JEE Chemistry Solutions MCQs Set 32

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Multiple Choice Questions

Question 1: The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively, at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are:
(a) 450 mmHg, 0.4, 0.6
(b) 500 mmHg, 0.5, 0.5
(c) 450 mmHg, 0.5, 0.5
(d) 500 mmHg, 0.4, 0.6
Show Answer & Explanation

Answer: (d) 500 mmHg, 0.4, 0.6

Explanation:
1. \( p_{total} = p^{\circ}_A x_A + p^{\circ}_B x_B = 400 \times 0.5 + 600 \times 0.5 = 500 \) mmHg.
2. \( y_A = p_A/p_{total} = 200/500 = 0.4 \), and \( y_B = 1 - 0.4 = 0.6 \).

Teacher's Note:
1. Apply Raoult's law for each component, then Dalton's law.
2. The vapour is richer in the more volatile component \( B \).

Question 2: The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01 M BaCl₂ in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY (in mol L⁻¹) in solution is:
(a) 4 × 10⁻²
(b) 16 × 10⁻⁴
(c) 4 × 10⁻⁴
(d) 6 × 10⁻²
Show Answer & Explanation

Answer: (d) 6 × 10⁻²

Explanation:
1. π = iCRT. For XY, \( i = 2 \); for BaCl₂, \( i = 3 \).
2. 2 × [XY] = 4 × 3 × 0.01, so [XY] = 0.06 mol \( L^{-1} = 6 \times 10^{-2} \).

Teacher's Note:
1. Osmotic pressure depends on the total number of particles.
2. Equate \( i_{1}C_{1} \) with \( 4 \times i_{2}C_{2} \).

Question 3: Liquid M and liquid N form an ideal solution. The vapour pressures of pure liquids M and N are 450 and 700 mmHg, respectively, at the same temperature. If x_M and x_N are the mole fractions in solution and y_M and y_N are the mole fractions in vapour phase, the correct statement is:
(a) x_M/x_N > y_M/y_N
(b) x_M/x_N = y_M/y_N
(c) x_M/x_N < y_M/y_N
(d) (x_M − y_M) < (x_N − y_N)
Show Answer & Explanation

Answer: (a) x_M/x_N > y_M/y_N

Explanation:
1. By Raoult's law \( y_M/y_N = (p^{\circ}_M x_M)/(p^{\circ}_N x_N) = (450/700)(x_M/x_N) \).
2. Since 450/700 < \( 1, y_M/y_N \) < \( x_M/x_N \), i.e. \( x_M/x_N \) > \( y_M/y_N \).

Teacher's Note:
1. The more volatile liquid \( N \) is enriched in the vapour phase.
2. Compare the ratio of mole fractions in liquid and vapour.

Question 4: What would be the molality of a 20% (mass/mass) aqueous solution of KI? (Molar mass of KI = 166 g mol⁻¹)
(a) 1.48
(b) 1.51
(c) 1.35
(d) 1.08
Show Answer & Explanation

Answer: (b) 1.51

Explanation:
1. 20% by mass means \( 20 g \) KI in \( 80 g \) of water.
2. Molality = \( (20/166) / 0.080 \) kg = 1.506 ≈ 1.51 mol kg⁻¹.

Teacher's Note:
1. Molality uses the mass of solvent only.
2. \( 100 g \) of solution contains \( 80 g \) of water.

Question 5: At room temperature, a dilute solution of urea is made by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, the lowering of vapour pressure will be (molar mass of urea = 60 g mol⁻¹):
(a) 0.027 mmHg
(b) 0.031 mmHg
(c) 0.017 mmHg
(d) 0.028 mmHg
Show Answer & Explanation

Answer: (c) 0.017 mmHg

Explanation:
1. Moles of urea = 0.01; moles of water = \( 360/18 = 20 \).
2. \( x_{urea} = 0.01/20.01 = 0.0005 \).
3. \( \Deltap = x_{solute} \times p^{\circ} = 0.0005 \times 35 = 0.0175 \approx 0.017 \) mmHg.

