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Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 13 Amines Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 13 Amines

PSEB 12th Class Chemistry Guide Amines InText Questions and Answers

Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.
(i) (CH₃)₂CHNH₂
(ii) CH₃(CH₂)₂NH₂
(iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃CNH₂
(v) C₆H₅NHCH₃
(vi) (CH₃CH₂)₂NCH₃
(vii) m-BrC₆H₄NH₂
Answer:
(i) Propan-2-amine (1° amine)
(ii) Propan-1-amine (1° amine)
(iii) N-Methylpropan-2-amine (2° amine)
(iv) 2-Methylpropan-2-amine (1° amine)
(v) AT-Methylbenzenamine or N-methylaniline (2° amine)
(vi) N-Ethyl-N-methyl ethan amine (3° amine)
(vii) 3-Bromobenzenamine or 3-bromoaniline (1° amine)

Question 2.
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylsunine and aniline
(iv) Aniline and benzylrnnlne
(v) Aniline and N-inethylaniline.
Answer:
(i) Methylamlne is 1° amine, therefore, it gives carbylamine test, i.e., when heated with an alcoholic solution of KOH and CHCl₃, it gives an offensive smell of methyl carbylamine. In contrast, dimethylamine is a secondary amine and hence does not give this test.

(ii) Secondary amines (both aliphatic and aromatic) react with nitrous acid to form nitrosamines (yellow oily liquid)

When the yellow oily liquid is warmed with a crystal of phenol and a few drops of conc. H₂SO₄, a greenish solution is formed. It changes to red on dilution with water but changes to deep blue on addition of aqueous NaOH solution. Tertiary amines do not give this test.

(iii) Ethylamine (primary aliphatic amine) and aniline (primary aromatic amine) can be distinguished by azo dye test.
Aniline responds to this test whereas ethyl amine does not.

(iv) Benzylantine reacts with nitrous acid to form a diazonium salt which being unstable even at low temperature, decomposes with evolution of N₂ gas.

Aniline reacts with HNO₂ to form benzene diazonium chloride which is stable at 273-278 K and hence does not decompose to evolve N₂ gas.

(v) Aniline being a primary amine gives carbylamine test, i.e., when heated with an alcoholic solution of KOH and CHCl₃, it gives an offensive smell of phenyl isocyanide. In contrast, N-methyl aniline, being secondary amine does not give this test.

Question 3.
Account for the following:
(i) pK_b of aniline is more than that of methylamine.
(ii) Ethylamme is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although the amino group is o-and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m -nitroaniline.
(v) Aniline does not undergo Friedel Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesizing primary amines.
Answer:
(i) In aniline due to resonance, the lone pair of electrons on the N-atom are delocalized over the benzene ring. Due to this, electron density on the nitrogen decreases. On the other hand, in CH₃NH₂, +I effect of CH₃ increases the electron density on the N-atom. Consequently, aniline is a weaker base than methylamine and hence its pK_b value is higher than that of methylamine.

(ii) Ethylamine dissolves in water because it forms H-bonds with water molecules as shown below :

In aniline, due to the large hydrocarbon part the extent of H-bonding decreases considerably and hence aniline is insoluble in water.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated fem oxide:

Due to the +I effect of – CH₃ group, methylamine is more basic than water. Therefore, in water, methylamine produces OH^– ions by accepting H⁺ ions from water.

Ferric chloride (FeCl₃) dissociates in water to form Fe³⁺ and Cl^– ions.
FeCl₃ → Fe³⁺+ +3Cl^–
Then, OH^– ion reacts with Fe³⁺ ion to form a precipitate of hydrated ferric oxide.
2Fe³⁺ +6OH^– → Fe₂O₃ -3H₂O Hydrated ferric oxide

(iv) Nitration is usually carried out with a mixture of cone. HNO₃ and cone. H₂SO₄.
In presence of these acids, most of aniline gets protonated to form anilinium ion. Thus, in presence of acids, the reaction mixture consists of aniline and anilinium ion. The -NH₂ group in aniline is o, p-directing and activating while the -NH₃ group in anilinium ion is m-directing and deactivating.
Nitration of aniline mainly gives p-nitroaniline. On the other hand, the nitration of anilinium ion gives m-nitroaniline.

Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of the amino group.

(v) Aniline being a Lewis base, reacts with Lewis acid AlCl₃ to form a salt.

As a result, N of aniline acquires positive charge and hence, it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Friedel Crafts’ reaction.
(vi) The diazonium ion undergoes resonance as shown below:

This resonance accounts for the stability of the diazonium ion. Hence, diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis results in the formation of 1° amine only. 2° or 3° amines are not formed in this synthesis. Thus, a pure 1° amine can be obtained. Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

Question 4.
Arrange the following:
(i) In decreasing order of the pK_b values :
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂
(ii) In increasing order of basic strength :
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂
(iii) In increasing order of basic strength :
(a) Aniline, p-nitroaniline andp-toluidine
(b) C₆H₅NH₂, C₆H5NHCH₃, C₆H₅CH₂NH₂.
(iv) In decreasing order of basic strength in gas phase:
C₂H₅NH₂,(C₂H₅)₂NH,(C₂H₅)₃N and NH₃
(v) In increasing order of boiling point:
C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
(vi) In increasing order of solubility in water :
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂.
Answer:
(i) C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH
(ii) C₆H₅NH₂C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH

(b) C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂.
(iv) (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃.
(v) (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH
(vi) C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂.

Question 5.
How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-amino pentane
(iii) Methanol to ethanoic acid
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid
Answer:

Question 6.
Describe a method for the identification of primary, secondary, and tertiary times. Also, write chemical equations of the reactions involved.
Answer:
Primary, secondary and tertiary amines can be identified and distinguished by Hmsberg’s test. In this test, the amines are allowed to react with Hinsberg’s reagent, benzene sulphonyl chloride (C₆H₅SO₂Cl). The three types of amines react differently with Hinsbergs reagent. Therefore, they can be easily identified using Hinsberg’s reagent. Primary amines react with benzene sulphonyl chloride to form N-allyl benzene sulphonyl amide which is soluble in alkali.

Due to the presence of a strong electron-withdrawing sulphonyl group in the sulphonamide, the H-atom attached to nitrogen can be easily released as proton. So, it is acidic and dissolves in alkali. Secondary amines react with Hinsberg’s reagent to give a sulphonamide which is insoluble in alkali.

There is no H-atom attached to the N-atom in the sulphonamide. Therefore, it is not acidic and insoluble in alkali.
On the other hand, tertiary amines do not react with Hinsberg’s reagent at all.

Question 7.
Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hofmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Acetylation
(vii) Gabriel phthalimide synthesis.
Answer:
(i) Carbylamine reaction: Carbylamine reaction is used as a test for the identification of primary amines. When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed. These carbylamines have very unpleasant odours. Secondary and tertiary amines do not respond to this test.

(ii) Diazotisation: Aromatic primary amines react with nitrous acid (prepared in situ from NaNO₂ and a mineral acid such as HCl) at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotization. For example, on treatment with NaNO₂ and HCl at 273 – 278 K, aniline produces benzene diazonium chloride, with NaCl and H₂O as by-products.

(iii) Hofmann’s bromamide reaction: When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. This degradation reaction is known as Hofmann bromamide reaction. This reaction involves the migration of an alkyl or aryl group from the carbonyl carbon atom of the amide to the nitrogen atom.

(iv) Coupling reaction: The reaction of joining two aromatic rings through the -N=N- bond is known as a coupling reaction. Arenediazonium salts such as benzene diazonium salts react with phenol or aromatic amines to form colored azo compounds.

It can be observed that the para-positions of phenol and aniline are coupled with the diazonium salt. This reaction proceeds through electrophilic substitution.

(v) Ammonolysis: When an alkyl or benzyl halide is allowed to react with an ethanolic solution of ammonia, it undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (- NH₂) group. This process of cleavage of the carbon-halogen bond is known as ammonolysis.

When this substituted ammonium salt is treated with a strong base such as sodium hydroxide, amine is obtained.

Though primary amine is produced as the major product, this process produces a mixture of primary, secondary and tertiary amines, and also a quaternary ammonium salt as shown.

(vi) Acetylation: The process in which acetyl (CH₃CO -) group is introduced in a molecule, is called acetylation, reagents used for this purpose are acetyl chloride or acetic anhydride.
(vii) Gabriel phthalimide synthesis: Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines.

It involves the treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide. This salt is further heated with alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

(vii) Gabriel phthalimide synthesis: Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines. It involves the treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide. This salt is further heated with an alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

Question 8.
Accomplish the following conversions:
(i) Nitrobenzene to benzolc acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2,4,6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanolamine
(vi) Chiorobenzene top-chloroaniline
(vii) Aniline top-bromoaniline
(viii) Benzainide to toluene
(ix) Aniline to benzyl alcohol.
Answer:

Question 9.
Give the structures of A, B and C In the following reactions:

Answer:

Question 10.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C.
Answer:
It is given that compound ‘C’ having the molecular formula, C₆H₇ N is formed by heating compound ‘B’ with Br₂ and KOH. This is a Hofmann bromamide degradation reaction. Therefore, compound ‘B’ is an amide, and compound ‘C’ is an amine. The only amine having the molecular formula, C₆H₇N is aniline, (C₆H₅NH₂).

Therefore, compound ‘B’ (from which ‘C’ is formed) must be benzamide, (C₆H₅CONH₂)

Further, benzamide is formed by heating compound ‘A’ with aqueous ammonia. Therefore, compound ‘A’ must be benzoic acid.

The given reactions can be explained with the help of the following equations:

Question 11.
Complete the following reactions:
(i) C₆H₅ NH₂ + CHCl₃ + alc.KOH →
(ii) C₆H₅N₂Cl+H₃PO₂+H₂O →
(iii) C₆H₅NH₂ + H₂SO₄(conc.) →
(iv) C₆H₅N₂Cl+C₂H₅OH →
(v) C₆H₅NH₂+Br₂(aq) ) →
(vi) C₆H₅NH₂ + (CH₃CO)₂O →

Answer:

Question 12.
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Answer:
Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (S_N2) of alkyl halides by the anion formed by the phthalimide.

But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

Hence, aromatic primary amines cannot be prepared by this process.

