CBSE Class 12 Chemistry Solutions NEET JEE Questions Set 03

Single Correct Answer

Question 1. An element X of atomic mass 25.0 exists as X₄ in benzene to the extent of 100%. When 10.30 g of saturated solution of X is added to 20.0 g of benzene, the depression in freezing point of the resulting solution is 0.51 K. If K_f for benzene is 5.1 K kg mol⁻¹, the solubility of X in 100 g of benzene will be
(a) 3.0 g
(b) 2.7 g
(c) 0.30 g
(d) 0.27 g
Answer:
Ans. (a)
Suppose saturated solution of X in benzene contains w g of X (present as X₄). Hence, benzene present = (10.30 − w) g
∴ Total benzene present = 20 + (10.30 − w) = (30.30 − w) g
\( \Delta T_f = \frac{1000 K_f w_2}{w_1 \times M_2} \) ∴ \( 0.51 = \frac{1000 \times 5.1 \times w}{(30.30 - w) \times 100} \) (M₂ of X₄ = 25 × 4 = 100)
or 51(30.3 − w) = 5100 w or 30.3 − w = 100 w or 101 w = 30.3 or w = 0.3 g
Thus, 10.30 g of saturated solution contains 0.3 g of X and 10.0 g of benzene.
∴ Solubility of X in 100 g of benzene = 3.0 g

Teacher's Note:
1. Remember that the benzene in the saturated solution adds to the 20 g taken; and use the molar mass of X₄, not of X.

Question 2. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol⁻¹. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T) [Molecular weight of ethanol is 46 g mol⁻¹]. Among the following, the option representing change in the freezing point is (JEE Advanced 2017)
(a) Plot (a)
(b) Plot (b)
(c) Plot (c)
(d) Plot (d)

Answer:
Ans. (a)
As T increases, V.P. increases. Hence, options (c) and (d) are wrong.
ΔT_f = K_f × m; \( T_f^\circ - T_f = 2 \times \frac{34.5/46}{0.5} \)
273 − T_f = 3 or T_f = 270
Hence, only option (a) is correct.

Teacher's Note:
1. The freezing point is where the solution's vapour pressure curve meets the ice curve; vapour pressure can only rise with temperature.

Question 3. 0.075 molal solution of sucrose containing 1 kg water is kept at a temperature of −0.4°C. If K_f(H₂O) = 1.86 K/m, how much ice in g will freeze out? (JEE Main 2021)
(a) 350 g
(b) 100 g
(c) 180 g
(d) 650 g
Answer:
Ans. (d)
\( \Delta T_f = K_f \times m = K_f \times \frac{w_2}{M_2} \times \frac{1}{w_1} \times 1000 = K_f \times n_2 \times \frac{1}{w_1} \times 1000 \)
\( 0.4 = 1.86 \times 0.075 \times \frac{1000}{w_1} \) (w₁ = water present)
or w₁ = 348.75 g
∴ Water frozen as ice = 1000 − 348.75 = 651.25 ≈ 650 g

Teacher's Note:
1. As ice separates, the solute stays in less water and the solution becomes more concentrated; find the water left that gives ΔT_f = 0.4.

Question 4. Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y respectively in 1 kg of water. The ratio of the depression in freezing point for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is (JEE Main 2022)
(a) 1 : 4
(b) 1 : 0.25
(c) 1 : 0.20
(d) 1 : 5
Answer:
Ans. (b)
ΔT_f = iK_fm = K_fm (i = 1 for both non-electrolytes)
\( \frac{\Delta T_f(A)}{\Delta T_f(B)} = \frac{m_A}{m_B} = \frac{1/M_X}{1/M_Y} = \frac{M_Y}{M_X} \)
But \( \frac{\Delta T_f(A)}{\Delta T_f(B)} = \frac{1}{4} = 0.25 \) (Given). ∴ \( \frac{M_Y}{M_X} = \frac{0.25}{1} \) or M_X : M_Y = 1 : 0.25

Teacher's Note:
1. For equal masses, depression is inversely proportional to molar mass: the solute giving 4 times the depression has 1/4 the molar mass.

Question 5. The correct equation for the degree of association 'α' of an associating solute, 'n' molecules of which undergo association in solution is (Kerala PMT 2015)
(a) \( \alpha = \frac{n(i-1)}{1-n} \)
(b) \( \alpha = \frac{i(n-1)}{1+n} \)
(c) \( \alpha = \frac{i(n+1)}{1-n} \)
(d) \( \alpha = \frac{i(n+1)}{n-1} \)
(e) \( \alpha = \frac{n(1-i)}{1-n} \)
Answer:
Ans. (a)
nA ⇄ A_n; 1 mole; after association: 1 − α, \( \frac{\alpha}{n} \); Total \( = 1 - \alpha + \frac{\alpha}{n} \)
∴ \( i = 1 - \alpha + \frac{\alpha}{n} \) or \( 1 - i = \alpha - \frac{\alpha}{n} = \alpha\left(1 - \frac{1}{n}\right) = \alpha\left(\frac{n-1}{n}\right) \)
or \( \alpha = \frac{n}{(n-1)}(1 - i) = \frac{n(i-1)}{(1-n)} \)

Teacher's Note:
1. Note that (a) and (e) look alike but have opposite signs; check with n = 2, i = 0.5, which must give α = 1.

