JEE Chemistry Solutions MCQs Set 04

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

Explanation:
1. \( Y_A/Y_B = (P^0_A X_A)/(P^0_B X_B) \).
2. Since \( P^0_A > P^0_B \), the ratio \( Y_A/Y_B \) exceeds \( X_A/X_B \).

Teacher's Note:
1. The vapour is richer in the more volatile component.
2. Only the ratio statement holds for all compositions.

Answer: (a) 0.25

Explanation:
1. \( p_A = 0.4\times100 = 40 \); \( p_B = 0.6\times200 = 120 \).
2. \( Y_A = 40/160 = 0.25 \).

Teacher's Note:
1. Vapour mole fraction = partial pressure/total pressure.
2. B is more volatile, so A is less than 0.4 in vapour.

Answer: (c) \( \dfrac{P^0-P_s}{P_s} = \dfrac{n}{N} \)

Explanation:
1. \( P_s = P^0\,N/(n+N) \), so \( P^0 - P_s = P^0\,n/(n+N) \).
2. Dividing, \( (P^0-P_s)/P_s = n/N \) exactly; here n is moles of solute and N moles of solvent.

Teacher's Note:
1. The relative lowering \( (P^0-P_s)/P^0 \) equals \( n/(n+N) \), not n/N.
2. The exact simple ratio n/N appears when the denominator is Ps.

Answer: (b) 85.88 mm Hg

Explanation:
1. Vapour above solution: \( p_A = 25 \), \( p_B = 60 \); \( y_A = 25/85 = 0.294 \), \( y_B = 0.706 \).
2. The condensate has this composition: \( P = 0.294\times100 + 0.706\times80 = 85.88 \) mm Hg.

Teacher's Note:
1. The distillate has the composition of the vapour, not of the original liquid.
2. Apply Raoult's law again to the condensate.

Answer: (a) 1 M C\(_{12}\)H\(_{22}\)O\(_{11}\)

Explanation:
1. Lowering of vapour pressure depends on the number of particles.
2. Sucrose does not dissociate (i = 1), so it lowers the vapour pressure least.

Teacher's Note:
1. Fewer particles give less lowering and hence higher vapour pressure.
2. Order of i: AlCl3 (4) > CaCl2 (3) > NaCl (2) > sucrose (1).

Answer: (c) 1.0

Explanation:
1. Glucose is a non-electrolyte and does not associate or dissociate.
2. So \( i = 1 \).

Teacher's Note:
1. i is 1 for non-electrolytes in ideal dilute solutions.
2. It is greater than 1 for dissociation and less than 1 for association.

Answer: (b) \( T_1 > T_2 \)

Explanation:
1. In very dilute solution, AlCl\(_3\) gives 4 ions (i = 4) and CaCl\(_2\) gives 3 ions (i = 3).
2. Elevation is proportional to i, so \( T_1 > T_2 \).

Teacher's Note:
1. More particles give a larger elevation of boiling point.
2. Compare i values for equal molalities.

Answer: (d) [MA\(_6\)]A\(_2\)

Explanation:
1. \( i = 0.0054/(1.86\times0.001) = 2.9 \approx 3 \).
2. Three ions form: [MA\(_6\)]\(^{2+}\) and 2 A\(^-\), so the formula is [MA\(_6\)]A\(_2\).

Teacher's Note:
1. Compute i from the freezing point depression and round to the nearest whole number of ions.
2. Counter ions outside the bracket are the ones that dissociate.

Answer: (c) 5.7

Explanation:
1. \( x_B = 0.05 \), so \( n_A/n_B = 19 \).
2. \( w_A/w_B = 19\,M_A/M_B = 19\times0.3 = 5.7 \).

Teacher's Note:
1. Convert mole ratio to weight ratio by multiplying by the ratio of molar masses.
2. The relative lowering is 5%, so the mole fraction of the solute is 0.05.

