JEE Chemistry Solutions MCQs Set 12

Answer: (c) 0.20

Explanation:
1. Moles of glycerine = 46/92 = 0.5; moles of water = 36/18 = 2.
2. Mole fraction = 0.5/2.5 = 0.20.

Teacher's Note:
1. Molar mass of glycerine = 3(12) + 8(1) + 3(16) = 92.
2. Mole fraction is dimensionless.

Answer: (b) 0.192

Explanation:
1. Moles of urea = 18/60 = 0.30.
2. Mass of solution = 1500 × 1.052 = 1578 g, so mass of water = 1578 − 18 = 1560 g = 1.56 kg.
3. Molality = 0.30/1.56 = 0.192 m.

Teacher's Note:
1. Molality needs the mass of the solvent, so subtract the solute from the solution mass.
2. Density converts volume of solution to mass.

Answer: (c) 18.4 M

Explanation:
1. In 100 g solution there are 98 g H₂SO₄ = 1 mol.
2. Volume of 100 g = 100/1.84 = 54.35 mL = 0.05435 L.
3. Molarity = 1/0.05435 = 18.4 M.

Teacher's Note:
1. \( M=\frac{10\times\%\times d}{\text{molar mass}} \) is a handy shortcut.
2. Here 10 × 98 × 1.84/98 = 18.4.

Answer: (c) 0.787 M

Explanation:
1. Moles from first solution = 3 × 0.025 = 0.075.
2. Moles from second = 0.05 × 0.075 = 0.00375.
3. Total = 0.07875 mol in 0.100 L, so M = 0.7875 M.

Teacher's Note:
1. When mixing solutions of the same solute, add moles and add volumes.
2. Use M₁V₁ + M₂V₂ = M V.

Answer: (b) 0.8 M

Explanation:
1. Cl⁻ from HCl = 0.2 mol; from CaCl₂ = 2 × 0.1 = 0.2 mol.
2. Total = 0.4 mol in 0.5 L, so [Cl⁻] = 0.8 M.

Teacher's Note:
1. CaCl₂ gives two chloride ions per formula unit.
2. Add the contributions from every source.

Answer: (d) 2.4 atm

Explanation:
1. Pressure of N₂ = 1 − 0.3 = 0.7 atm. On reducing the volume to 1/3, N₂ pressure becomes 3 × 0.7 = 2.1 atm (Boyle's law).
2. The vapour pressure of water is fixed by temperature, so some vapour condenses and the water vapour pressure stays 0.3 atm.
3. Total = 2.1 + 0.3 = 2.4 atm.

Teacher's Note:
1. Vapour pressure of a liquid depends only on temperature, not on the volume of the container.
2. Apply Boyle's law only to the non-condensing gas.

Answer: (b) 85.88 mm Hg

Explanation:
1. Vapour pressure of the original solution: P = 0.25 × 100 + 0.75 × 80 = 85 mm Hg.
2. Mole fraction of A in the vapour = 25/85 = 0.2941, so the distillate has x(A) = 0.2941 and x(B) = 0.7059.
3. Vapour pressure of the distillate = 0.2941 × 100 + 0.7059 × 80 = 85.88 mm Hg.

Teacher's Note:
1. The first condensate has the composition of the vapour, which is richer in the more volatile component.
2. Apply Raoult's law twice: once to the liquid, then to the distillate.

Answer: (c) 135, 254

Explanation:
1. For an ideal mixture P = P°(B) + x(A)[P°(A) − P°(B)].
2. Comparing with 254 − 119 x(A): P°(B) = 254 and P°(A) − 254 = −119, so P°(A) = 135.

Teacher's Note:
1. The intercept at x(A) = 0 is the vapour pressure of pure B.
2. The value at x(A) = 1 is P°(A): 254 − 119 = 135.

Answer: (a) 7 : 1

Explanation:
1. Let x be the liquid mole fraction of benzene: 150x + 50(1 − x) = 120 gives x = 0.7.
2. Partial pressures: benzene 105 torr, toluene 15 torr.
3. Vapour mole fraction ratio = 105 : 15 = 7 : 1.

Teacher's Note:
1. In the vapour phase the mole ratio equals the ratio of partial pressures.
2. Find the liquid composition first.

Answer: (c) 90 torr

Explanation:
1. First solution (5 mol): \( \frac35P_A^\circ+\frac25P_B^\circ=600 \), that is 3P°(A) + 2P°(B) = 3000 … (1).
2. New solution: A = 4.5 mol, B = 2 mol, C = 0.5 mol (non-volatile, but it counts in the total of 7 mol). The vapour pressure is 630 torr: \( \frac{4.5}{7}P_A^\circ+\frac{2}{7}P_B^\circ=630 \), that is 4.5P°(A) + 2P°(B) = 4410 … (2).
3. (2) − (1): 1.5 P°(A) = 1410, so P°(A) = 940 torr.
4. Substituting in (1): 2P°(B) = 3000 − 2820 = 180, so P°(B) = 90 torr.

