Given : Mass of solute (glucose) = 25 g
Volume of solution = 1.0 L
Osmotic pressure = π₁ = 3.4 atm
After dilution osmotic pressure = π₂ = 2 atm
Final concentration = c₂ = ?
The number of moles of glucose = n = \(\frac{25}{180}\)
= 0.1389 mol
Initial concentration = c₁ = \(\frac{n}{V}\)
= \(\frac{0.1389}{1}\)
At constant temperature, by van’t Hoff-Boyle’s law,

Ans. Concentration of solution = 0.0817 mol L^-1

(7) A solution of a substance having mass 1.8 × 10^-3 kg has the osmotic pressure of 0.52 atm at 280 K. Calculate the molar mass of the substance used.
[Volume = 1 dm³, R = 8.314 JK^-1 mol^-1]
Solution :
Given : W= 1.8 × 10^-3 kg = 1.8 g
π = 0.52 atm; V = 1 dm³
T = 280 K;
Molar mass = M = ?

Ans. Molar mass = M = 79.53 g mol^-1

(8) An organic substance (M = 169 gram mol^-1) is dissolved in 2000 cm³ of water. Its osmotic pressure at 12 °C was found to be 0.54 atm. If R = 0.0821 L atm K^-1 mol^-1, calculate the mass of the solute.
Solution :
Given : M = 169 g mol^-1
V = 200 cm3 = 0.2 dm³
T = 273 + 125 = 285 K
π = 0.54 atm
R = 0.0821 L atm K^-1 mol^-1; W = ?

Ans. Mass of solute = 0.78 g

(9) A solution containing 10 g glucose has osmotic pressure 3.84 atm. If 10 g more glucose is added to the same solution, what will be its osmotic pressure. (Temperature remains constant)
Solution :
Given : Mass of glucose = W₁ = 10 g
Mass of more glucose added = W₂ = 10 g
Initial osmotic pressure = π₁ = 3.84 atm
Final osmotic pressure = π₂ = ?
Total mass of glucose in final solution = W₁ + W₂
W’ = 10 + 10
= 20 g

∴ π₂ = \(\pi_{1} \times \frac{W^{\prime}}{W_{1}}\)
= \(3.84 \times \frac{20}{10}\) = 7.68 atm
Ans. Osmotic pressure of the solution = 7.68 atm

(10) A 0.1 m solution of K₂SO₄ in water has freezing point of -0.43 °C. What is the value of van’t Hoff factor if K_f for water is 1.86 K kg mol^-1?
Solution :
Given : m = 0.1 m, ΔT_f = 0 – (-0.43) = 0.43 °C
K_f = 1.86 K kg mol^-1, i = ?
ΔT_f = i × K_f × m
∴ i = \(\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}=\frac{0.43}{1.86 \times 0.1}=2.312\)
Ans. van’t Hoff factor = i = 2.312

Multiple Choice Questions

Question 65.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. A molal solution is one that contains one mole of solute in
(a) one litre of the solvent
(b) 1000 g of the solvent
(c) one litre of the solution
(d) 22.4 litre of solution
Answer:
(b) 1000 g of the solvent

2. 10.0 grams of caustic soda when dissolved in 250 cm³ of water, the resultant molarity of solution is
(a) 0.25 M
(b) 0.5 M
(c) 1.0 M
(d) 0.1 M
Answer:
(c) 1.0 M

3. 5 × 10-3 kg of urea is dissolved in 2 × 10^-2 kg of water. The percentage by weight of urea is
(a) 15%
(b) 20%
(c) 25%
(d) 30%
Answer:
(b) 20%

4. Vapour pressure of solution of a nonvolatile solute is always
(a) equal to the vapour pressure of pure solvent
(b) higher than vapour pressure of pure solvent
(c) lower than vapour pressure of pure solvent
(d) constant
Answer:
(c) lower than vapour pressure of pure solvent

5. According to the Raoult’s law, the relative lowering of vapour pressure is equal to the
(a) mole fraction of solvent
(b) mole fraction of solute
(c) independent of mole fraction of solute
(d) molality of solution
Answer:
(b) mole fraction of solute

6. Partial pressure of solvent in solution of non-volatile solute is given by equation,
(a) P = x₂P⁰
(b) P⁰ = xP
(c) P = x₁P⁰
(d) P⁰ = x₁P
Answer:
(c) P = x₁P⁰

7. When partial pressure of solvent in solution of nonvolatile solute is plotted against its mole fraction, nature of graph is
(a) a straight line passing through origin
(b) a straight line parallel to mole fraction of solvent
(c) a straight line parallel to vapour pressure of solvent
(d) a straight line intersecting vapour pressure axis
Answer:
(a) a straight line passing through origin

8. Colligative property depends only on ………………. in a solution.
(a) Number of solute particles
(b) Number of solvent particles
(c) Nature of solute particles
(d) Nature of solvent particles
Answer:
(a) Number of solute particles