Answer: (a) 120 and 45

Explanation:
1. At \( X_B = 0 \), \( P = P^0_A = 120 \).
2. At \( X_B = 1 \), \( P = P^0_B = 120-75 = 45 \).

Teacher's Note:
1. Pure-component values are found at the end points of the composition range.
2. A straight line of P against X is the ideal behaviour.

Answer: (d) Decimolar Al\(_2\)(SO\(_4\))\(_3\)

Explanation:
1. Al\(_2\)(SO\(_4\))\(_3\): \( i = 1+0.9\times4 = 4.6 \).
2. BaCl\(_2\) and Na\(_2\)SO\(_4\): \( i = 1+0.9\times2 = 2.8 \).
3. The mixture is not simply additive (BaSO\(_4\) would precipitate) and has lower particle concentration than 4.6 × 0.1.

Teacher's Note:
1. Boiling point elevation is proportional to i × molality.
2. The mixture's particle count is at most that of 2.8 × 0.1 = 0.28 per litre.

Answer: (b) 0.6

Explanation:
1. Decrease in vapour pressure is proportional to the mole fraction of solute.
2. Doubling the decrease doubles \( x_{solute} \) to 0.4, so \( x_{solvent} = 0.6 \).

Teacher's Note:
1. \( \Delta P = P^0 x_{solute} \) with P0 constant.
2. The question asks for the solvent fraction, not the solute.

Answer: (b) 0.98 \( K_b \)

Explanation:
1. 1 L of solution weighs 1200 g and contains 180 g of glucose, so water = 1020 g.
2. Molality = 1/1.020 = 0.98 m, so \( \Delta T_b = 0.98K_b \).

Teacher's Note:
1. Molality is moles per kg of solvent, not solution.
2. The solution is denser than water, so molality is below molarity.

Answer: (d) less than 111

Explanation:
1. CaCl\(_2\) dissociates, so the number of particles exceeds the number of formula units (i > 1).
2. The apparent molar mass is 111/i, which is less than 111.

Teacher's Note:
1. Dissociation raises the colligative effect and reduces the calculated molar mass.
2. Association would have the opposite effect.

Answer: (a) 374.76 K

Explanation:
1. A\(_2\)B\(_3\) gives 5 ions, so \( i = 1+0.6\times4 = 3.4 \).
2. \( \Delta T_b = 3.4\times0.52\times1 = 1.77 \) K, so boiling point = 373 + 1.77 = 374.77 K ≈ 374.76 K.

Teacher's Note:
1. Use i = 1 + (n-1)α, where n is the number of ions.
2. Water boils at 373 K.

Answer: (c) Plot of \( 1/P_{total} \) versus \( Y_A \) is linear

Explanation:
1. In vapour, \( p_A = Y_AP \) so \( X_A = Y_AP/P^0_A \) and \( X_B = Y_BP/P^0_B \); adding gives \( 1 = P(Y_A/P^0_A + Y_B/P^0_B) \).
2. So \( 1/P = Y_A/P^0_A + (1-Y_A)/P^0_B \), which is linear in \( Y_A \) (and in \( Y_B \)).

Teacher's Note:
1. For ideal solutions P is linear in the liquid mole fraction, but 1/P is linear in the vapour mole fraction.
2. Option D is wrong because 1/P is linear in Y_B as well.

Answer: (d) 141.4 mm Hg

Explanation:
1. Total moles = 20; 10 mol are in vapour, 10 in liquid. Let the liquid have \( x_A \).
2. A balance: \( 10 = 10x_A + 10y_A \) with \( y_A = 200x_A/P \) and \( P = 100(1+x_A) \).
3. This gives \( x_A = \sqrt2 - 1 = 0.414 \), so \( P = 100(1.414) = 141.4 \) mm Hg.

Teacher's Note:
1. Write mass balance for A between liquid and vapour.
2. The solution is a quadratic in x_A with one physical root.

Answer: (a) None of these (the value is 220 torr)

Explanation:
1. \( x_A = 0.4,\ x_B = 0.6 \).
2. \( P = 0.4\times100 + 0.6\times300 = 220 \) torr.