Question 6. 1 g of a monobasic acid in 100 g of water lowers the freezing point by 0.168°. If 0.2 g of the same acid requires 15.1 mL of N/10 alkali for complete neutralisation, the degree of dissociation of the acid is (K_f for water = 1.86 K kg mol⁻¹)
(a) 9.8%
(b) 19.6%
(c) 4.9%
(d) 1.68%
Answer:
Ans. (b)
M₂ (mol mass of acid) \( = \frac{1000 \times K_f \times w_2}{w_1 \times \Delta T_f} = \frac{1000 \times 1.86 \times 1}{100 \times 0.168} \) = 110.71, i.e., Observed molecular mass = 110.71
15.1 mL of N/10 alkali neutralise acid = 0.2 g
∴ 1000 mL of 1 N alkali will neutralise acid \( = \frac{0.2}{15.1} \times 1000 \times 10 \) = 132.45 g
∴ Eq. wt. of the acid = 132.45. Mol. mass of the acid = 132.45 (as it is monobasic). This is calculated (normal) mol. mass
\( i = \frac{\text{Normal mol. mass}}{\text{Observed mol. mass}} = \frac{132.45}{110.71} = 1.196 \)
HA ⇄ H⁺ + A⁻; Initial 1, 0, 0; After disso. 1 − α, α, α; Total = 1 + α
∴ i = 1 + α or α = i − 1 = 0.196 = 19.6%

Teacher's Note:
1. Titration gives the true (normal) molar mass; the freezing point gives the observed one; their ratio is i.

Question 7. The freezing point (in °C) of a solution containing 0.1 g of K₃[Fe(CN)₆] (Mol. Wt. 329) in 100 g of water (K_f = 1.86 K kg mol⁻¹) is
(a) −2.3 × 10⁻²
(b) −5.7 × 10⁻²
(c) −5.7 × 10⁻³
(d) −1.2 × 10⁻²
Answer:
Ans. (a)
K₃[Fe(CN)₆] ⇄ 3K⁺ + [Fe(CN)₆]³⁻ ∴ i = 4
\( \Delta T_f = iK_f m = i \times K_f \times \frac{w_2}{M} \times \frac{1}{w_1} \times 1000 = 4 \times 1.86 \times \frac{0.1}{329} \times \frac{1}{100} \times 1000 \)
= 0.023 = 2.3 × 10⁻² °C or K
T_f = 0 − 2.3 × 10⁻² °C = −2.3 × 10⁻² °C

Teacher's Note:
1. Count ions carefully: K₃[Fe(CN)₆] gives 3 K⁺ and one complex ion, so i = 4 (the complex ion itself does not break up).

Question 8. The freezing point depression constant for water is −1.86°C m⁻¹. If 5.00 g Na₂SO₄ is dissolved in 45.0 g H₂O, the freezing point is changed by −3.82°C. Calculate the van't Hoff factor for Na₂SO₄ (AIPMT Prelim 2011)
(a) 0.381
(b) 2.05
(c) 2.63
(d) 3.11
Answer:
Ans. (c)
\( \Delta T_f = i \times K_f \times m = i \times K_f \times \frac{w_2}{M_2} \times \frac{1}{w_1} \times 1000 \)
\( 3.82 = i \times 1.86 \times \frac{5}{142} \times \frac{1}{45} \times 1000 \) (Molar mass of Na₂SO₄ = 142)
or i = 2.62

Teacher's Note:
1. i below the ideal value of 3 for Na₂SO₄ shows incomplete dissociation (ion pairing) in a concentrated solution.

Question 9. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If K_f for water is 1.86° C/m, the freezing point of the solution will be (AIPMT Main 2011)
(a) −0.18°C
(b) −0.54°C
(c) −0.36°C
(d) −0.24°C
Answer:
Ans. (d)
HA → H⁺ + A⁻; 1 − α, α, α; 1 − 0.3, 0.3, 0.3
∴ i = (1 − 0.3) + 0.3 + 0.3 = 1.3
ΔT_f = iK_fm = 1.3 × 1.86 × 0.1 = 0.2418
T_f = 0 − 0.2418°C = −0.2418°C ≈ −0.24°C.