Question 13.
Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Answer:
(i) Aromatic amines react with nitrous acid (prepared in situ from NaNO₂ and a mineral acid such as HCl) at 273 – 278 K to form stable aromatic diazonium salts i.e., NaCl and H₂O.

(ii) Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO₂ and a mineral acid such as HCl) to form unstable aliphatic diazonium salts, which further produce alcohol and HCl with the evolution of N₂ gas.

Question 14.
Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Answer:
(i) Amines are less acidic than alcohols of comparable molecular masses because N – H bond is less polar than O – H bond. Hence, amines release H⁺ ion with more difficulty as compared to alcohol.

(ii) Intermolecular hydrogen bonding is present in primary amines but not in tertiary amines (H-atom absent in amino group) so primary amines have higher boiling point than tertiary amines.

(iii) Aliphatic amines are stronger bases than aromatic amines due to following reasons :
(a) Electron pair on the nitrogen atom of aromatic amines is involved in conjugation with the π-electron pairs of the ring as follows :

(b) Anilinium ion obtained by accepting a proton is less stabilized by resonance.

So, aniline is a weaker base than alkyl amines, in which the +1 effect increases the electron density on the nitrogen atom.

Chemistry Guide for Class 12 PSEB Amines Textbook Questions and Answers

Question 1.
Classify the following amines as primary, secondary or tertiary:

(iii) (C₂H₅)₂CHNH₂
(iv) (C₂H₅)₂NH
Answer:
(i) Primary
(ii) Tertiary
(iii) Primary
(iv) Secondary.

Question 2.
(i) Write structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(ii) Write IUPAC names of all the isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?
Answer:
(i) and (ii)
Primary amines :

Secondary amines:

Tertiary amine:

(iii) Chain isomers : (a) and (d), (b) and (c)
Position isomers : (a) and (b), (f) and (g)
Metamers : (e) and (f)
Functional isomers: All primary amines are functional isomers of secondary and tertiary amines and vice-versa.

Question 3.
How will you convert:
(i) Benzene into Aniline ?
(ii) Benzene into N, N-Dimethylaniline
(iii) Cl-(CH₂)₄ -Cl into hexane-1, 6-diamine?
Answer:

(ii) Benzene is converted into aniline which can be subsequently heated with excess of methyl iodide under pressure to obtain N, N-dimethylaniline.

Question 4.
Arrange the following in increasing order of their basic strength:
(i) C₂H₅NH₂,C₆H₆NH₂,NH₃,C₆H₅CH₂NH₂,(C₂H₅)₂NH
(ii) C₂H₅NH₂, (C₂H₆)₂NH, (C₂H₅)₃N, C₆H₅NH₂
(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂
Answer:
(i) C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂< (C₂H₅)₂NH
(ii) C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH
(iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH

Question 5.
Complete the following acid-base reactions and name the products:
(i) CH₃CH₂CH₂NH₂ + HCl →
(ii) (C₂H₅)₃N + HCl →
Answer:

Question 6.
Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Answer:

Question 7.
Write the chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Answer:

Question 8.
Write structures of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
Answer:
In all, four structural isomers are possible. These are as follows:
Primary amines :

Secondary amInes :

Tertiary amines :

Only primary amines react with HNO₂ to liberate N₂ gas.

Question 9.
Convert:
(i) 3-Methylaniline into 3-nitrotoluene
(ii) Aniline into 1,3,5-tribromobenzene.
Answer:

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 2 Solutions Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 2 Solutions

PSEB 12th Class Chemistry Guide Solutions InText Questions and Answers

Question 1.
Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer:
Homogeneous mixtures of two or more than two components are known as solutions. Solute and solvent are two components of a solution.
There are three types of solutions.
(i) Gaseous solution: The solution in which the solvent is a gas is known as a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.

(ii) Liquid solution : The solution in which the solvent is a liquid is known as a liquid solution. In these solutions, the solute may be gas, liquid, or solid. For example, a solution of ethanol in water is a liquid solution.

(iii) Solid solution : The solution in which the solvent is a solid is known as a solid solution. In these solution, the solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution.

Question 2.
Give an example of a solid solution in which the solute is a gas.
Answer:
In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid Solution will be formed. For example, a solution of hydrogen in palladium is a solid solution in which the solute is a gas.

Question 3.
Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Answer:
(i) Mole fraction: The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component (solute or solvent) to the total number of moles of all the components in the mixture.
i.e., Mole fraction of a component

Mole fraction is denoted by ‘χ’
If in a binary solution, the number of moles of the solute and the solvent are nA and nB respectively, then the mole fraction of the solute in the solution is given by,
χ_A = \(\frac{n_{A}}{n_{A}+n_{B}}\)
Similarly, the mole fraction of the solvent in the solution is given as:
χ_A = \(\frac{n_{B}}{n_{A}+n_{B}}\)

(ii) Molality: Molality (m) is defined as the number of moles of the solute per kilogram of the solvent. It is expressed as:

Unit of molality is mol kg^-1 or m (molal).

(iii) Molarity: Molarity (M) is defined as the number of moles of the solute dissolved in one litre of the solution.

(iv) Mass percentage: The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100 g of the solution.

Question 4.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504gmL ^-1?
Solution:
68% HNO₃ by mass means
Mass of HNO₃ (w) = 68 g
Mass of solution = 100 g
Molar mass of nitric acid (HN0),
(M) = 1 × 1 + 1 × 14 + 3 × l6 = 63g mol^-1
Then, number of moles of HNO₃ = \(\frac{W}{M}\) = \(\frac{68}{63}\) mol
= 1.079 mol
Density of solution = 1.504 g mL^-1
∴ Volume of solution = \(\frac{\text { Mass }}{\text { Density }}\) = \(\frac{100}{1.504}\) mL
66.49 mL
66.49 × 10^-3 L

= \(\frac{1.079 \mathrm{~mol}}{66.49 \times 10^{-3} \mathrm{~L}}\) = 16.23 mol L^-1

Question 5.
A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 gmL^-1, then what shall be the molarity of the solution?
Solution:
10% w/w solution of glucose means 10 g of glucose is present in 100 g of the solution i.e., 90 g of water.
Molar mass of glucose
(C₆H₁₂O₆) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol^-1
Then, number of moles of glucose = \(\frac{10}{180}\)mol = 0.056 mol

Question 6.
How many mL of 0.1 M HCl are required to react completely with lg mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?
Solution:
Let the mass of Na₂CO₃ = x g
Mass of mixture = 1.0 g
Then, the mass of NaHCO₃ = (1 – x) g
Molar mass of Na₂CO₃ = 2 × 23 + 1 × 12 + 3 × 16 =106 g mol^-1
∴ Number of moles of Na₂CO₃ = \(\frac{\text { Mass }}{\text { Molar mass }}\) = \(\frac{x}{106}\) = mol
Molar mass of NaHCO₃ = 1 × 23 + 1 × 1 + 1 × 12 + 3 × 16 = 84g mol^-1
∴ Number of moles of NaHCO₃ = \(\frac{\text { Mass }}{\text { Molar mass }}\) = \(\frac{1-x}{84}\) mol
According to the question, the mixture contains equimolar amounts of Na₂CO3and NaHCO₃
Moles of Na₂CO₃ = Moles of NaHCO₃
\(\frac{x}{106}=\frac{1-x}{84}\)
⇒ 84x =106 -106x
⇒ 190x =106
⇒ x = 0.5579 g
Thus, number of moles of Na₂CO₃ = \(\frac{0.5579}{106}\) mol = 0.0053 mol
Number of moles of NaHCO₃ = \(\frac{1-0.5579}{84}\)mol = 0.0053 mol

To calculate the moles of HCl required
HCl reacts with Na₂CO₃ and NaHCO₃ according to the following equations
2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂
HCl + NaHCO₃ → NaCl + H₂O + CO₂;
1 mol of Na₂CO₃ reacts with 2 mol of HCl.
Therefore, 0.0053 mol of Na₂CO₃ reacts with 2 × 0.0053 mol = 0.0106 mol
Similarly, 1 mol of NaHCO₃ reacts with 1 mol of HCl.
Therefore, 0.0053 mol of NaHCO₃ reacts with 0.0053 mol of HCl.
Total moles of HCl required = (0.0106 + 0.0053) mol
= 0.0159 mol

To calculate the volume of 0.1 MHC1
0.1 mol of 0.1 M HCl present in 1000 mL of HCl
Therefore, 0.0159 mol of HCl will be present in HCl
= \(\frac{1000 \times 0.0159}{0.1}\) mol
= 159 mL

Question 7.
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Solution:
Mass of solute in I solution = \(\frac{25}{100}\) × 300 g = 75 g
Mass of solute in II solution = \(\frac{40}{100}\) × 400 g = 160 g
After mixing both solutions
Total mass of solute = 75 + 160 = 235 g
Total mass of solution = 300 + 400 = 700 g
Therefore, mass percentage (w/w) of the solute in the resulting solution
= \(\frac{235}{700}\) × 100%
= 33.57%
And, mass percentage (w/w) of the solvent in the resulting solution.
= (100 – 33.57)%
= 66.43%

Question 8.
An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂ ) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL ‘1, then what shall be the molarity of the solution?
Solution:
To calculate the molality of the solution
Molar mass of ethylene glycol (C₂H₆O₂)
= 2 x 12 + 6 x 1 + 2 x 16 = 62 g mol^-1
Mass of ethylene glycol = 222.6 g
Number of moles of ethylene glycol = \(\frac{\text { Mass of ethylene glycol }}{\text { Molar mass of ethylene glycol }}\)
= \(\frac{222.6 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 3.59 mol
Mass of water = 200 g = 0.2 kg
Therefore, molality of the solution \(=\frac{\text { No. of moles of ethylene glycol }}{\text { Mass of solvent in kg }}\)
= \(\frac{3.59 \mathrm{~mol}}{0.200 \mathrm{~kg}}\) = 17.95m

To calculate the molarity of the solution
Density of the solution = 1.072 g mL^-1
Mass of solution = Mass of solute + Mass of solvent
222.6 g + 200 g = 422.6 g
∴ Volume of the solution = \(\frac{\text { Mass }}{\text { Density }}\) = \(\frac{422.6 \mathrm{~g}}{1.072 \mathrm{~g} \mathrm{~mL}^{-1}}\)
= 394.22 mL = 0.3942 x 10^-3 L
⇒ Molarity of the solution = \(\frac{\text { Moles of ethylene glycol }}{\text { Volume of solution in litre }}\)
= \(\frac{3.59 \mathrm{~mol}}{0.3942 \times 10^{-3} \mathrm{~L}}\) = 9.1 M

Question 9.
A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.
Solution:
Let the mass of solution be 10⁶ g.
Mass of solute, CHCl₃ = 15 g

(i) Therefore, percent by mass of CHCl₃ = \(\frac{\text { Mass of } \mathrm{CHCl}_{3}}{\text { Mass of solution }}\) × 100%
\(\frac{15}{10^{6}}\) × 100%
= 1.5 × 10^-3%

(ii) Molar mass of chloroform (CHCl₃ ) = 1 × 12 + 1 × 1 + 3 × 35.5
= 119.5g mol^-1

= \(\frac{\frac{15}{119.5} \mathrm{~mol}}{10^{6} \times 10^{-3} \mathrm{~kg}}\) = 1.26 × 10 4 mol^-4 kg^-1
1.26 × 10^-4 m

Question 10.
What role does the molecular interaction play in a solution of alcohol and water?
Answer:
In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol-alcohol and water-water interactions. As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.