Teacher's Note:
1. Use mole fractions in the liquid phase.
2. The three other options come from wrong mole fraction assignments.

Answer: (c) 4.04 × 10\(^{-2}\)

Explanation:
1. Relative lowering = 0.05/17.25 = 2.9×10\(^{-3}\) = mole fraction of ions.
2. Moles of ions per litre ≈ 2.9×10\(^{-3}\)/(1-2.9×10\(^{-3}\)) × 55.55 = 0.1615.
3. XCl\(_3\) gives 4 ions, so solubility = 0.1615/4 = 4.04×10\(^{-2}\) M.

Teacher's Note:
1. Count all ions when relating particles to formula units.
2. Take 55.55 mol of water per litre of dilute solution.

Answer: (b) 405 torr

Explanation:
1. First solution: \( 0.6P^0_A + 0.4P^0_B = 600 \).
2. Second: A = 4.5 mol, B = 2 mol, C = 0.5 mol, so \( (4.5/7)P^0_A + (2/7)P^0_B = 630 \).
3. Solving, \( P^0_A = 730 \) torr and \( P^0_B = 405 \) torr.

Teacher's Note:
1. Non-volatile C counts in the total moles but contributes no vapour pressure.
2. Verify your answer in both equations.

Answer: (a) 1.39 × 10\(^{-3}\)

Explanation:
1. \( i = 0.20/(1.8\times0.1) = 1.111 \), so \( \alpha = 0.111 \).
2. \( K = C\alpha^2/(1-\alpha) = 0.1\times0.01235/0.8889 = 1.39\times10^{-3} \).

Teacher's Note:
1. For a weak electrolyte, i = 1 + α.
2. Then apply Ostwald's expression for K.

Answer: (c) 87%

Explanation:
1. Ba(NO\(_3\))\(_2\) gives 3 ions, so \( i = 1+2\alpha \).
2. \( \alpha = (2.74-1)/2 = 0.87 \).

Teacher's Note:
1. Use i = 1 + (n-1)α.
2. Here n = 3.

Answer: (b) 777 torr

Explanation:
1. \( m = 1.04/0.52 = 2 \); mole fraction of solute = 2/(2+55.56) = 0.03476.
2. \( 750 = P_{atm}(1-0.03476) \), so \( P_{atm} = 777 \) torr.

Teacher's Note:
1. At the boiling point of pure water the pure solvent vapour pressure equals the atmospheric pressure.
2. The solution has a lower vapour pressure at that temperature.

Answer: (d) 1, 3 and 4 only

Explanation:
1. Positive deviation gives a minimum-boiling azeotrope, whose boiling point is below both pure components, so statement 2 is correct.
2. The mixture boils below acetone only over a range of compositions, not always, so statement 1 is incorrect.
3. Adding CS\(_2\) to acetone raises vapour pressure and lowers the boiling point, so 3 is incorrect.
4. An azeotrope cannot be separated completely by fractional distillation, so 4 is incorrect.

Teacher's Note:
1. A minimum-boiling azeotrope forms for positive deviation.
2. Only the three statements 1, 3, 4 are wrong.

Answer: (c) Statement 1 is false and statement 2 is true

Explanation:
1. Freezing releases the latent heat of fusion, so it is exothermic.
2. Heat must indeed be removed from water for it to freeze, so statement 2 is true.

Teacher's Note:
1. Freezing is exothermic and melting is endothermic.
2. Removal of heat is needed for the process, because the process itself gives out heat.

Answer: (a) Statement 1 is true and statement 2 is false

Explanation:
1. Non-volatile solutes lower the freezing point of the solvent, so statement 1 is true.
2. Statement 2 is too general: substances that dissolve and remain in solution lower it, but, for example, an insoluble substance does not, and a substance that forms a solid solution can raise or not change it.

Teacher's Note:
1. Freezing point depression holds for non-volatile solutes that do not separate with the solvent.
2. General claims with 'any substance' are usually false.