9. Lowering of vapour pressure of solution
(a) is a property of solute
(b) is a property of solute as well as solvent
(c) is a property of solvent
(d) is a colligative property
Answer:
(d) is a colligative property

10. Raoult’s law is not applicable to
(a) solutions of volatile solutes
(b) solutions of electrolytes
(c) dilute solutions
(d) concentrated solutions
Answer:
(d) concentrated solutions

11. The relative lowering of vapour pressure of a solution is proportional to the
(a) mole fraction of the solvent
(b) mole fraction of the solute
(c) amount of the substance
(d) volume of the solvent
Answer:
(b) mole fraction of the solute

12. The vapour pressure of an aqueous solution of glucose at 100 °C is 710 mm Hg. Hence the molality of the solution is
(a) 2.83 m
(b) 3.65 m
(c) 16.47 m
(d) 12.5 m
Answer:
(b) 3.65 m

13. Relative vapour pressure lowering depends only on
(a) Mole fraction of solute
(b) Nature of solvent
(c) Nature of solute
(d) Nature of solute and solvent
Answer:
(a) Mole fraction of solute

14. A solution having mole fraction of a solute equal to 0.05 has vapour pressure 20 × 10³ Nm^-2. Hence the vapour pressure of a pure solvent is
(a) 21.05 × 10³ Nm^-2
(b) 4 × 10⁵ Nm^-2
(c) 1 × 10³ Nm^-2
(d) 2 × 10⁴ Nm^-2
Answer:
(a) 21.05 × 10³ Nm^-2

15. The addition of the nonvolatile solute into the pure solvent ……………..
(a) increases the vapour pressure of solvent
(b) decreases the boiling point of solvent
(c) decreases the freezing point of solvent
(d) increases the freezing point of solvent
Answer:
(c) decreases the freezing point of solvent

16. A solution having the highest vapour pressure is
(a) 1 M Al₂(SO₄)₃
(b) 0.1 M NaNO₃
(c) 1 M BaCl₂
(d) 1 M Ca(NO₃)₂
Answer:
(b) 0.1 M NaNO₃

17. Molal elevation constant is elevation in boiling point produced by
(a) 1 g of solute in 100 g of solvent
(b) 100 g of solute in 1000 g of solvent
(c) 1 mole of solute in one litre of solvent
(d) 1 mole of solute in one kg of solvent
Answer:
(d) 1 mole of solute in one kg of solvent

18. The determination of molar mass from elevation in boiling point is called
(a) cryoscopy
(b) osmometry
(c) ebullioscopy
(d) spectroscopy
Answer:
(c) ebullioscopy

19. Which of the following aqueous solutions will have minimum elevation in boiling point ?
(a) 0.1 M KCl
(b) 0.05 M NaCl
(c) 1 M AIPO₄
(d) 0.1 M MgSO₄
Answer:
(b) 0.05 M NaCl

20. Which of the following solutions shows maximum depression in freezing point ?
(a) 0.5 M Li₂SO₄
(b) 1 M NaCl
(c) 0.5 M Al₂ (SO₄)₃
(d) 0.5 M BaCl₂
Answer:
(c) 0.5 M Al₂ (SO₄)₃

21. If the freezing point of 0.1 m solution is 272.814 K, then the freezing point of 0.2 m solution will be
(a) 545.628 K
(b) 265.64 K
(c) 272.628 K
(d) 0.482 K
Answer:
(c) 272.628 K

22. At a freezing point,
(a) Vapour pressure of a solution = Vapour pressure of a solid
(b) Vapour pressure of a liquid = Vapour pressure of a solid
(c) Vapour pressure of a liquid > Vapour pressure of a solid
(d) Vapour pressure of a solid > Vapour pressure of a liquid
Answer:
(b) Vapour pressure of a liquid = Vapour pressure of a solid

23. In osmosis
(a) solvent molecules pass from high concentration of solute to low concentration
(b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute
(c) solute molecules pass from low concentration to high concentration
(d) solute molecules pass from high concentration to low concentration
Answer:
(b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute

24. As temperature increases
(a) Osmotic pressure and vapour pressure decrease
(b) Vapour pressure and osmotic pressure increase
(c) Vapour pressure increases but osmotic pressure decreases
(d) Osmotic pressure increases but vapour pressure decreases
Answer:
(b) Vapour pressure and osmotic pressure increase

25. A mango kept in a salt solution shrinks. Hence the liquid content in mango with respect to the salt solution is
(a) isotonic
(b) hypertonic
(c) hypotonic
(d) equimolar
Answer:
(c) hypotonic

26. The osmotic pressure of 5% glucose solution at 300 K is
(a) 6.93 × 10⁵ Nm^-2
(b) 6.93 × 10² Nm^-2
(c) 6.83 Pa
(d) 6.00 × 10³ Pa
Answer:
(a) 6.93 × 10⁵ Nm^-2