Question 11.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Solubility of gases in liquids decreases with an increase in temperature. This is because dissolution of gases in liquids is an exothermic process.
Gas + Liquid → Solution + Heat
Therefore, when the temperature is increased, heat is supplied and the equilibrium shifts backwards, thereby decreasing the solubility of gases.

Question 12.
State Henry’s law and mention some important applications?
Answer:
Henry’s law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and % is the mole fraction of the gas, then Henry’s law can be expressed as:
P = K_Hχ
Where, K_H is Henry’s law constant

Some important applications of Henry’s law are mentioned below:
(i) Bottles are sealed under high pressure to increase the solubility of C02 in soft drinks and soda water.

(ii) Henry’s law states that the solubility of gases increases with an increase in pressure. Therefore, when a scuba diver dives deep into the sea, the increased sea pressure causes the nitrogen present in air to dissolve in his blood in great amounts. As a result, when he comes back to the surface, the solubility of nitrogen again decreases and the dissolved gas is released, leading to the formation of nitrogen bubbles in the blood. This results in the blockage of capillaries and leads to a medical condition known as “bends’ or ‘decompression sickness’. Hence, the oxygen tanks used by scuba divers are filled with air and diluted with helium to avoid bends.

(iii) The concentration of oxygen is low in the blood and tissues of people living at high altitudes such as climbers. This is because at high altitudes, partial pressure of oxygen is less than that at ground level. Low-blood oxygen causes climbers to become weak and disables them from thinking clearly. These are symptoms of anoxia.

Question 13.
The partial pressure of ethane over a solution containing 6.56 × 10^-3g of ethane is 1 bar. If the solution contains 5.00 × 10^-2 g of ethane, then what shall be the partial pressure of the gas?
Solution:
According to Henry’s law the mass of the gas dissolved in solution x Partial pressure (p) (At constant temperature)
(6.56 × 10^-3g) ∝ 1 bar
(5.00 × 10^-2g) ∝ p
or p = \(\frac{5.0 \times 10^{-2} \mathrm{~g}}{6.56 \times 10^{-3} \mathrm{~g}}\) × 1 bar = 7.62 bar

Question 14.
What is meant by positive and negative deviations from Raoult’s law and how is the sign of Δ_mix H related to positive and negative deviations from Raoult’s law?
Answer:
Positive deviation: ‘When the vapour pressure of a solution is higher than the predicted value by Raoult’s law, it is called positive deviation’. In such cases intermolecular interactions between solute and solvent particles (A and B) are weaker than those between solute-solute (A – A) and solvent-solvent (B – B). Hence, the molecules of (A or B) will escape more easily from the surface of solution than in their pure state. Therefore, the vapour pressure of the solution will be higher. Characteristics of a solution showing positive deviation :
(i) P_A > \(p_{A}^{0}\) χ_A; P_B > \(p_{B}^{0}\) χ_B
(ii) ΔH_mix >0;i.e., + ve
(iii) ΔV_mix > 0, i.e., + ve

Examples of solutions showing positive deviation:
(i) Ethyl alcohol and water
(ii) Acetone and carbon disulphide
(iii) Carbon tetrachloride and benzene
(iv) Acetone and benzene

Negative deviation: “When the vapour pressure of a solution is lower than the predicted value by Raoult’s law, it is called negative deviation.’ In case of negative deviation the intermolecular attractive forces between A – A and B – B are weaker than those between A – B. It leads to decrease in vapour pressure resulting in negative deviation.

Characteristics of a solution showing negative deviation:
(i) P_A < \(p_{A}^{0}\) χ_B; P_B < \(p_{B}^{0}\) χ_B
(ii) ΔV_mix < 0; i. e., – ve; because weak A – A and B – B bonds are broken
and strong A – B bonds are formed. Heat is consequently released.
(iii) _mix<0;i.e.,-ve

Examples of solutions showing negative deviation:
(i) HNO₃ and water
(ii) Chloroform and acetone
(iii) Acetic acid and pyridine
(iv) Hydrochloric acid and water

Question 15.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Solution:
Here,
Vapour pressure of the solution at normal boiling point (p₁) = 1.004 bar
Vapour pressure of pure water at normal boiling point (\(p_{1}^{0}\)) = 1.013 bar
Mass of solute, (w₂) = 2 g
Mass of solvent (water), (w₁ ) = 100 – 2 = 98g
Molar mass of solvent (water), (M₁) = 18 g mol^-1
Molar mass of solute (M₂) = ?
According to Raoult’s law,

Hence, the molar mass of the solute is 41.35 g mol^-1.

Question 16.
Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Solution:
Vapour pressure of heptane (\(p_{1}^{0}\)) = 105.2 kPa
Vapour pressure of octane (p\(p_{2}^{0}\)) = 46.8 kPa
Mass of heptane = 26.0 g
Mass of octane = 35 g
Molar mass of heptane (C₇H₁₆) = 7 × 12 + 16 × 1 = 100 g mol^-1;
∴ Number of moles of heptane = \(\frac{26}{100}\) mol = 0.26 mol
Molar mass of octane (C₈H₁₈) = 8 × 12 + 18 × 1 = 114g mol^-1
∴ Number of moles of octane = \(\frac{35}{114}\) mol = 0.31 mol
Mole fraction of heptane, χ₁ = \(\frac{0.26}{0.26+0.31}\) = 0.456
Mole fraction of octane, χ₂ = 1 – 0.456 = 0.544
Now, partial pressure of heptane, p₁ = χ₁ \(p_{1}^{0}\)
= 0.456 × 105.2 = 47.97 kPa
Partial pressure of octane, p₂ = χ₂\(p_{2}^{0}\)
= 0.544 × 46.8 = 25.46 kPa
Hence, vapour pressure of solution, p_total = P₁ + p₂
= 47.97 + 25.46 = 73.43 kPa

Question 17.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Solution:
1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water = 18 g mol^-1
> Number of moles present in 1000 g of water = \(\frac{1000}{18}\) = 55.56 mol
Therefore, mole fraction of the solute
χ₂ = \(\frac{1}{1+55.56}\) = 0.0177
It is given that,
Vapour pressure of water, \(p_{1}^{0}\) = 12.3 kPa
∴ \(\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}\) = χ₂
Applying the relation,
⇒ \(\frac{12.3-p_{1}}{12.3}\) = 0.0177
⇒ 12.3 – p₁ = 0.0177 × 12.3
⇒ 12.3 -P₁ = 0.2177
⇒ P₁ = 12.3 – 0.2177
⇒ p₁ = 12.0823
= 12.08 kPa
Hence, the vapour pressure of the solution is 12.08 kPa.

Question 18.
Calculate the mass of a non-volatile solute (molar mass 40 g mol^-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Solution:

Question 19.
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate :
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K.
Solution:
(i) Let, the molar mass of the solute be M g mol^-1
Now, the no. of moles of solvent (water), = n₁ = \(\frac{\text { Mass }}{\text { Molar mass }}\)
= \(\frac{90 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 5mol
And, the no. of moles of solute, n₂ = \(\frac{\text { Mass }}{\text { Molar mass }}\)
= \(\frac{30 \mathrm{~g}}{\mathrm{Mg} \mathrm{mol}^{-1}}=\frac{30}{M}\) mol
Vapour pressure of I solution
p₁ = 2.8 kPa
According to Raoult’s law

Question 20.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Solution:
Let the mass of solution = 100 g
∴ Mass of cane sugar (w₂) = 5 g
ΔT_f = (273.15 – 271) K = 2.15 K
Molar mass of cane sugar (C₁₂H₂₂O₁₁), (M₂) = 12 x 12 + 22 x 1 +11 x 16
= 342 g mol^-1
Mass of solvent (water), (w₁) = 100 – 5 = 95 g

(Molar mass of glucose = 180 g mol^-1)
ΔT_f = 4.08 K
Freezing point of glucose solution
T_f = \(T_{f}^{0}\) – ΔT_f = 273.15 – 4.08 = 269.07 K

Question 21.
Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆, 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.
Solution:
We know,
M₂ = \(\frac{1000 \times w_{2} \times K_{f}}{\Delta T_{f} \times w_{1}}\)
Then M_AB2 = \(\frac{1000 \times 1 \times 5.1}{2.3 \times 20}\) = 110.87 g mol^-1
Then M_AB2 = \(\frac{1000 \times 1 \times 5.1}{1.3 \times 20}\) = 196.15 g mol^-1
Let the atomic masses of A and B be x and y respectively.
Molar mass of AB2 = x + 2 y = 110.87 …………… (i)
Molar mass of AB4 = x + 4y = 196.15 ………… (ii)
Subtracting equation (i) from (ii) , we get
2y=85.28 y = 42.64
Putting the value of ‘/ in equation (i), we get
x + 2 x 42.64 = 110.87
⇒ x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.

Question 22.
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Solution:
Here, T = 300 K, π = 1.52 bar, R = 0.083 bar L K^-1mol^-1
Applying the relation, π = CRT
⇒ C = \(\frac{\pi}{R T}\)
= \(\frac{1.52 \mathrm{bar}}{0.083 \mathrm{bar} \mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}}\)
= 0.061 mol
Since, the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.