Teacher's Note:
1. For a binary electrolyte i = 1 + α; don't use the fully dissociated value i = 2 for a weak acid.

Question 10. Benzoic acid undergoes dimerisation in benzene solution. The van't Hoff factor (i) is related to the degree of association 'x' of the acid as
(a) i = (1 − x)
(b) i = (1 + x)
(c) i = (1 − x/2)
(d) i = (1 + x/2)
Answer:
Ans. (c)
2C₆H₅COOH ⇄ (C₆H₅COOH)₂
Before asso. 1 mol; After asso. 1 − x, \( \frac{x}{2} \); Total \( = 1 - x + \frac{x}{2} = 1 - \frac{x}{2} \)
∴ \( i = \frac{1 - x/2}{1} = 1 - \frac{x}{2} \).

Teacher's Note:
1. Association always makes i less than 1, so options with a plus sign can be ruled out at once.

Question 11. The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be (K_f for benzene = 5.12 K kg mol⁻¹) (JEE Main 2017)
(a) 76.6%
(b) 94.6%
(c) 64.6%
(d) 80.4%
Answer:
Ans. (b)
ΔT_{f (observed)} = 0.45°
\( \Delta T_{f\,(calculated)} = \frac{1000 K_f w_2}{w_1 \times M_2} = \frac{1000 \times 5.12 \times 0.2}{20 \times 60} \) [M₂(CH₃COOH) = 60] = 0.853°
∴ \( i = \frac{\Delta T_{f\,(observed)}}{\Delta T_{f\,(calculated)}} = \frac{0.45}{0.853} = 0.527 \)
2CH₃COOH ⇄ (CH₃COOH)₂; 1 − α, \( \frac{\alpha}{2} \); Total moles \( = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2} \)
\( i = 1 - \frac{\alpha}{2} = 0.527 \) or \( \frac{\alpha}{2} = 0.473 \) or α = 0.946
% association = 94.6%

Teacher's Note:
1. For dimerisation, α = 2(1 − i).

Question 12. A 0.004 M solution of Na₂SO₄ is isotonic with a 0.010 M solution of glucose at the temperature. The apparent degree of dissociation of Na₂SO₄ is
(a) 25%
(b) 50%
(c) 75%
(d) 85%
Answer:
Ans. (c)
π(Na₂SO₄) = iCRT = i(0.004)RT; π(Glucose) = CRT = 0.010 RT
As solutions are isotonic, i(0.004)RT = 0.01 RT. This gives i = 2.5
Now, Na₂SO₄ ⇄ 2Na⁺ + SO₄²⁻; 1 mole, 0, 0; 1 − α, 2α, α; Total = 1 + 2α
∴ i = 1 + 2α or \( \alpha = \frac{i-1}{2} = \frac{2.5-1}{2} \) = 0.75 = 75%.

Teacher's Note:
1. General formula: α = (i − 1)/(n − 1), where n is the number of ions from one formula unit (n = 3 for Na₂SO₄).

Question 13. For a weak monobasic acid, if pK_a = 4, then at a concentration of 0.01 M of the acid solution, the van't Hoff factor is
(a) 1.01
(b) 1.02
(c) 1.10
(d) 1.20
Answer:
Ans. (c)
pK_a = 4 means K_a for HA = 10⁻⁴
For weak acid, HA ⇄ H⁺ + A⁻, K_a = Cα² (Ostwald's dilution law)
∴ \( \alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{10^{-4}}{0.01}} = 10^{-1} = 0.10 \)
HA ⇄ H⁺ + A⁻; Initial 1 mole; Moles after dissoc. 1 − α, α, α; Total = 1 + α
i = 1 + α = 1 + 0.10 = 1.10

Teacher's Note:
1. Get α from K_a = Cα² (valid when α is small), then i = 1 + α.

Question 14. The pH of 1 M solution of a weak monobasic acid (HA) is 2. Then, the van't Hoff factor is
(a) 1.01
(b) 1.02
(c) 1.10
(d) 1.20
Answer:
Ans. (a)
pH = 2 means [H⁺] = 10⁻² M
HA ⇄ H⁺ + A⁻; Initial C mol L⁻¹, 0, 0; After disso. C − Cα, Cα, Cα; Total = C(1 + α)
Thus, [H⁺] = Cα, i.e., 10⁻² = 1 × α or α = 10⁻²
i = 1 + α = 1 + 0.01 = 1.01

Teacher's Note:
1. [H⁺] = Cα for a monobasic acid; with C = 1 M, α equals [H⁺] directly.