27. 0.1 M solution of A has osmotic pressure xNm^-2 at 300 K. If 200 ml of A and 100 ml of 0.2 M solution of nonreactive solute B are mixed then the osmotic pressure will be
(a) 3x
(b) 0.05 x
(c) 1.33 x
(d) 0.75 x
Answer:
(c) 1.33 x

28. The solution A is twice hypertonic to the solution B at a given temperature. If the solution A contains 8.6 × 10²² molecules, then the number of molecules present in B are,
(a) 8.6 × 10²²
(b) 1.73 × 10²³
(c) 3.24 × 10²²
(d) 4.3 × 10²²
Answer:
(d) 4.3 × 10²²

29. Isotonic solutions are the solutions having the same
(a) surface tension
(b) vapour pressure
(c) osmotic pressure
(d) viscosity
Answer:
(c) osmotic pressure

30. If Kb for water is 0.52 Km^-1, the boiling point of 0.2 m solution of a nonvolatile solute will be
(a) 371.96 K
(b) 373.104 K
(c) 373.52 K
(d) 374.0 K
Answer:
(b) 373.104 K

31. The vapour pressure of water is 15.5 mm at 20 °C. The lowering of vapour pressure of 0.02 m K Br solution will be
(a) 0.0112 mm Hg
(b) 0.0056 mm Hg
(c) 0.056 mm Hg
(d) 0.31 mm Hg
Answer:
(a) 0.0112 mm Hg

32. Which of the following 0.1 M aqueous solutions will exert highest osmotic pressure ?
(a) Al₂(SO₄)₃
(b) Na₂SO₄
(c) MgCl₂
(d) KCl
Answer:
(a) Al₂(SO₄)₃

33. If equimolar solutions of urea, NaCl, sucrose and BaCl₂ have boiling points A, B, C and D, then
(a) A = C < B < D
(b) A = D < B < C
(c) A > B > C < D
(d) A < B < C < D
Answer:
(a) A = C < B < D

34. The molar mass of acetic acid obtained by measuring depression in freezing point is 115.8 gmol^-1. Hence the degree of association is
(a) 0.482
(b) 0.964
(c) 0.883
(d) 1.12
Answer:
(b) 0.964

35. Δ T_b/m for NaBr solution will have value (K_b = 0.52 K kg mol^-1)
(a) 0.52 K mol^-1
(b) 0.104 K kg mol^-1
(c) 1.24 kg mol^-1
(d) 1.04 K kg mol^-1
Answer:
(d) 1.04 K kg mol^-1

36. 0.2 M urea solution can be isotonic with
(a) 0.1 M KBr solution
(b) 0.1 M glucose solution
(c) 0.15 m NaCl solution
(d) 0.2 M KBr solution
Answer:
(a) 0.1 M KBr solution

37. The osmotic pressure of 0.1 M HCl solution at 27 °C will be
(a) 2.46 atm
(b) 0.164 atm
(c) 4.92 atm
(d) 0.0082 atm
Answer:
(c) 4.92 atm

38. If a, b, c and d are the van’t Hoff factors for Na₂SO₄, glucose and K₄[Fe(CN)₆] then
(a) a > b > c
(b) a < b < c
(c) b < a < c
(d) c < a < b
Answer:
(c) b < a < c

39. The osmotic pressure of 0.2 M KCl solution at 310 K is
(a) 10.17 atm
(b) 5.084 atm
(c) 8.36 atm
(d) 12.2 atm
Answer:
(a) 10.17 atm

40. A temperature at which 0.1 M KCl solution will have osmotic pressure 10 atm will be
(a) 408 °C
(b) 263 °C
(c) 310 °C
(d) 337 °C
Answer:
(d) 337 °C

41. 5 % solution of glucose is isotonic with a solution of urea (M = 60). Hence the weight of urea present in the solution is
(a) 1.67 g
(b) 6.0 g
(c) 18.6 g
(d) 1.2 g
Answer:
(a) 1.67 g

42. Abnormal molar mass is produced by
(a) association of solute
(b) dissociation of solute
(c) both association and dissociation of solute
(d) separation by semipermeable membrane
Answer:
(c) both association and dissociation of solute

43. The van’t Hoff factor for an aqueous solution of an electrolyte is
(a) less than 1
(b) zero
(c) greater than 1
(d) equal to 1
Answer:
(c) greater than 1

44. The value of van’t Hoff factor will be minimum for
(a) 0.05 M AlCl₃
(b) 0.2 M NaNO₃
(c) 5.0 M glucose
(d) 0.1 M H₂SO₄
Answer:
(c) 5.0 M glucose

45. van’t Haff factor for K₄[FeC(N)₆] dissociated 10% is
(a) 1.1
(b) 1.4
(c) 0.86
(d) 1.6
Answer:
(b) 1.4