Question 23.
Suggest the most important type of intermolecular attractive interaction in the following pairs. .
(i) n-hexane and n-octane
(ii) I₂ and CCl₄
(iii) NaClO₄ and water
(iv) methanol and acetone
(v) acetonitrile (CH₃CN) and acetone (C₃H₆O).
Answer:
(i) Van der Wall’s forces of attraction. (London forces)
(ii) Van der Wall’s forces of attraction. (London forces) .
(iii) Ion-dipole interaction. ‘
(iv) Dipole-dipole interaction.
(v) Dipole-dipole interaction.

Question 24.
Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.
Answer:
(i) Cyclohexane and n-octane both are non-polar.
So, they will mix completely in all proportions.
(ii) KCl is an ionic compound, but n-octane is non-polar.
So, KCl will not dissolve in n-octane.
(iii) CH₃OH and CH₃CN both are polar but CH₃CN is less polar than CH3OH. As the solvent is non-polar CH3CN will dissolve more than CH₃OH in n-octane.
Therefore, the order of solubility in n-octane will be KCl < CH₃OH < CH₃CN < Cyclohexane

Question 25.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol.
Answer:
(i) Phenol (C₆H₅OH) has the polar group -OH and non-polar group -C₆H₅. Thus, phenol is partially soluble in water.

(ii) Toluene (C₆H₅ – CH₃) has no polar groups. Thus, toluene is insoluble in water.

(iii) Formic acid (HCOOH) has the polar group -OH and can form H-bond with water. Thus, formic acid is highly soluble in water.

(iv) Ethylene glycol has polar -OH group and can form H-bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water because it cannot form hydrogen bonds with water.

(vi) Pentanol (C₅H₁₁OH) has polar -OH group, but it also contains a very bulky non-polar -C₅H₁₁ group. Thus, pentanol is partially soluble in water.

Question 26.
If the density of some lake water is 1.25 g mL^-1 and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na+ ions in the lake.
Solution:
Number of moles present in 92 g of Na⁺ ions
= \(\frac{92 \mathrm{~g}}{23 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 4 mol

= \(\frac{4 \mathrm{~mol}}{1 \mathrm{~kg}}\) = m

Question 27.
If the solubility product of CuS is 6 × 10¹⁶, calculate the maximum molarity of CuS in aqueous solution.
Solution:
Solubility product of CuS, K_sp = 6 × 10^-16 .
Let s be the solubility of CuS in mol L^-1. Then,

= s × s
= s²
Then we have, K _sp s² = 6 × 10^-16
⇒ s = \(\sqrt{6 \times 10^{-16}}\)
= 2.45 x 10^-8 mol^-1
Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × ^-8 mol^-1.

Question 28.
Calculate the mass percentage of aspirin (C₉H₈O₄) in
acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.
Solution:
Mass of aspirin = 6.5 g
Mass of acetonitrile = 450 g
Then, total mass of the solution = (6.5 + 450) g = 456.5 g
Therefore, mass percentage of C₉H₈O₄ = \(\frac{6.5}{456.5}\) × 100%
= 1.424%

Question 29.
Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10^-3 m aqueous solution required for the above dose.
Solution:
1.5 × 10^-3 m solution means that 1.5 × 10^-3 mole of nalorphene is dissolved in 1 kg of water.
Molar mass of C₁₉H₂₁N0₃
= 19 × 12 + 21 + 14 + 48
= 311 g mol^-1
∴ 1.5 × 10^-3 mole of nalorphene
= 1.5 × 10^-3 × 311 g = 0.467 g
= 467 mg
Mass of solution
= 1000 g + 0.467 g
= 1000.467 g
Thus, for 467 mg of nalorphene solution required 1000.467.
For 1.5 mg nalorphene = \(\frac{1000.467 \times 1.5}{467}\) = 3.21 g

Question 30.
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Solution:
Molarity = 0.15 M or 0.15 mol L^-1
Volume of solution = 250 mL = 0.25 L
Molar mass of benzoic acid (C₆H₅COOH) = 7 × 12 + 6 × 1 + 2 × 16
= 122 g mol^-1
Molality \(\frac{\text { Mass }}{\text { Molar mass }}\) × \(\frac{1}{\text { Volume (L) }}\)
0.15 mol L^-1 = \(\frac{w}{122 \mathrm{~g} \mathrm{~mol}^{-1}}\) × \(\frac{1}{0.25 \mathrm{~L}}\)
Mass of solute = (0.15 × 122 × 0.25) g = 4.575 g

Question 31.
The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer:

Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H⁺ ions i.e., trifluoroacetic acid ionises to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:
Acetic acid < trichloroacetic acid < trifluoroacetic acid

Question 32.
Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water, K_a = 1.4 × 10^-3, K_f = 1.86 K kg mol^-1.
Solution:
Mass of solute (CH₃CH₂CHClCOOH) = 10 g
Molar mass of
CH₃CH₂CHClC00H = 4 × 12 + 7 × 1 + 1 × 35.5 + 2 × 16 = 48 + 7 + 35.5 + 32
= 122.5 g mol^-1
\frac{\text { Mass / Molar mass }}{\text { Mass of solvent (Kg) }} = \(\frac{\text { Mass / Molar mass }}{\text { Mass of solvent (Kg) }}[latex/latex]
= [latex]\frac{10 \mathrm{~g}}{\left(122.5 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0.25 \mathrm{Kg})}\)
= 0.326 m
Let α be the degree of dissociation of CH₃CH₂CHClCOOH then


Total no. of moles after dissociation = 1 – α + α + α
= 1 + α
Van’t Hoff factor
Total no. of moles after dissociation

∴ i = \(\frac{1+\alpha}{1}\)
= 1 + α
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:
ΔT_f = i.K_fm
= 1.0655 × 1.86 kg mol^-1 × 0.326 mol kg^-1
= 0.65K

Question 33.
19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Solution:
Calculation of Van’t Hoff factor (i)
Given, w₁ = 500 g = 0.5 kg, w₂ = 19.5 g, K_f = 1.86 K kg mol^-1, ΔT_f = 1 K
Molar mass of CH₂FCOOH (M₂)
= 2 × 12 + 3 × 1 + 1 × 19 + 2 × 16
= 24 + 3 + 19 + 32
= 78 g mol^-1
ΔT_f = i K_f m

Question 34.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Solution:
Vapour pressure of water, \(\) = 17.535 mm Hg
Mass of glucose, w₂ = 25 g
Mass of water, w₁ = 450 g
Molar mass of glucose (C₆H₁₂O₆), M₂ = 6 × 12 + 12 × 1 + 6 × 16
= 180 g mol^-1

we know that
\(\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}\) = \(\frac{n_{2}}{n_{2}+n_{1}}\)
⇒ \(\frac{17.535-p_{1}}{17.535}\) = \(\frac{0.139}{0.139+25}\)
⇒ 17.535 – p₁ = \(\frac{0.139 \times 17.535}{25.139}\)
⇒ 17.535 – p₁ = 0.097
⇒ P₁ = 17.44 mm Hg
Hence, the vapour pressure of water is 17.44 mm Hg.

Question 35.
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Solution:
Here, p = 760 mm Hg, K_H = 4.27 × 10⁵ mm Hg (at 298 K)
According to Henry’s law, p = K_Hχ
χ = \(\frac{p}{k_{\mathrm{H}}}\)
= \(\frac{760 \mathrm{~mm} \mathrm{Hg}}{4.27 \times 10^{5} \mathrm{~mm} \mathrm{Hg}}\)
= 177.99 × 10^-5
= 178 × 10^-5
Hence, the mole fraction of methane in benzene is 178 × 10^-5.

Question 36.
100 g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Solution:
Number of moles of liquid A, n_A = \(\frac{w_{1}}{M_{1}}\) = \(\frac{100}{140}\) mol = 0.714 mol
Number of moles of liquid B,n_B = \(\frac{w_{2}}{M_{2}}\) = \(\frac{1000}{180}\) mol = 5.556 mol
Then, mole fraction of A, χ_A = \(\frac{n_{A}}{n_{A}+n_{B}}\)
= \(\frac{0.714 \mathrm{~mol}}{(0.714+5.556) \mathrm{mol}}\) = 0.114
Mole fraction of B, χ_B = 1 – 0.114 = 0.886
Vapour pressure of pure liquid B, \(p_{B}^{0}\) = 500 torr
Therefore, vapour pressure of liquid B in the solution,
P_B = \(p_{B}^{0}\)χ_B
= 500 × 0.886
= 443 torr
Total vapour pressure of the solution, p_total = 475 torr
∴ Vapour pressure of liquid A in the solution,
P_A = P_total – P_B
= 475 – 443 = 32 torr
Now, P_A = \(p_{A}^{0}\)χ_A
⇒ \(\frac{p_{A}}{\chi_{A}}\) = \(\frac{32}{0.114}\)
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.

Question 37.
Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 nun Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot
P_total’ P_chloroform and P_acetone as a function of x_acetone experimental data observed for different compositions of mixture is

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal
solution.

It can be observed from the graph that the plot for the ptotai of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

Question 38.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively.
Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Solution:
Molar mass of benzene (C₆H₆) = 6 × 12 + 6 × 1 = 78g mol^-1
Molar mass of toluene (C₆H₅CH₃ ) = 7 × 12 + 8 × 1 = 92 g mol^-1
No. of moles present in 80 g of benzene = \(\frac{80}{78}\) mol = 1.026 mol
No. of moles present in 100 g of toluene = \(\frac{100}{92}\) mol = 1.087 mol
Mole fraction of benzene, χ_C6H6, = \(\frac{1.026}{1.026+1.087}\) = 0.486
∴ Mole fraction of toluene,χ_C6H5CH35013 = 1 – 0.486 = 0.514
It is given that vapour pressure of pure benzene, \(p_{\mathrm{C}_{6} \mathrm{H}_{6}}^{0}\) = 50.71 mm Hg
Vapour pressure of pure toluene, \(p_{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}}^{0}\) = 32.06 mm Hg
Therefore, partial vapour pressure of benzene,
P_total = χ_C6H6 × \(p_{\mathrm{C}_{6} \mathrm{H}_{6}}^{0}\)
= 0.486 × 50.71
= 24.645 mm Hg
Partial vapour pressure of toluene, P_C6H5CH3 = χ_C6H5CH3 × \(P_{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}}^{0}\)
= 0.514 × 32.06
= 16.479 mm Hg
Total vapour pressure of solution (p) = 24.645 + 16.479
= 41.124 mm Hg
Mole fraction of benzene in vapour phase
= \(\frac{\chi_{\mathrm{C}_{6} \mathrm{H}_{6}} \times p_{\mathrm{C}_{6} \mathrm{H}_{6}}^{0}}{p_{\text {total }}}\)
= \(\frac{0.486 \times(50.71) \mathrm{mm}}{(41.124) \mathrm{mm}}\)
= 0.599 ≅ 0.6

Question 39.
The air is a mixture of a number of gases. The mqjor components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen \md nitrogen are 330 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.
Solution:
Percentage of oxygen (O₂) in air = 20%
Percentage of nitrogen (N₂) in air = 79%
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm that is, (10 × 760) mm = 7600 mm
Therefore, partial pressure of oxygen,
PO₂ = \(\frac{20}{100}\) x 7600 mm
= 1520 mm Hg
Partial pressure of nitrogen, pN₂ = \(\frac{79}{100}\) x 7600 mm
= 6004 mm Hg
Now, according to Henry’s law,
p = K_H.χ
For oxygen:

\(\frac{6004 \mathrm{~mm}}{6.51 \times 10^{7} \mathrm{~mm}}\)
(Given K_H = 6.51 × 10⁷ mm)
= 9.22 × 10^-5
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10^-5 and 9.22 × 10^-5 respectively.