Question 15. At a certain Hill station, water boils at 96°C. The amount of NaCl that should be added to one litre of water so that it boils at 100°C will be (K_b for H₂O = 0.52 K/m)
(a) 450 g
(b) 225 g
(c) 125 g
(d) 250 g
Answer:
Ans. (b)
Required ΔT_b = 100 − 96 = 4°
\( \Delta T_b = iK_b m = iK_b \frac{w_2}{M_2} \times \frac{1}{w_1} \times 1000 \), i.e., \( 4 = 2 \times 0.52 \times \frac{w_2}{58.5} \times \frac{1}{1000} \times 1000 \)
or w₂ = 225 g (1 L H₂O = 1000 g)

Teacher's Note:
1. Take i = 2 for NaCl; forgetting the van't Hoff factor gives 450 g, which is option (a).

Question 16. One molal solution of a complex of cobalt chloride with NH₃ in water showed an elevation in boiling point equal to 2.08°. Assuming that the complex is completely ionized in the solution, the complex is (K_b for water = 0.52 K kg mol⁻¹)
(a) [Co(NH₃)₆]Cl₃
(b) [Co(NH₃)₅Cl]Cl₂
(c) [Co(NH₃)₄Cl₂]Cl
(d) None of these
Answer:
Ans. (a)
ΔT_b = iK_bm; 2.08 = i × 0.52 × 1 or i = 4
This means that the complex produces 4 ions in the solution.
Hence, the complex is [Co(NH₃)₆]Cl₃.

Teacher's Note:
1. Only chlorides outside the coordination sphere ionise; i = 1 + number of ionisable Cl⁻.

Question 17. Depression in freezing point of 0.01 m aqueous acetic acid solution is found to be 0.02046 K. One molal urea solution freezes at −1.86°C. Assuming molarity equal to molality, pH of acetic acid solution is
(a) 2
(b) 3
(c) 3.2
(d) 4.2
Answer:
Ans. (b)
For 1 m urea solution, ΔT_f = K_fm gives K_f = 1.86° C/m
For acetic acid, ΔT_f = iK_fm; 0.02046 = i × 1.86 × 0.01 or i = 1.1
CH₃COOH ⇄ CH₃COO⁻ + H⁺; Initial C mol L⁻¹; After disso. C − Cα, Cα, Cα; Total = C + Cα = C(1 + α)
∴ \( i = \frac{C(1+\alpha)}{C} = 1 + \alpha \) or α = i − 1 = 0.1
[H⁺] = Cα = (0.01)(0.1) = 10⁻³ M. Hence, pH = 3

Teacher's Note:
1. The urea data are given only to find K_f; then i → α → [H⁺] → pH.

Question 18. The average osmotic pressure of human blood is 7.8 bar at 37°C. What is the concentration of an aqueous NaCl solution that could be used in the blood stream?
(a) 0.15 mol/L
(b) 0.30 mol/L
(c) 0.60 mol/L
(d) 0.45 mol/L
Answer:
Ans. (a)
NaCl sol. used should be isotonic with blood stream. For NaCl, i = 2. π = iCRT
\( C = \frac{\pi}{iRT} = \frac{7.8\ \text{bar}}{2 \times 0.083\ \text{bar L K}^{-1}\text{mol}^{-1} \times 310\ \text{K}} \) = 0.15 mol L⁻¹.

Teacher's Note:
1. A solution injected into blood must be isotonic with it; with i = 2, the NaCl molarity is half the total particle concentration.

Teacher's Note:
1. A semipermeable membrane lets only solvent molecules through; solute particles (ions here) never cross it.

Question 20. Pure benzene freezes at 5.3°C. A solution of 0.223 g of phenylacetic acid (C₆H₅CH₂COOH) in 4.4 g of benzene (K_f = 5.12 K kg mol⁻¹) freezes at 4.47°C. From this observation, one can conclude that
(a) phenylacetic acid exists as such in benzene
(b) phenylacetic acid undergoes partial ionization in benzene
(c) phenylacetic acid undergoes complete ionization in benzene
(d) phenylacetic acid dimerizes in benzene
Answer:
Ans. (d)
Observed molecular mass of phenylacetic acid \( = \frac{1000 \times 5.12 \times 0.223}{(5.3 - 4.47) \times 4.4} \) = 312.6
Calculated molecular mass of C₆H₅CH₂COOH = 72 + 5 + 12 + 2 + 12 + 32 + 1 = 136
As observed molecular mass is nearly double of the theoretical value, it dimerizes in benzene.

Teacher's Note:
1. Carboxylic acids dimerise through hydrogen bonding in non-polar solvents such as benzene, so their observed molar mass is about double.

