JEE Chemistry Solutions MCQs Set 26

Answer: (d) -3.53 C

Explanation:
1. Moles of HBr = 8.1/81 = 0.1; molality = 1.
2. i = 1 + alpha = 1.9.
3. \( \Delta T_f=1.9\times1.86\times1=3.53 \) K, so freezing point = -3.53 C.

Teacher's Note:
1. HBr gives 2 ions, so i = 1 + (2-1) x 0.9.
2. The molality is 0.1 mol / 0.1 kg.

Answer: (c) I and II

Explanation:
1. I: lower vapour pressure of the BaCl2 solution (3 ions versus 2) is a colligative effect.
2. II: depression of freezing point is colligative.
3. III: compares two pure liquids, so it is not a colligative property.

Teacher's Note:
1. Colligative properties need a solute dissolved in a solvent.
2. They depend on the number of particles of solute.

Answer: (a) 85 g/mol

Explanation:
1. \( \dfrac{p^0-p}{p^0}=X_2=\dfrac{1.8}{53.3}=0.03377 \).
2. Moles of benzene = 500/78 = 6.410; \( n_2=X_2n_1/(1-X_2)=0.2240 \).
3. M = 19/0.2240 = 84.8, about 85 g/mol.

Teacher's Note:
1. Using the dilute approximation \( X_2\approx n_2/n_1 \) would give 88 g/mol.
2. Use the exact mole fraction when the lowering is as large as 3%.

Answer: (d) 0.66

Explanation:
1. \( p_X=0.5\times200=100 \); \( p_Y=0.5\times100=50 \).
2. \( y_X=\dfrac{100}{150}=0.667\approx0.66 \).

Teacher's Note:
1. Vapour is richer in the more volatile liquid X.
2. y_X + y_Y = 1.

Answer: (b) 1165 g

Explanation:
1. \( \Delta T_f=7=1.86\,m \), so m = 3.763 mol/kg.
2. Moles for 5 kg water = 18.82 mol.
3. Mass = 18.82 x 62 = 1167 g, about 1165 g.

Teacher's Note:
1. Multiply molality by the mass of water in kg.
2. Ethylene glycol is a non-electrolyte.

Answer: (b) I and II only

Explanation:
1. Dissolved salts depress the freezing point and elevate the boiling point.
2. As water evaporates the solution becomes more concentrated, so its boiling point rises further.
3. III and IV are false: sea water boils higher and is denser.

Teacher's Note:
1. Colligative properties follow the concentration of the solution.
2. Sea water is about 3.5% salt.

Answer: (d) Osmotic pressure

Explanation:
1. Polymer solutions are of very low molar concentration, so the boiling-point and freezing-point changes are too small to measure.
2. Osmotic pressure is large enough even for dilute solutions and is measured at room temperature.

Teacher's Note:
1. Osmotic pressure is also used for proteins and other macromolecules.
2. Pressure is measured without heating, protecting the polymer.

Answer: (c) Solubility

Explanation:
1. Colligative properties depend only on the number of solute particles.
2. Solubility depends on the nature of the solute and solvent.

Teacher's Note:
1. There are four colligative properties: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.
2. Solubility changes with the identity of the gas or solid.

Answer: (a) 3

Explanation:
1. Gibbs phase rule: \( F=C-P+2=3-P \).
2. F cannot be negative, so P is at most 3 (ice, liquid and vapour at the triple point).

Teacher's Note:
1. At the triple point F = 0.
2. Along a boundary line two phases coexist.

Answer: (d) 70 g/mol

Explanation:
1. \( \dfrac{p^0-p_s}{p_s}=\dfrac{0.04}{2.5}=0.016 \) (loss of pure solvent over loss of solution).
2. For a dilute solution this equals \( \dfrac{n}{N}=\dfrac{5/M}{80/18} \).
3. \( 5/M=0.016\times\dfrac{80}{18}=0.0711 \), so M = 70.3 g/mol.

Teacher's Note:
1. The Ostwald-Walker method measures relative lowering of vapour pressure.
2. Gain in weight of the drying agent measures p0.

Answer: (b) 300 torr

Explanation:
1. Let the liquid left have mole fraction x of A. Then the vapour has \( y_A=1-x \) (1 mol vapour, 1 mol liquid).
2. \( P=100x+900(1-x)=900-800x \) and \( y_A=\dfrac{100x}{P} \).
3. \( 100x=(1-x)(900-800x) \) gives \( 8x^2-18x+9=0 \), so x = 0.75.
4. P = 900 - 600 = 300 torr.

Teacher's Note:
1. Check: y_A = 0.25 and 100 x 0.75 / 300 = 0.25.
2. The other root x = 1.5 is impossible.

Answer: (c) Vaporisation of the pure solvent

Explanation:
1. A solution has higher entropy than the pure solvent, while the vapour is the same.
2. So the entropy increase on vaporisation is smaller for a solution; it is smaller still for an electrolyte solution (more particles, more disorder).

Teacher's Note:
1. The same argument explains the elevation of boiling point.
2. Entropy of vapour is unchanged.

Answer: (b) Both I and II are correct

Explanation:
1. Osmosis is the flow of solvent through a semipermeable membrane.
2. Reverse osmosis: pressure greater than the osmotic pressure is applied on the salt solution, forcing pure water out.

Teacher's Note:
1. Reverse osmosis is used for desalination.
2. The applied pressure must exceed pi.

Answer: (a) 1.85 atm

Explanation:
1. Osmotic pressure is proportional to molarity. Mixing equal volumes halves each concentration.
2. \( \pi=\dfrac{1.20+2.5}{2}=1.85 \) atm.