Question 40.
Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
Solutio:
We know that,
π = i \(\)RT
⇒ π = i \(\)RT
⇒ w = \(\)
Given,
π = 0.75 atm
V = 2.5L
i = 2.47
T = (27 + 273)K = 300K
R = 0.0821 L atm K^-1mol^–
Molar mass of CaCl₂(M) = 1 × 40 + 2 × 35.5 = 111 g mol ^-1
Therefore, w = \(\frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300}\) = 3.42g
Hence the required amount of CaCl₂ is3.42g

Question 41.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25° C, assuming that it is completely dissociated.
Solution:
When K₂SO₄ is dissolved in water, K⁺ and \(\mathrm{SO}_{4}^{2-}\) ions are produced.

Total number of ions produced = 3
∴ i = 3
Given, w = 25 mg = 0.025 g, V = 2 L
T = 25°C = (25 + 273) K = 298 K
Also, we know that R = 0.0821 L atm K^-1 mol^-1
Molar Mass of K₂SO₄ (M) = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol^-1
Applying the following relation,
π = i \(\frac{n}{V}\) RT = i \(\frac{w}{M} \frac{1}{V}\) RT
= 3 × \(\frac{0.025}{174}\) × 1 × 0.0821 × 298 = 5.27 × 10^-3 atm

Chemistry Guide for Class 12 PSEB Solutions Textbook Questions and Answers

Question 1.
Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Solution:

Alternatively,
Mass percentage of CCl₄ = (100 – 15.28)% = 84.72%

Question 2.
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Solution:
Let the total mass of the solution be 100 g and the mass of benzene be 30 g.
∴ Mass of carbon tetrachloride = (100 – 30) g = 70 g
Molar mass of benzene (C₆H₆) = (6 × 12 + 6 × 1) g
∴ mol^-1 = 78 g mol^-1

Question 3.
Calculate the molarity of each of the following solutions:
(a) 30 g of CO(NO₃) ₂.6H₂O in 4.3 L of solution
(b) 30 mL of 0.5 M H₂SO₄diluted to 500 mL.
Solution:
Molarity is given by

(a) Molar mass of CO (NO₃)₂.6H₂O = 59 + 2 (14 + 3 × 16) + 6 × 18
= 291 g mol^-1
∴ Moles of Co (NO₃)₂.6H₂O = \(\frac{30}{291}\) mol = 0.103 mol
Volume of solution = 4.3 L
Therefore, molarity = \(\frac{0.103 \mathrm{~mol}}{4.3 \mathrm{~L}}\) = 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H₂SO₄ = 0.5 mol
∴ Number of moles present in 30 mL of 0.5 M H₂SO₄
= \(\frac{0.5 \times 30}{1000}\) mol
= 0.015 mol
Volume of solution = 500 mL = 0.5 L
Therefore, molarity = \(\frac{0.015}{0.5 \mathrm{~L}}\) mol = 0.03M

Question 4.
Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Solution:
Mass of required aqueous solution = 2.5 kg = 2500 g
Molar mass of urea (NH₂CONH₂) = 2 (1 × 14 + 2 × 1) + 1 × 12 +1 × 16
= 60 g mol^-1
0.25 molal aqueous solution of urea means 0.25 mole of urea is dissolved in 1000 g of water.
Mass of water = 1000 g
Moles of urea = 0.25 mol
Mass of urea = No. of moles of urea × Molar mass of urea
∴ Mass of 0.25 moles of urea = 0.25 × 60 = 15 g
Mass of solution = 1000 + 15 = 1015 g
1015 g of aqueous solution contains urea = 15 g
∴ 2500 g of aqueous solution will require urea
= \(\frac{15 g}{1015 g}\) × 2500 g = 36.95 g

Question 5.
Calculate (a) molality (b) molarity and (c) mole fraction of KI. If the density of 20% (mass/mass) aqueous KI is 1.202 g mL^-1.
Solution:
(a) Molar mass of KI = 39 +127 = 166 g mol^-1
20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.
∴ 20 g of KI is present in (100 – 20) g of water = 80 g of water

Question 6.
H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.
Solution:
It is given that the solubility of H₂S in water at STP is 0.195 m, i.e., 0.195 mol of H₂S is dissolved in 1000 g of water.

Question 7.
Henry’s law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K.
Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm COa pressure at 298 K.
Solution:
Given, K_H = 1.67 × 10⁸ Pa, p_CO2 = 2.5 atm = 2.5 × 10⁵ Pa
According to Henry’s law
p_CO2 = K_HχCO2 = \(\frac{P_{\mathrm{CO}_{2}}}{K_{\mathrm{H}}}\)
\(\frac{2.5 \times 10^{5}}{1.67 \times 10^{8}}\) = 1.5 × 10^-3 ……….. (i)
Mass of water = Density of water × Volume of water
= 1 g/mL × 500 mL = 500 g
Number of moles of water, (n_H2O)
= \(\frac{\text { Mass of water }}{\text { Molar mass }}=\frac{500 \mathrm{~g}}{18 \mathrm{~g} / \mathrm{mol}}\)
= 27.78 mol
χ_CO2 = \(\frac{n_{\mathrm{CO}_{2}}}{n_{\mathrm{H}_{2} \mathrm{O}}+n_{\mathrm{CO}_{2}}}=\frac{n_{\mathrm{CO}_{2}}}{n_{\mathrm{H}_{2} \mathrm{O}}}\)
⇒ n_CO2 = χ_CO2n_H2O
= 1.5 × 10^-3 × 27.78 mol
= 41.67 × 10^-3 mol
Mass of CO₂ = No. of moles of CO₂ × Molar mass
= 41.67 × 10^-3 × 44 =1.834 g

Question 8.
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Solution:
Given, p\(p_{A}^{0}\) = 450 mm Hg, \(p_{B}^{0}\) = 700 mm Hg,
P_total = 600 mm Hg
According to Raoult’s law
P_A = \(p_{A}^{0}\)χ_A
P_B = \(p_{B}^{0}\)χ_B – \(p_{A}^{0}\) (1 – χ_A )

Therefore, total pressure, p_total = p_A + p_B
⇒ p_total = \(p_{A}^{0}\)χ_A + \(p_{B}^{0}\) (1 – χ_A)
⇒ p_total = \(p_{A}^{0}\)χ_A + \(p_{B}^{0}\) – \(p_{B}^{0}\)χ_A
⇒ p_total = \(p_{A}^{0}\) – \(p_{B}^{0}\) χ_A + \(p_{B}^{0}\)
⇒ 600 = (450 – 700)χ_A + 700
⇒ -100 = -250χ_A
⇒ χ_A = 0.4
Mole fraction of A (χ_A) = 0.4
Mole fraction of B (χ_B) = 1 – 0.4 = 0.6
Now, P_A = P_Aχ_A = 450 × 0.4
= 180 mm Hg
P_B = \(p_{B}^{0}\)χ_B
= 700 × 0.6
= 420 mm Hg
Now, in the vapour phase:
Mole fraction of liquid A = \(\frac{p_{A}}{p_{A}+p_{B}}\)
\(\frac{180}{180+420}=\frac{180}{600}\) = 0.30
And, mole fraction of liquid B = 1 – 0.30 = 0.70

Question 9.
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution:
Given, vapour pressure of water, \(p_{1}^{0}\) = 23.8 mm Hg
Weight of water w₁, = 850 g
Weight of urea, w₂ = 50 g
Molecular weight of water, M₁ = 18 g mol^-1
Molecular weight of urea, M₂ = 60 g mol^-1

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p₁.
Now, from Raoult’s law, we have

Question 10.
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
Solution:
Here, elevation in boiling point ΔT_b = (100 + 273) – (99.63 + 273)
= 0.37 K
Mass of water, w₁ = 500 g
Molar mass of sucrose (C₁₂H₂₂O₁₁), M₂ = 12 × 12 + 22 × 1 + 11 × 16
= 342 g mol^-1
Molal elevation constant, K_b = 0.52 K kg mol^-1
We know that,
ΔT_b = \(\frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}\)
⇒ w₂ = \(\frac{\Delta T_{b} \times M_{2} \times w_{1}}{K_{b} \times 1000}\)
= \(\frac{0.37 \times 342 \times 500}{0.52 \times 1000}\)
= 121.67 g
Hence, 121.67 g of sucrose is to be added. ’

Question 11.
Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_f = 3.9 Kkg mol^-1.
Solution:
Mass of acetic acid, w₁ = 75 g
Molar mass of ascorbic acid (C₆H₈O₆), M = 6 × 12 + 8 × 1 + 6 × 16
= 176 g mol^-1
Depression in melting point (AT_f ) = 1.5 K
Molal depression constant (K_f ) = 3.9 K kg mol^-1
We know that,
ΔT_b = \(\frac{K_{f} \times w_{2} \times 1000}{M_{2} \times w_{1}}\)
⇒ w₂ = \(\frac{\Delta T_{f} \times M_{2} \times w_{1}}{K_{f} \times 1000}\)
= \(\frac{1.5 \times 176 \times 75}{3.9 \times 1000}\)
= 5.08 g
Hence, 5.08 g of ascorbic acid is needed to be dissolved.