Question 21. Consider separate solutions of 0.500 M C₂H₅OH (aq), 0.100 M Mg₃(PO₄)₂ (aq), 0.250 M KBr (aq) and 0.125 M Na₃PO₄ (aq) at 25°C. Which statement is true about these solutions, assuming all salts to be strong electrolytes? (JEE Main 2014)
(a) 0.500 M C₂H₅OH (aq) has the highest osmotic pressure
(b) They all have the same osmotic pressure
(c) 0.100 M Mg₃(PO₄)₂ (aq) has the highest osmotic pressure
(d) 0.125 M Na₃PO₄ (aq) has the highest osmotic pressure
Answer:
Ans. (b)
Osmotic pressure for electrolytes, π = iCRT
π(C₂H₅OH) = 1 × 0.5 RT = 0.5 RT; π(Mg₃(PO₄)₂) = 5 × 0.100 RT = 0.5 RT
π(KBr) = 2 × 0.25 RT = 0.5 RT; π(Na₃PO₄) = 4 × 0.125 RT = 0.5 RT
Thus, all solutions have the same osmotic pressure.

Teacher's Note:
1. Compare i × C (total particle concentration), not C alone.

Question 22. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point? (JEE Main 2018)
(a) [Co(H₂O)₆]Cl₃
(b) [Co(H₂O)₅Cl]Cl₂.H₂O
(c) [Co(H₂O)₄Cl₂]Cl.2H₂O
(d) [Co(H₂O)₃Cl₃].3H₂O
Answer:
Ans. (d)
Maximum freezing point means minimum depression in freezing point. Less the number of particles produced, less is the depression in freezing point. We have:
(a) [Co(H₂O)₆]Cl₃ → [Co(H₂O)₆]³⁺ + 3Cl⁻, i = 4
(b) [Co(H₂O)₅Cl]Cl₂.H₂O → [Co(H₂O)₅Cl]²⁺ + 2Cl⁻, i = 3
(c) [Co(H₂O)₄Cl₂]Cl.2H₂O → [Co(H₂O)₄Cl₂]⁺ + Cl⁻, i = 2
(d) [Co(H₂O)₃Cl₃].3H₂O → No ionization, i = 1
Thus, [Co(H₂O)₃Cl₃].3H₂O produces minimum number of particles. Hence, it has highest freezing point.

Teacher's Note:
1. Highest freezing point = smallest ΔT_f = fewest particles; water of crystallisation does not count as a particle.

Question 23. One gram of silver gets distributed between 10 cm³ of molten zinc and 100 cm³ of molten lead at 800°C. The distribution coefficient of silver between molten zinc and molten lead is 300. The percentage of silver still left in the lead layer is approximately (Karnataka CET 2011)
(a) 2
(b) 5
(c) 3
(d) 1
Answer:
Ans. (c)
According to Nernst distribution law, \( \frac{\text{Conc. (\%) of Ag in molten Zn}}{\text{Conc. (\%) of Ag in molten Pb}} = K \) (distribution coeff.) = 300
If at equilibrium, x g of Ag is left in 100 cm³ of molten lead, Ag that goes into 10 cm³ zinc = (1 − x) g
∴ \( \frac{10(1-x)}{x} = 300 \) or 10 − 10x = 300x or 310x = 10 or \( x = \frac{1}{31} \) g
Thus, out of 1 g of Ag, Ag present in Pb layer \( = \frac{1}{31} \) g. ∴ % of Ag left in the Pb layer \( = \frac{1}{31} \times 100 \) ≈ 3%

Teacher's Note:
1. In distribution problems compare concentrations (amount per unit volume), not the amounts, in the two layers.

One or More Correct Answers

Question 24. Mixture(s) showing positive deviations from Raoult's law at 35°C is (are) (JEE Advanced 2016)
(a) carbon tetrachloride + methanol
(b) carbon disulphide + acetone
(c) benzene + toluene
(d) phenol + aniline
Answer:
Ans. (a, b)
CCl₄ + CH₃OH = Positive deviation from Raoult's law
CS₂ + O=C(CH₃)₂ = Positive deviation from Raoult's law
Benzene + Toluene = Ideal solution
C₆H₅OH + C₆H₅NH₂ = Negative deviation from Raoult's law

Teacher's Note:
1. Positive deviation arises when A–B attractions are weaker than A–A/B–B (e.g., H-bonds of methanol broken by CCl₄); phenol–aniline form new H-bonds, giving negative deviation.

Question 25. For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here x_L and x_M represent mole fractions of L and M respectively, in the solution. The correct statement(s) applicable to this system is (are) (JEE Advanced 2017)
(a) The point Z represents vapour pressure of pure M and Raoult's law is obeyed from x_L = 0 to x_L = 1
(b) Attractive intermolecular interactions between L–L in pure liquid L and M–M in pure liquid M are stronger than those between L–M when mixed in solution
(c) The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed when x_L → 0
(d) The point Z represents vapour pressure of pure liquid L and Raoult's law is obeyed when x_L → 1

Answer:
Ans. (b, d)
Point Z represents the point when mole fraction of M is zero, i.e., for pure L and not for pure M. Hence, (a) and (c) are wrong.
When x_L → 1, solution is very dilute (containing very very small amount of M). Hence, it obeys Raoult's law, i.e., (d) is correct.
As the given plot shows +ve deviation from Raoult's law, L–M interactions are weaker than L–L and M–M interactions. Hence, (b) is correct.