Teacher's Note:
1. The non-volatile solute adds to the total moles in the denominator of each mole fraction but contributes no vapour pressure.
2. Two unknowns need two equations: one before and one after the addition.

Answer: (b) ΔH(mix) < 0

Explanation:
1. Ideal value: P = 0.08 × 300 + 0.92 × 800 = 24 + 736 = 760 torr = 1 atm.
2. Observed: 0.95 atm = 722 torr, which is below the ideal value, so the deviation is negative.
3. Negative deviation means stronger A–B attractions, so heat is released (ΔH(mix) < 0) and the volume decreases (ΔV(mix) < 0).

Teacher's Note:
1. Observed pressure below the Raoult value means negative deviation.
2. Ideal solutions have ΔH(mix) = 0 and ΔV(mix) = 0.

Answer: (a) CHCl₃ + CH₃COCH₃

Explanation:
1. Ideal value: 0.4 × 300 + 0.6 × 800 = 120 + 480 = 600 torr.
2. Observed 580 torr is lower, so the solution shows negative deviation.
3. Chloroform and acetone form hydrogen bonds with each other, giving negative deviation. The other pairs are nearly ideal.

Teacher's Note:
1. Compare the observed pressure with the Raoult's law value first.
2. Pairs that form new hydrogen bonds on mixing show negative deviation.

Answer: (d) 1.01

Explanation:
1. pH = 2 gives [H⁺] = 0.01 M.
2. For 1 M HA, α = 0.01/1 = 0.01.
3. i = 1 + α = 1.01.

Teacher's Note:
1. Weak acid HA gives two particles when ionised: i = 1 + α.
2. Degree of ionisation α = [H⁺]/C.

Answer: (c) FeCl₃, 30% ionised

Explanation:
1. i = 1 + (n − 1)α.
2. KCl: 1 + 0.5 = 1.5. K₂SO₄: 1 + 2(0.4) = 1.8. FeCl₃: 1 + 3(0.3) = 1.9. SnCl₄: 1 + 4(0.2) = 1.8.
3. The maximum is 1.9 for FeCl₃.

Teacher's Note:
1. n is the number of ions produced by one formula unit.
2. Compare i values, not percentages.

Answer: (d) 3

Explanation:
1. Relative lowering = 0.05 = x(solute), so n₂/(n₁ + n₂) = 0.05 and n₂/n₁ = 0.05/0.95 = 0.0526.
2. For 1 kg water, n₁ = 55.5 mol, so n₂ = 0.0526 × 55.5 = 2.92 mol.
3. Molality ≈ 2.9, that is about 3.

Teacher's Note:
1. Convert the mole ratio into moles per kg of water using n₁ = 1000/18 = 55.5.
2. The result is approximate because the solution is not dilute.

Answer: (b) 87%

Explanation:
1. Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻, so n = 3 and i = 1 + 2α.
2. 2.74 = 1 + 2α gives α = 0.87, that is 87%.

Teacher's Note:
1. i = 1 + (n − 1)α.
2. Do not use n = 2 for this salt.

Answer: (b) 90 amu

Explanation:
1. Relative lowering = (10 − 9)/10 = 0.1 = x(B).
2. n(A) = 20/200 = 0.1 mol. x(B) = n(B)/(n(B) + 0.1) = 0.1 gives n(B) = 0.1/9 = 0.01111 mol.
3. M(B) = 1/0.01111 = 90 amu.

Teacher's Note:
1. Use the exact mole fraction expression rather than assuming dilution.
2. Molar mass = mass/moles.

Answer: (b) 5.7

Explanation:
1. P/P° = x(A) = 0.95, so x(B) = 0.05 and n(A)/n(B) = 19.
2. w(A)/w(B) = [n(A) M(A)]/[n(B) M(B)] = 19 × 0.3 = 5.7.

Teacher's Note:
1. The mole ratio times the molar mass ratio gives the mass ratio.
2. P/P° equals the mole fraction of the solvent.

Answer: (c) 0.6

Explanation:
1. ΔP = P° x(solute), so 10 = P° × 0.2 and P° = 50 mm Hg.
2. For ΔP = 20: x(solute) = 20/50 = 0.4.
3. x(solvent) = 1 − 0.4 = 0.6.

Teacher's Note:
1. The relative lowering of vapour pressure equals the mole fraction of solute.
2. Use the first case to find P°.

Answer: (b) 5 ppm

Explanation:
1. ppm = (mass of solute/mass of solution) × 10⁶ = (5/10⁶) × 10⁶ = 5 ppm.

Teacher's Note:
1. ppm means parts per million by mass.
2. The solution is so dilute that M and g L⁻¹ would be tiny numbers.