Question 12.
Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C. ,
Solution:
It is given that,
Volume of water, V = 450 mL = 0.45 L
Temperature, T = (37 + 273) K = 310 K
R = 8.314 K Pa L K^-1 mol^-1
= 8.314 × 10³ Pa LK^-1 mol^-1
Number of moles of the polymer, n = \(\frac{1}{185000}\) mol
We know that,
Osmotic pressure, n = \(\frac{n}{V}\)RT
= \(\frac{1}{185000}\) mol × \(\frac{1}{0.45 \mathrm{~L}}\) × 8.314 × 10³ Pa LK^-1 mol^-1 × 310 K
= 30.98 Pa

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 14 Biomolecules Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 14 Biomolecules

PSEB 12th Class Chemistry Guide Biomolecules InText Questions and Answers

Question 1.
What are monosaccharides?
Answer:
Monosaccharides are carbohydrates which cannot be further hydrolysed to simpler molecules. The general formula of monosaccharides is (CH₂O)_n where n = 3-7.

Question 2.
What are reducing sugars?
Answer:
Carbohydrates which reduce Fehling’s solution to red precipitate of Cu₂O or Tollen’s reagent to metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars.

Question 3.
Write two main functions of carbohydrates in plants.
Answer:
Two main functions of carbohydrates in plants are:

1. Polysaccharides such as starch serve as storage molecules.
2. Cellulose, a polysaccharide, is used to build the cell wall.

Question 4.
Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose
Answer:
Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
Disaccharides: Maltose, lactose.

Question 5.
What do you understand by the term glycosidic linkage?
Answer:
Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule. For example, in a sucrose molecule, two monosaccharide units, α-D-glucose and β-D-fructose, are joined together by a glycosidic linkage.

Question 6.
What is glycogen? How is it different from starch?
Answer:
Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components-amylose (15 – 20%) and amylopectin (80 – 85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin. Glycogen chain consists of 10-14 glucose units whereas amylopectin chains consist of 20-25 glucose units.

Question 7.
What are the hydrolysis products of (i) sucrose and (ii) lactose?
Answer:
(i) On hydrolysis, sucrose gives one molecule of α-D glucose and one molecule of β-D-fructose.

(ii) On hydrolysis, lactose gives β-D-galactose and β-D-glucose.

Question 8.
What is the basic structural difference between starch and cellulose?
Answer:
Starch consists of two components-amylose and amylopectin. Amylose is a long linear chain of ≅α-D-(+)-glucose units joined by C₁ -C₄ glycosidic linkage (α-link).

Amylopectin is a branched-chain polymer of []α-D-glucose units, in which the chain is formed by C₁ -C₄ glycosidic linkage and the branching occurs by C₁ – C₆ glycosidic linkage.

On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C₁– C₄ glycosidic linkage (β-link).

Question 9.
What happens when D-glucose is treated with the following reagents?
(i) HI
(ii) Bromine water
(iii) HNO₃
Answer:
(i) HI s On prolonged heating with HI, glucose forms n-hexane.

(ii) Br₂ water: Glucose gets oxidised (gluconic acid) on reaction with a mild oxidising agent like bromine water.

(iii) HNO₃: On oxidation with nitric acid, glucose yields a dicarboxylic acid, that is saccharic acid.

Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open-chain structure.
Answer:

• Aldehydes give 2, 4-DNP test, Schiff’s test, and react with NaHSO₄ to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
• The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free -CHO group is absent from glucose.
• Glucose exists in two crystalline forms-α and β. The α-form (m.p. = 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p = 423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.

Question 11.
What are essential and non-essential amino acids? Give two examples of each type.
Answer:
Essential amino acids are required by the human body, but they cannot be synthesised in the body. They must be taken through food, e.g., valine and leucine. Non-essential amino acids are also required by the human body, but they can be synthesised in the body e.g., glycine and alanine.

Question 12.
Define the following as related to proteins
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) Peptide linkage: The amide formed between —COOH group of one molecule of an amino acid and -NH₂ group of another molecule of the amino acid by the elimination of a water molecule is called a peptide linkage.


(ii) Primary structure: The primary structure of protein refers to the specific sequence in which various amino acids are present in it, i.e., the sequence of linkages between amino acids in a polypeptide chain. The sequence in which amino acids are arranged is different in each protein. A change in the sequence creates a different protein.

(iii) Denaturation: In a biological system, a protein is found to have a unique 3-dimensional structure and a unique biological activity. In such a situation, the protein is called native protein. However, when the native protein is subjected to physical changes such as change in temperature or chemical changes such as change in pH, its H-bonds are disturbed.

This disturbance unfolds the globules and uncoils the helix. As a result, the protein loses its biological activity. This loss of biological activity by the protein is called denaturation. During denaturation, the secondary and the tertiary structures of the protein get destroyed, but the primary structure remains unaltered. One of the examples of denaturation of proteins is the coagulation of egg white when an egg is boiled.

Question 13.
What are the common types of secondary structures of proteins?
Answer:
There are two common types of secondary structure of proteins:
(i) α- helix structure: In this structure, the -NH group of an amino acid residue forms H-bond with the img group of the adjacent turn of the right-handed screw (α-helix).

(ii) β-pleated sheet structure: This structure is called so because it looks like the pleated folds of drapery. In this structure, all the peptide chains are stretched out to nearly the maximum extension and then laid side by side. These peptide chains are held together by intermolecular hydrogen bonds.

Question 14.
What type of bonding helps in stabilising the α-helix structure of proteins?
Answer:
The α-helix structure of proteins is stabilised by intramolecular H-bonding between C-O of one amino acid residue and the N-H of the fourth amino acid residue in the chain.

Question 15.
Differentiate between globular and fibrous proteins.
Answer:

Globular proteins	Fibrous proteins
1. These are water-soluble proteins.	These are water-insoluble proteins.
2. These are spherical in shape.	These are linear in shape.
3. Globular proteins are highly unstable.	Fibrous proteins are stable to moderate changes in temperature and pH.

Question 16.
How do you explain the amphoteric behaviour of amino acids?
Answer:
In aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as zwitterion.

Therefore, in zwitterionic form, the amino acid can act both as an acid and as a base.

Thus, amino acids show amphoteric behaviour.

Question 17.
What are enzymes?
Answer:
Enzymes are proteins that catalyse biological reactions. They are very specific in nature and catalyse only a particular reaction for a particular substrate. Enzymes are usually named after the particular substrate or class of substrate and sometimes after the particular reaction. For example, the enzyme used to catalyse the hydrolysis of maltose into glucose is named as maltase.

Again, the enzymes used to catalyse the oxidation of one substrate with the simultaneous reduction of another substrate are named as oxidoreductase enzymes. The name of an enzyme ends with ‘ – ase’.

Question 18.
What is the effect of denaturation on the structure of proteins?
Answer:
During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. Due to denaturation, the globular proteins (soluble in H20) are converted into fibrous proteins (insoluble in H₂O) and their biological activity is lost. For example, boiled egg which contains coagulated proteins cannot be hatched.

Question 19.
How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Answer:
Vitamins are classified into two groups depending upon their solubility in water or fat.

1. Water-soluble vitamins: These include vitamin B-complex (B₁, B₂, B₅, i.e., nicotinic acid, B₆, B₁₂, pantothenic acid, biotin, i.e., vitamin H and folic acid) and vitamin C.
2. Fat-soluble vitamins: These include vitamins A, D, E and K. These are stored in liver and adipose tissues (fat-storing tissues). Vitamin K is responsible for coagulation of blood.

Question 20.
Why are vitamin A and vitamin C essential to us? Give their important sources.
Answer:
Vitamin A: Vitamin A is essential for us because its deficiency can cause xerophthalmia (hardening of cornea of eye) and night blindness. Sources: Carrots, fish liver oil, butter and milk.
Vitamin C: Vitamin C is essential for us because its deficiency causes scurvy (bleeding gums) and pyorrhea (loosening and bleeding of teeth).
Sources: Amla, citrus fruits and green leafy vegetables.

Question 21.
What are nucleic acids? Mention their two important functions.
Answer:
Nucleic acids are biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins or chromosomes (proteins containing nuclei acids as the prosthetic group).

These are of two types: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).
The two main functions of nucleic acids are :

1. DNA is responsible for transmission of hereditary effects from one generation to another. This is because of the unique property of replication during cell division and the transfer of two identical DNA strands to the daughter cells.
2. DNA and RNA are responsible for synthesis of all proteins essential for the growth and maintenance of our body. Actually, the proteins are synthesised by various RNA molecules (rRNA, mRNA and tRNA) in the cell but the message for the synthesis of a particular protein is present in DNA.

Question 22.
What is the difference between a nucleoside and a nucleotide?
Answer:
A nucleoside is formed when 1-position of pyrimidine (cytosine, thymine or uracil) or 9-position of purine (guanine or adenine) base is connected to C-1 of sugar (ribose or deoxyribose) by a β-linkage. Hence, in general, nucleosides may be represented as: Sugar-Base.

A nucleotide contains all the three basic compounds of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. These are obtained by esterification of C’₅– OH group of the pentose sugar by phosphoric acid.

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Answer:
In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms hydrogen bond with guanine, while adenine forms hydrogen bond with thymine. As a result, the two strands are complementary to each other.

Question 24.
Write the important structural and functional differences between DNA and RNA.
Answer:
Structural differences

DNA	RNA
(i) The sugar present in DNA is 2-deoxy-D-(-)-ribose.	The sugar present in RNA is D-(-)-ribose.
(ii) DNA contains cytosine and thymine as pyrimidine bases.	RNA contains cytosine and uracil as pyrimidine bases.
(iii) DNA has a double-stranded a-helix structure.	RNA has a single-stranded a-helix structure.
(iv) DNA molecules are very large; their molecular mass may vary from 6 x 106 -16 x 106 u.	RNA molecules are much smaller with molecular mass ranging from 20,000 to 40,000 u.

Functional differences:

(i) DNA has unique property of replication.	RNA usually does not replicate.
(ii) DNA controls the transmission of hereditary effects.	RNA controls the synthesis of proteins.