Teacher's Note:
1. The solvent (component in large excess) obeys Raoult's law near its own pure end, even in a non-ideal solution.

Passage-Based Questions

Question 26. Comprehension 1: An ideal solution of two liquids is a solution in which each component obeys Raoult's law which states that the vapour pressure of any component in the solution depends on the mole fraction of that component in the solution and the vapour pressure of that component in the pure state. However, there are many solutions which do not obey Raoult's law. In other words, they show deviations from ideal behaviour which may be positive or negative. However, in either case, corresponding to a particular composition, they form a constant boiling mixtures called azeotropes.
A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20°C are 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in the vapour phase would be
(a) 0.200
(b) 0.478
(c) 0.549
(d) 0.786
Answer:
Ans. (b)
\( x_{pentane} = \frac{1}{5} = 0.2,\ x_{hexane} = 0.8 \)
p_pentane = 0.2 × 440 mm = 88 mm; p_hexane = 0.8 × 120 mm = 96 mm
∴ Mole fraction of pentane in vapour phase \( = \frac{88}{88 + 96} \) = 0.478

Teacher's Note:
1. A mole ratio of 1 : 4 means a mole fraction of 1/5, not 1/4.

Question 27. Comprehension 2: Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.
A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.
Given: Freezing point depression constant of water (K_f^water) = 1.86 K kg mol⁻¹; Freezing point depression constant of ethanol (K_f^ethanol) = 2.0 K kg mol⁻¹; Boiling point elevation constant of water (K_b^water) = 0.52 K kg mol⁻¹; Boiling point elevation constant of ethanol (K_b^ethanol) = 1.2 K kg mol⁻¹; Standard freezing point of water = 273 K; Standard freezing point of ethanol = 155.7 K; Standard boiling point of water = 373 K; Standard boiling point of ethanol = 351.5 K; Vapour pressure of pure water = 32.8 mm Hg; Vapour pressure of pure ethanol = 40 mm Hg; Molecular weight of water = 18 g mol⁻¹; Molecular weight of ethanol = 46 g mol⁻¹.
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. (IIT 2008)
The freezing point of solution M is
(a) 268.7 K
(b) 268.5 K
(c) 234.2 K
(d) 150.9 K
Answer:
Ans. (d)
Mole fraction of ethanol in the mixture = 0.9, i.e., 0.9 mole of ethanol is mixed with 0.1 mole of water, i.e., water is the solute and ethanol is the solvent.
∴ Molality of water in the solution \( = \frac{0.1}{0.9 \times 46} \times 1000 \) = 2.415
ΔT_f = K_f(ethanol) × m = 2.0 × 2.415 = 4.83 K
∴ Freezing point of the solution (ethanol) = T_f°(ethanol) − ΔT_f = 155.7 − 4.83 = 150.87 K = 150.9 K

Teacher's Note:
1. Decide the solvent first: the component with the larger mole fraction (ethanol, 0.9) is the solvent, so use ethanol's K_f and freezing point.

Matrix Match Questions

Teacher's Note:
1. Positive deviation goes with ΔH_mix > 0 and ΔV_mix > 0; negative deviation with ΔH_mix < 0 and ΔV_mix < 0.

Teacher's Note:
1. At equal molarity, compare i × C: the larger it is, the lower the freezing point and the higher the osmotic pressure.

Integer and Numerical Answer Questions

Question 30. Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions x_A and x_B respectively, has a vapour pressure of 22.5 Torr. The value of x_A/x_B in the new solution is ................ (Given that the vapour pressure of pure liquid A is 20 Torr at temperature T) (JEE Advanced 2018)
Answer:
Ans. 19.00
In 1st case: x_A = 0.5, x_B = 0.5, P_T = 45 torr. Applying Raoult's law, P_T = x_Ap_A° + x_Bp_B°
∴ 45 = 0.5 × 20 + 0.5 × p_B° or 0.5 p_B° = 45 − 10 = 35 or p_B° = 70 torr
In 2nd case: P_T = 22.5 torr, x_A = ?, x_B = ? ∴ P_T = x_Ap_A° + (1 − x_A)p_B°
22.5 = x_A × 20 + (1 − x_A) × 70 or 22.5 = 20x_A + 70 − 70x_A or 50x_A = 47.5
or \( x_A = \frac{47.5}{50} = \frac{19}{20} \); \( x_B = 1 - \frac{19}{20} = \frac{1}{20} \). ∴ \( \frac{x_A}{x_B} = \frac{19/20}{1/20} \) = 19

Teacher's Note:
1. Use the first (equimolar) data only to find p_B°; then solve the second case for x_A.