Teacher's Note:
1. Total moles stay the same while the volume doubles.
2. Both solutes are non-electrolytes.

Answer: (c) \( \ce{Hg2Cl2} \) molecules

Explanation:
1. \( m=1.24/34.3=0.03615 \) mol/kg, so moles = 0.03615 x 0.05 = 1.808 x 10^-3.
2. Molar mass = 0.849 / 1.808 x 10^-3 = 469.7 g/mol.
3. Hg2Cl2 has M = 471: the compound is undissociated, as Hg2Cl2 molecules.

Teacher's Note:
1. HgCl would have M = 235.5 and ions would give a lower observed mass.
2. Molten HgCl2 is a poor ionising solvent.

Answer: (d) 0.124 mol/L

Explanation:
1. \( C=k_Hp=3.1\times10^{-2}\times4=0.124 \) mol/L.

Teacher's Note:
1. Here the constant is in solubility form (C = kp).
2. On opening, the pressure falls to about 0.0004 atm and gas escapes.

Answer: (b) 3

Explanation:
1. Start with 1 mol H3A: it gives 1 H2A- + 1 H+ completely.
2. Of H2A-, 50% dissociates: 0.5 H+ + 0.5 HA(2-) formed, leaving 0.5 H2A-.
3. Of HA(2-) (0.5 mol), 20% dissociates: 0.1 H+ and 0.1 A(3-), leaving 0.4 HA(2-).
4. Total particles = H+ 1.6 + H2A- 0.5 + HA(2-) 0.4 + A(3-) 0.1 = 2.6, i = 2.6, about 3.

Teacher's Note:
1. i is the ratio of total particles after dissociation to the initial moles.
2. Each step only affects the species formed in the previous step.

Answer: (c) (A) Decimolar \( \ce{Al2(SO4)3} \)

Explanation:
1. \( i=1+(n-1)\alpha \): Al2(SO4)3 (n = 5): 4.6; BaCl2 (n = 3): 2.8; Na2SO4 (n = 3): 2.8.
2. Effective concentrations: A 0.46 M, B and C 0.28 M.
3. In the mixture BaSO4 precipitates and the particle concentration falls to about 0.3 M, below (A).

Teacher's Note:
1. Mixing three equal volumes dilutes each to one third, and BaSO4 removes particles.
2. So the Al2(SO4)3 solution is highest.

Answer: (b) Ice is less dense than water

Explanation:
1. Ice has a larger volume than the same mass of water.
2. By Le Chatelier's principle, raising the pressure favours the smaller-volume phase (water), so ice melts at a lower temperature.

Teacher's Note:
1. The solid-liquid line of water has a negative slope.
2. Most substances show the opposite, with melting points rising with pressure.

Answer: (a) 2.5 L

Explanation:
1. NaCl 0.1 M gives [Cl-] = 0.1 M; CaCl2 0.2 M gives [Cl-] = 0.4 M.
2. Mix x L of NaCl and y L of CaCl2: \( 0.1x+0.4y=0.34(x+y) \), so \( y=4x \).
3. y cannot exceed 2 L, so y = 2 and x = 0.5. Total volume = 2.5 L.

Teacher's Note:
1. Check: Cl- moles = 0.05 + 0.8 = 0.85; 0.85/2.5 = 0.34.
2. The limit comes from the available volume of CaCl2.

Answer: (d) \( \dfrac{X_A}{X_B}<\dfrac{Y_A}{Y_B} \)

Explanation:
1. \( \dfrac{Y_A}{Y_B}=\dfrac{p_A^0X_A}{p_B^0X_B} \).
2. Since \( p_A^0/p_B^0>1 \), \( Y_A/Y_B>X_A/X_B \): the vapour is richer in the more volatile A (so \( Y_A>X_A \)).

Teacher's Note:
1. Lever rule results from Raoult's and Dalton's laws.
2. This is why fractional distillation works.

Answer: (c) All four are correct

Explanation:
1. \( p=K_HX \) with K_H specific to the gas, the solvent and the temperature.
2. Solubility is inversely proportional to K_H.
3. Gases become less soluble on heating, so K_H rises with temperature.

Teacher's Note:
1. The values of K_H for N2 and O2 differ, giving different solubilities in water.
2. Carbonated drinks fizz more when warm.

Answer: (a) 3

Explanation:
1. For equal masses of solute and solvent, \( \Delta T_b\propto 1/M \).
2. \( M_P/M_Q=0.4/0.2=2 \), so M_Q = 50.
3. x : y = 2 : 1, x + y = 3.

Teacher's Note:
1. Larger molar mass gives smaller elevation.
2. The solvent mass and Kb cancel in the ratio.

Answer: (b) 9

Explanation:
1. \( i=\dfrac{7.2\times10^{-3}}{1.8\times10^{-3}}=4 \), so 1 + x = 4 and x = 3: \( \ce{K3[Fe(CN)6]} \).
2. Primary valency = oxidation state of Fe = +3; secondary valency = coordination number = 6.
3. Sum = 9.

Teacher's Note:
1. Complex ion plus x potassium ions gives i = x + 1.
2. Charge: x = 3 since [Fe(CN)6] carries 3-.

Answer: (b) 4

Explanation:
1. Partial pressure of A = 0.25 x 4 = 1 atm.
2. \( X_A=\dfrac{p}{K_H}=\dfrac{1}{2500}=4\times10^{-4} \).
3. Moles of water = 36/18 = 2, so \( n_A\approx4\times10^{-4}\times2=8\times10^{-4} \). x = 4.

Teacher's Note:
1. Use partial pressure, not the total pressure.
2. n_A is negligible against 2 mol of water.