Question 25.
What are the different types of RNA found in the cell?
Answer:
Three types of RNA present in the cell are :

1. Messenger RNA (m-RNA)
2. Ribosomal RNA (r-RNA)
3. Transfer RNA (t-RNA)

Chemistry Guide for Class 12 PSEB Biomolecules Textbook Questions and Answers

Question 1.
Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer:
Glucose and sucrose have hydroxyl (-OH) groups (5 in glucose and 8 in sucrose) which form strong hydrogen bonds with water molecules and hence these are soluble in water. Cyclohexane and benzene are non-polar compounds and do not have hydroxyl groups and hence they are not able to form hydrogen bonding with water molecules and, therefore, these are not soluble in water.

Question 2.
What are the expected products of hydrolysis of lactose?
Answer:
Lactose is composed of β-D galactose and β-D glucose. Thus, on hydrolysis, it gives D-(+)- galactose and D-(+)- glucose.

Question 3.
How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Answer:
D-glucose reacts with hydroxylamine (NH₂OH) to form an oxime because of the presence of aldehydic (-CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open-chain structure in an aqueous medium, which then reacts with NH₂OH to give an oxime.

But pentaacetate of D-glucose does not react with NH₂OH. This is because pentaacetate does not form an open-chain structure.

Question 4.
The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
Answer:
Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept a proton, thus giving rise to a dipolar ion known as a zwitterion.

Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. But halo-acids do not exhibit such dipolar behaviour. For this reason, the melting points and the solubility of amino acids in water is higher than those of the corresponding halo-acids.

Question 5.
Where does the water present in the egg go after boiling the egg?
Answer:
When an egg is boiled, the proteins present inside the egg get denatured and coagulated. After boiling the egg, the water present in it is absorbed by the coagulated protein through H-bonding.

Question 6.
Why cannot vitamin C be stored in our body?
Answer:
Because vitamin C are water-soluble vitamins and so these are readily excreted through urine and hence cannot be stored in our body.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer:
When a nucleotide from the DNA containing thymine is hydrolysed, thymine β-D-2-deoxyribose and phosphoric acid are obtained as products.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer:
On complete hydrolysis of RNA six compounds are isolated. These compounds are adenine, guanine, cytosine, uracil, D-ribose and phosphoric acid. There is, no relationship among the quantities of different bases obtained on hydrolysis of RNA suggests that RNA is single-stranded. If it would have been double-stranded like DNA the complementary pairing of bases would have given equal proportion of complementary bases.

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 15 Polymers Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 15 Polymers

PSEB 12th Class Chemistry Guide Polymers InText Questions and Answers

Question 1.
Explain the terms polymer and monomer.
Answer:

• Polymer is a high molecular mass macromolecule consisting of repeating structural units derived from monomers.
• A monomer is a simple molecule capable of undergoing polymerisation and leading to the formation of the corresponding polymer.

Question 2.
What are natural and synthetic polymers? Give two examples of each type.
Answer:

1. Natural polymers are high molecular mass macromolecules and are found in plants and animals. For example, proteins and nucleic acids,
2. Synthetic polymers are man-made high molecular mass macromolecules. These include synthetic plastics, fibres and rubbers. For example, polythene and dacron.

Question 3.
Distinguish between the terms homopolymer and copolymer and give an example of each.
Answer:
Homopolymer: Polymers whose repeating structural units are derived from only one type of monomer units are called homopolymers. For example, polythene, PAN, Teflon, nylon-6, etc.

Copolymer: Polymers whose repeating structural units are derived from two or more types of monomer molecules are called copolymers. For example, Buna-S, Buna-N, nylon-6,6 polyester, bakelite, etc.

Question 4.
How do you explain the functionality of a monomer?
Answer:
The functionality of a monomer is the number of binding sites in a molecule. For example, the functionality of ethene, propene, styrene, acrylonitrile is one and that of 1, 3-butadiene, adipic acid, terephthalic acid, hexamethylenediamine is two.

Question 5.
Define the term polymerisation.
Answer:
The polymerisation is a process of formation of a high molecular mass polymer from one or more monomers by linking together a large number of repeating structural units through covalent bonds.

Question 6.

Answer:
is a homopolymer because it is obtained from a single monomer unit, NH₂-CHR-COOH.

Question 7.
In which classes, the polymers are classified on the basis of molecular forces?
Answer:
On the basis of molecular forces of attraction polymers are classified into the following classes :

• Elastomers
• Fibres
• Thermoplastic polymers and
• Thermosetting polymers.

Question 8.
How can you differentiate between addition and condensation polymerisation?
Answer:
(i) Addition polymerisation: In this, polymers are formed by the repeated addition of monomers molecules possessing double or triple bonds.
For example :

(ii) Condensation polymerisation: It is a process in which two or more bi-functional molecules undergo a series of condensation reactions with the elimination of some simple molecules and leading to the formation of polymers.

Question 9.
Explain the term copolymerisation and give two examples.
Answer:
When a mixture containing more than one monomeric species is allowed to polymerise, the product obtained is called a copolymer and the process is called copolymerisation. For example, Buna-S, a copolymer of 1,3-butadiene and styrene and Buna-N, a copolymer of 1,3-butadiene and acrylonitrile.

Question 10.
Write the free radical mechanism for the polymerisation of ethene.
Answer:
The mechanism of chain growth polymerisation of ethene of free radical mechanism is given below :
Step I. Chain initiation step :

Step II. Chain propagation step :

Step III. Chain terminating step :

Question 11.
Define thermoplastics and thermosetting polymers with two examples Of each.
Answer:
Thermoplastic polymers: These polymers are the linear or slightly. branched long-chain molecules capable of repeatedly softening on heating and hardening on cooling. These polymers possess intermolecular forces of attraction intermediate between elastomers and fibres. Some common examples are polythene, polystyrene, polyvinyl, etc.

Thermosetting polymers: These polymers are cross-linked or heavily branched molecules, which on heating undergo extensive cross-linking in moulds and again become infusible. These polymers cannot be reused. Some common examples are bakelite, urea-formaldehyde resins, etc.

Question 12.
Write the monomers used for getting the following polymers :
(i) Polyvinyl chloride
(ii) Teflon
(iii) Bakelite
Answer:

Question 13.
Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Answer:
One common initiator used in free radical addition polymerisation is benzoyl peroxide. Its structure is given below:

Question 14.
How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Answer:

• In natural rubber (cis- 1,4-polyisoprene), double bonds are located between C₂ and C₃ of isoprene units.
• The double bonds are reactive sites and helps in vulcanisation of natural rubber forming S-S linkages between chains.
• The cis-configuration about double bonds do not allow the chains to come closer and so there are weak intermolecular attractions between the chains. This leads to coiled structure and elasticity for natural rubber.

Question 15.
Discuss the main purpose of vulcanisation of rubber.
Answer:
Natural rubber becomes soft at high temperatures (> 335K) and brittle at low temperatures (< 283 K) and shows high water absorption capacity. It is soluble in non-polar solvents and is non-resistant to attack by oxidising agents. To improve upon these physical properties, a process of vulcanisation is carried out. This process consists of heating a mixture of raw rubber with sulphur and an appropriate additive at a temperature range between 373 K and 415 K. On vulcanisation, sulphur forms cross-links at the reactive sites of double bonds and thus the rubber gets stiffened.

Question 16.
What are the monomeric repeating units of Nylon-6 and Nylon-6,6?
Answer:

(ii) Nylon-6,6: 1,6-Hexamethylenediamine and adipic acid.

Question 17.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S,
(ii) Buna-N,
(iii) Dacron,
(iv) Neoprene
Answer:
The names and structures of monomers are :

Question 18.
Identify the monomer in the following polymeric structures :

Answer:
The monomers forming the polymer are :
(i) Decanoic acid HOOC-(CH₂)₈-COOH and Hexamethylenediamine H₂N-(CH₂)₆-NH₂.
(ii)

Question 19.
How is dacron obtained from ethylene glycol and terephthalic acid?
Answer:
Dacron is a synthetic condensation polymer which has ester group in the polymer chain. Terylene was also known by the name.

Question 20.
What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester.
Answer:
A polymer that can be decomposed by bacteria is called a biodegradable polymer.
Poly-β-hydroxybutyrate-CO-β-hydroxy valerate (PHBV) is a biodegradable aliphatic polyester.

Chemistry Guide for Class 12 PSEB Polymers Textbook Questions and Answers

Question 1.
What are polymers?
Answer:
Polymers are high molecular mass macromolecules, which consist of repeating structural units derived from monomers. Polymers have a high molecular mass (103 – 107 u). In a polymer, various monomer units are joined by strong covalent bonds. These polymers can be natural as well as synthetic. Polythene, rubber, and nylon 6, 6 are examples of polymers.

Question 2.
How are polymers classified on the basis of structure?
Answer:
Polymers are classified on the basis of structure as follows:
1. Linear polymers: These polymers are formed of long straight chains. They can be depicted as :

e.g., high-density polythene (HDP), polyvinyl chloride PVC, etc.

2. Branched-chain polymers: These polymers are basically linear chain polymers with some branches. These polymers are represented as:

e.g., low-density polythene (LDP), amylopectin, etc.

3. Cross-linked or Network polymers: These polymers have many cross-linking bonds that give rise to a network-like structure. These polymers contain bi-functional and tri-functional monomers and strong covalent bonds between various linear polymer chains. Examples of such polymers include bakelite and Melmac.

Question 3.
Write the names of monomers of the following polymers:

Answer:
(i) Hexamethylenediamine [H₂N—(CH₂)₆—NH₂] and adipic acid [HOOC—(CH₂)₄ —COOH].
(ii) Caprolactam

(iii) Tetrafluoroethene, (F₂C – CF₂).

Question 4.
Classify the following as addition and condensation polymers: Terylene, Bakelite, Polyvinyl chloride, Polythene.
Answer:
Addition polymers: Polyvinyl chloride, polythene.
Condensation polymers: Terylene, bakelite.

Question 5.
Explain the difference between Buna-N and Buna-S.
Answer:
Both are copolymers. Buna-N is a copolymer of 1, 3-butadiene and acrylonitrile while Buna-S is a copolymer of 1,3-butadiene and styrene.