Question 31. The plot given in the figure shows P–T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.
On addition of equal number of moles of a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerisation in these solvents. If the degree of dimerisation is 0.7 in solvent Y, the degree of dimerisation in solvent X is ...... (JEE Advanced 2018)

Answer:
Ans. 0.05
From curves 1 and 2, elevation in boiling point of solvent X = 2°. For NaCl, i = 2
∴ (ΔT_b)_x = i(K_b)_xm or 2 = 2(K_b)_xm ...(i)
From curves 3 and 4, elevation in boiling point of solvent Y = 1°. (ΔT_b)_y = i(K_b)_ym; 1 = 2(K_b)_ym ...(ii)
Dividing eqn. (i) by eqn. (ii), \( \frac{(K_b)_x}{(K_b)_y} = \frac{2}{1} \) ...(iii). Also, we are given: \( \frac{(\Delta T_b)_x}{(\Delta T_b)_y} = \frac{3}{1} \)
As solute S undergoes dimerisation in solvent X as well as solvent Y: In solvent Y, 2S → S₂; Initial 1 mole, 0; After dimerisation (1 − α), \( \frac{\alpha}{2} \); Total \( = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2} \)
\( (i)_y = 1 - \frac{\alpha}{2} = 1 - \frac{0.7}{2} = \frac{1.3}{2} = 0.65 \)
In solvent X, 2S → S₂; Initial 1 mole, 0; After dimerisation 1 − β, \( \frac{\beta}{2} \); Total \( = 1 - \frac{\beta}{2} \); \( (i)_x = 1 - \frac{\beta}{2} \)
(ΔT_b)_x = (i)_x(K_b)_xm; (ΔT_b)_y = (i)_y(K_b)_ym. ∴ \( \frac{(\Delta T_b)_x}{(\Delta T_b)_y} = \frac{(i)_x}{(i)_y}\cdot\frac{(K_b)_x}{(K_b)_y} \)
\( 3 = \frac{(1 - \beta/2)}{0.65} \times 2 \) or \( 1 - \frac{\beta}{2} = \frac{3 \times 0.65}{2} = 0.975 \) or \( \frac{\beta}{2} = 0.025 \) or β = 0.05

Teacher's Note:
1. Use the NaCl curves only to get the ratio of K_b values of the two solvents; then compare the van't Hoff factors for the dimerising solute.

Question 32. On dissolving 0.5 g of a non-volatile non-ionic solute in 39 g of benzene, its vapour pressure decreases from 650 mm Hg to 640 mm Hg. The depression of freezing point of benzene (in K) upon addition of the solute is ............ (Given data: Molar mass and the molal freezing point depression constant of benzene are 78 g mol⁻¹ and 5.12 K kg mol⁻¹ respectively) (JEE Advanced 2019)
Answer:
Ans. 1.03
\( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1 + n_2} \) (n₂ = moles of solute, n₁ = moles of solvent)
\( \frac{650 - 640}{650} = \frac{n_2}{0.5 + n_2} \) (39 g C₆H₆ \( = \frac{39}{78} \) mole = 0.5 mole)
or \( \frac{1}{65} = \frac{n_2}{0.5 + n_2} \) or 0.5 + n₂ = 65n₂ or 64n₂ = 0.5 or \( n_2 = \frac{5}{640} \) mole
Molality of the solution \( = \frac{5}{640} \times \frac{1}{39} \times 1000 \)
\( \Delta T_f = K_f \times m = 5.12 \times \frac{5}{640} \times \frac{1}{39} \times 1000 \) = 1.0256 ≈ 1.03 K

Teacher's Note:
1. The solute mass (0.5 g) is not needed once the moles of solute are found from the vapour pressure data.

Question 33. The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm⁻³, the molarity of urea solution is ............ (Given data: Molar masses of urea and water are 60 g mol⁻¹ and 18 g mol⁻¹ respectively) (JEE Advanced 2019)
Answer:
Ans. 2.98
\( x_{urea} = \frac{n_{urea}}{n_{urea} + n_{H_2O}} \); \( 0.05 = \frac{n_{urea}}{n_{urea} + 50} \) (900 g H₂O \( = \frac{900}{18} \) moles = 50 moles)
or n_urea = 0.05 n_urea + 0.05 × 50 or 0.95 n_urea = 2.5 or \( n_{urea} = \frac{2.5}{0.95} \) = 2.63
Mass of solution = Mass of urea + Mass of water = 2.63 × 60 + 900 = 157.8 + 900 = 1057.8 g
Volume of solution \( = \frac{\text{Mass}}{\text{Density}} = \frac{1057.8}{1.2} \) = 881.5 mL
Molarity of solution \( = \frac{2.63}{881.5} \times 1000 \) = 2.98 mol L⁻¹

Teacher's Note:
1. Volume of solution comes from the total mass (solute + solvent) divided by density, not from the water volume.

Question 34. A 1 molal K₄Fe(CN)₆ solution has a degree of dissociation 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non-electrolyte solution A. The molar mass of A is ......... g/mol (Round off to the nearest integer) (JEE Main 2021)
Answer:
Ans. 85
ΔT_b[K₄Fe(CN)₆] = ΔT_b(A), (i × K_b × m)[K₄Fe(CN)₆] = (i × K_b × m)(A)
or (i × m)[K₄Fe(CN)₆] = (i × m)(A) ...(i)
K₄Fe(CN)₆ → 4K⁺ + Fe(CN)₆⁴⁻; 1 − α, 4α, α; i = 1 + 4α = 1 + 4 × 0.4 = 2.6
Substituting in eqn. (i), \( 2.6 \times 1 = 1 \times m = \frac{18.1}{M} \times \frac{1}{100 - 18.1} \times 1000 \)
\( 2.6 = \frac{18100}{81.9M} \) or M = 85

Teacher's Note:
1. 18.1 weight per cent means 18.1 g of A in 81.9 g of water (not in 100 g of water).

Question 35. 2.5 g of protein containing only glycine (C₂H₅NO₂) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be 5.03 × 10⁻³ bar. The total number of glycine units present in the protein is ......... (Given R = 0.083 L bar K⁻¹ mol⁻¹) (JEE Main 2022)
Answer:
Ans. 330
π = iCRT (i = 1); \( 5.03 \times 10^{-3} = \frac{2.5}{M} \times \frac{1}{500} \times 1000 \times 0.083 \times 300 \)
M (molar mass of protein) = 24751.5 g mol⁻¹; Molar mass of glycine (C₂H₅NO₂) = 75 g mol⁻¹
∴ No. of glycine units in protein \( = \frac{24751.5}{75} \) = 330

Teacher's Note:
1. Strictly, each peptide bond loses one H₂O (residue mass 57), but this exam question divides by 75, the molar mass of glycine.

Assertion–Reason Questions

Question 36. Assertion: Addition of a non-volatile solute to a volatile solvent increases the boiling point.
Reason: Addition of a non-volatile solute results in lowering of vapour pressure. (AIIMS 2014)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
Lowering of vapour pressure increases the boiling point as more heat is required to make vapour pressure equal to atmospheric pressure. Hence, R is the correct explanation of A.

Teacher's Note:
1. Elevation of boiling point is a direct result of lowering of vapour pressure.

Question 37. Assertion: The depression in freezing point depends on the amount of the solute dissolved and not on the nature of the solute or solvent.
Reason: For aqueous solutions of different electrolytes, molal depression constant will have different values.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct A. The depression in freezing point depends upon the nature of the solvent (because different solvents have different value for K_f).
Correct R. For aqueous solutions of different electrolytes, K_f is same because solvent is same, i.e., water.

Teacher's Note:
1. K_f is a property of the solvent only; it does not change with the solute.

Question 38. Assertion: 0.1 M solution of glucose has higher increment in the freezing point than 0.1 M solution of urea.
Reason: K_f for both has different values.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct A. Both being non-electrolytes with same molar concentration will have same decrease (not increment) in freezing point.
Correct R. Both being aqueous solution, i.e., solvent is same (water), K_f will be same for both.

Teacher's Note:
1. Dissolving a solute always lowers the freezing point; equal concentrations of non-electrolytes in the same solvent give equal lowering.

Question 39. Assertion: The molecular weight of acetic acid determined by depression in freezing point method in benzene and water is found to be different.
Reason: Water is polar and benzene is non-polar. (AIIMS 2005)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
R is the correct explanation of A. (In water, acetic acid undergoes dissociation while in benzene, it undergoes association.)

Teacher's Note:
1. In benzene, acetic acid dimerises (observed M about 120); in water it ionises slightly (observed M a little below 60).

Question 40. Assertion: Higher the molal depression constant of the solvent used, higher the freezing point of the solution.
Reason: Depression in freezing point does not depend on the nature of the solvent. (AIIMS 2011)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct A. Higher the molal depression constant of the solvent used, higher is the depression in freezing point.
Correct R. Depression in freezing point depends upon the nature of the solvent.

Teacher's Note:
1. Larger K_f means a larger ΔT_f, i.e., a lower (not higher) freezing point of the solution.