Question 6.
Arrange the following polymers in increasing order of their intermolecular forces:
(i) Nylon-6,6, Buna-S, Polythene
(ii) Nylon-6, Neoprene, Polyvinylchloride.
Answer:
(i) Buna-S < Polythene < Nylon-6,6
(ii) Neoprene < Polyvinyl chloride < Nylon-6

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 16 Chemistry in Everyday Life Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 16 Chemistry in Everyday Life

PSEB 12th Class Chemistry Guide Chemistry in Everyday Life InText Questions and Answers

Question 1.
Why do we need to classify drugs in different ways?
Answer:
Different ways of classification of drugs and the usefulness of such classification are as follows :

1. Classification on the basis of pharmacological effect is useful for doctors because it provides them the whole range of drugs available for the treatment of a particular type of problem.
2. Classification on the basis of drug action on a particular biochemical process is useful for choosing the correct lead compound for designing the synthesis of a desired drug.
3. Classification on the basis of molecular targets is useful for medicinal chemists so that they can design a drug which is most effective for a particular receptor site.
4. Classification on the basis of chemical structure is useful for doctors to design the synthesis of a number of structurally similar compounds having different substituents and then choosing the drug having the least toxicity.

Question 2.
Explain the term target molecules or drug targets as used in medicinal chemistry.
Answer:
Drugs interact with macromolecules such as proteins, carbohydrates, lipids, enzymes and nucleic acids. Hence, these are called drug targets. Drugs possessing some common structural features may have the same mechanism of action on targets.

Question 3.
Name the macro-molecules that are chosen as drug targets.
Answer:
Nucleic acids, proteins, carbohydrates, lipids, enzymes are chosen as drug targets.

Question 4.
Why should not medicines be taken without consulting doctors?
Answer:
Side effects are caused when a drug binds to more than one receptor site. So, a doctor must be consulted to choose the right drug which has the maximum affinity for a particular receptor site to have the desired effect. The dose of the drug is also crucial because some drugs like opiates in higher doses act as poisons and may cause death.

Question 5.
Define the term chemotherapy.
Answer:
The branch of chemistry which deals with the treatment of diseases using chemicals is called chemotherapy.

Question 6.
Which forces are involved in holding the drugs to the active site of enzymes?
Answer:
Ionic bonding, hydrogen bonding, van der Waals’ interaction, dipole-dipole interaction etc., are involved in holding the drugs to the active site of enzymes.

Question 7.
While antacids and antiallergic drugs interfere with the function of histamines, why do these not interfere with the function of each other?
Answer:
These (antacids and antiallergic drugs) do not interfere with the function of each other because they work on different receptors in the body.

Question 8.
Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs.
Answer:
In event of low level of neurotransmitters, noradrenaline, antidepressant drugs are required. These drugs inhibit the enzymes which catalyse the degradation of noradrenaline. If the enzyme is inhibited, noradrenaline is slowly metabolised and thus activates its receptor for longer periods of time thereby reducing depression. Two important drugs are iproniazid and phenelzine.

Question 9.
What is meant by the term ‘broad-spectrum antibiotics? Explain.
Answer:
Broad-spectrum antibiotics are effective against several different types of harmful bacteria. Examples are tetracycline, ofloxacin, chloramphenicol, etc. Chloramphenicol can be used in case of typhoid, acute fever, dysentery, urinary infections, meningitis and pneumonia.

Question 10.
How do antiseptics differ from disinfectants? Give one example of each.
Answer:
Differences between antiseptics and disinfectants are as follows :
Antiseptics

• Antiseptics are chemical substances which prevent the growth of microorganisms and may even kill them but are not harmful to living tissues.
• Antiseptics are generally applied to living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
• Dettol, furnace, soframicine are antiseptics.

Disinfectants

• Disinfectants are chemical substances which kill microorganisms or stop their growth but are harmful to human tissues.
• Disinfectants are applied to inanimate objects such as floor, drainage system, instruments, etc.
• Chlorine in the concentration of 0.2 to 0.4 ppm in aqueous solution and SO 2 in very low concentration are disinfectants.

Question 11.
Why are cimetidine and ranitidine better antacids than sodium hydrogen carbonate or magnesium or aluminium hydroxide?
Answer:
Excessive use of sodium hydrogen carbonate or a mixture of aluminium and magnesium hydroxide can make the stomach alkaline and trigger the production of even more acid. On the other hand, ranitidine and cimetidine prevent the interaction of histamine with the receptors present in the stomach wall. This results in release of lesser amount of acid. Thus, these are better antacids.

Question 12.
Name a substance which can be used as an antiseptic as well as disinfectant.
Answer:
0.2% solution of phenol acts as an antiseptic while 1% of the solution acts as a disinfectant.

Question 13.
What are the main constituents of Dettol?
Answer:
Chloroxylenol and a-terpineol in a suitable solvent.

Question 14.
What is tincture of iodine? What is its use?
Answer:
A 2-3 per cent solution of iodine in alcohol-water mixture is known as tincture of iodine. It is used as an antiseptic.

Question 15.
What are food preservatives?
Answer:
Chemical substances which are used to protect food against bacteria, yeasts and moulds are called food preservatives. For example, sodium metabisulphite, sodium benzoate, etc.

Question 16.
Why is the use of aspartame limited to cold foods and drinks?
Answer:
Aspartame decomposes on heating and may not work well. So, its use as an artificial sweetener is limited to foods and drinks at low temperatures.

Question 17.
What are artificial sweetening agents? Give two examples.
Answer:
Artificial sweetening agents are the chemical substances which provide sweetness to the food without increasing the calories to the body. For example, saccharin, aspartame, sucralose etc.

Question 18.
Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
Saccharin.

Question 19.
What problem arises in using alitame as artificial sweetener?
Answer:
Alitame is a high potency artificial sweetener. Therefore, it becomes difficult to control die level of sweetness while using it.

Question 20.
How are synthetic detergents better than soaps?
Answer:
Advantages of synthetic detergents over soaps :

• Detergents can work with hard water too while soaps cannot.
• They can work even in an acidic medium while soaps cannot.
• Synthetic detergents are stronger cleansing agents than soaps.
• Their solubility is higher than that of soaps.
• They are prepared from hydrocarbons (petroleum) so their use is to save vegetable oils which are used during the preparation of soaps.

Question 21.
Explain the following terms with suitable examples :
(i) Cationic detergents
(ii) Anionic detergents and
(iii) Non-ionic detergents
Answer:
(i) Cationic detergents: These are quaternary ammonium salts of amines with acetates, chlorides or bromides.
Example: Cetyl trimethyl ammonium bromide
(ii) Anionic detergents: These detergents have large anionic part in their molecules. These are of two types :
(a) Sodium alkyl sulphates: For example, Sodium lauryl sulphate CH₃ (CH₂)₁₀ CH₂OSO₃^–Na⁺.
(b) Sodium alkyl benzene sulphonates: For example, sodium dodecylbenzene sulphonate

(iii) Non-ionic detergents: These are esters of high molecular mass alcohols with fatty acids. For example, Polyethylene glycol stearate.
CH₃ (CH₂)₁₆ COO(CH₂CH₂O)_nCH₂CH₂OH.

Question 22.
What are biodegradable and non-biodegradable detergents? Give one example of each.
Answer:
Detergents having straight hydrocarbon chains are easily degraded by microorganisms and hence are called biodegradable detergents, whereas detergents containing branched hydrocarbon chains are not easily degraded by the microorganisms and hence are called non-biodegradable detergents. As a result, non-biodegradable detergents accumulate in rivers and waterways thereby causing severe water pollution.

Examples of biodegradable detergents are : sodium lauryl sulphate, sodium 4-(l-dodecyl) benzene-sulphonate and sodium 4-(2-dodecyl) benzenesulphonate. An example of non-biodegradable detergent is sodium 4-(l, 3, 5, 7-tetramethyloctyl) benzenesulphonate.

Question 23.
Why do soaps not work with hard water?
Answer:
Calcium and magnesium salts present in hard water react with soaps to form insoluble compounds, which form curdy white precipitates and are difficult to remove from the clothes.

Question 24.
Can you use soaps and synthetic detergents to check the hardness of water?
Answer:
We can use soaps to check the hardness of water because with hard water, soaps give a gummy mass (sticky precipitate) but we cannot use detergents for this purpose because they give foam with both hard and soft water.

Question 25.
Explain the cleansing action of soaps.
Answer:
The cleansing action of soap is due to the fact that soap molecules, such as sodium stearate form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the oil droplet like the bristles.

Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus, soap helps in emulsification and washing away of oils and fats. The negatively charged sheath around the globules prevents them from coming together and forming aggregates.

Question 26.
If the water contains dissolved calcium hydrogen carbonate, out of soaps and synthetic detergents which one will you use for cleaning clothes?
Answer:
Synthetic detergents.

Question 27.
Label the hydrophilic and hydrophobic parts in the following compounds.
(i) CH₃(CH₂)₁₀CH₂OSO₃^–Na⁺
(ii) CH₃(CH₂)₁₅N⁺(CH₃)₃ Br^–
(iii) CH₃(CH₂)₁₆COO(CH₂CH₂O)_n CH₂CH₂OH
Answer:

Chemistry Guide for Class 12 PSEB Chemistry in Everyday Life Textbook Questions and Answers

Question 1.
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but It is not advisable to take Its doses without consultation with the doctor. Why?
Answer:
Sleeping pills contain drugs that may be tranquillizers or antidepressants. They affect the nervous system, relieve anxiety, stress, irritability or excitement. But they should strictly be used under the supervision of a doctor. If not, the uncontrolled and overdosage can cause harm to the body and mind because in higher doses these drugs act as poisons.

Question 2.
With reference to which classification has the statement, “ranitidine is an antacid” been given?
Answer:
This statement refers to the classification according to the pharmacological effect of the drug because any drug which will be used to counteract the effect of excess acid in the stomach will be called antacid.

Question 3.
Why do we require artifical sweetening agents?
Answer:
Natural sweeteners (sucrose etc.) provide calories to the body. Taking extra calories is harmful for diabetic patients. So, artificial sweeteners are used (i) to control intake of calories and (ii) as a substitute of sugar for diabetics.

Question 4.
Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structural formulae of these compounds are given below.
(i) (C₁₅H₃₁COO)₃C₃H₅ – Glyceryl pfi]mitate
(ii) (C₁₇H₃₂COO)₃C₃H₅ – Glyceryl oleate
Answer:

Question 5.
Following types of non-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic parts In the molecule. Identify the functional group(s) present In the molecule.

Answer:

(b) Functional groups: Ether and alcohol.