Teacher's Note:
1. Lowering of vapour pressure equals the mole fraction of solute times \( p^{\circ} \).
2. Urea is a non-electrolyte, so \( i = 1 \).

Question 6: The mole fraction of a solvent in an aqueous solution of a solute is 0.8. The molality (in mol kg⁻¹) of the aqueous solution is:
(a) 13.88 × 10⁻²
(b) 13.88 × 10⁻¹
(c) 13.88
(d) 13.88 × 10⁻³
Show Answer & Explanation

Answer: (c) 13.88

Explanation:
1. Take 1 mol of solution: 0.8 mol water and 0.2 mol solute.
2. Mass of water = \( 0.8 \times 18 = 14.4 g = 0.0144 \) kg.
3. Molality = \( 0.2/0.0144 = 13.88 \) mol kg⁻¹.

Teacher's Note:
1. Pick a convenient basis such as one mole of solution.
2. Convert grams of solvent to kilograms.

Question 7: A solution is made by dissolving 0.6 g of urea (molar mass = 60 g mol⁻¹) and 1.8 g of glucose (molar mass = 180 g mol⁻¹) in 100 mL of water at 27°C. The osmotic pressure of the solution is (R = 0.08206 L atm K⁻¹ mol⁻¹):
(a) 8.2 atm
(b) 2.46 atm
(c) 4.92 atm
(d) 1.64 atm
Show Answer & Explanation

Answer: (c) 4.92 atm

Explanation:
1. Moles of urea = 0.01 and of glucose = 0.01, total 0.02 mol in \( 0.1 L \).
2. π = CRT = \( (0.02/0.1) \times 0.08206 \times 300 = 4.92 \) atm.

Teacher's Note:
1. Osmotic pressure depends on total moles of solute particles.
2. Convert \( 27^{\circ}C \) to \( 300 K \).

Question 8: Which one of the following statements regarding Henry's law is not correct?
(a) Different gases have different K_H (Henry's law constant) values at the same temperature
(b) Higher the value of K_H at a given pressure, higher is the solubility of the gas in the liquids
(c) The value of K_H increases with increase of temperature and K_H is a function of the nature of the gas
(d) The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution
Show Answer & Explanation

Answer: (b) Higher the value of K_H at a given pressure, higher is the solubility of the gas in the liquids

Explanation:
1. Henry's law: \( p = K_H x \), so \( x = p/K_H \).
2. At a given pressure a higher \( K_H \) means a lower mole fraction of dissolved gas, i.e. lower solubility. Statement \( (b) \) is wrong.

Teacher's Note:
1. Solubility is inversely related to \( K_H \).
2. \( K_H \) rises with temperature, so gases are less soluble in hot water.

Question 9: A solution containing 62 g of ethylene glycol in 250 g of water is cooled to −10°C. If K_f for water is 1.86 K kg mol⁻¹, then the amount of water (in g) separated as ice is:
(a) 32
(b) 48
(c) 64
(d) 16
Show Answer & Explanation

Answer: (c) 64

Explanation:
1. Ethylene glycol is 1 mol (62 g/mol) and \( \DeltaT_f = 10 K = 1.86 \times m \), so \( m = 5.376 \) mol kg⁻¹.
2. Water remaining in solution = 1 mol / 5.376 mol kg⁻¹ = 0.186 kg = \( 186 g \).
3. Ice separated = \( 250 - 186 = 64 g \).

Teacher's Note:
1. As ice separates, the solution becomes more concentrated.
2. Find the mass of water still liquid at \( -10^{\circ}C \).

Question 10: The amount of sugar (C₁₂H₂₂O₁₁) required to prepare 2 L of its 0.1 M aqueous solution is:
(a) 17.1 g
(b) 68.4 g
(c) 136.8 g
(d) 34.2 g
Show Answer & Explanation

Answer: (b) 68.4 g

Explanation:
1. Moles = \( 0.1 \times 2 = 0.2 \) mol.
2. Mass = \( 0.2 \times 342 = 68.4 g \).

Teacher's Note:
1. Molar mass of sucrose is 342 g/mol.
2. Moles = molarity × volume \( (L) \).

Question 11: The freezing point of a diluted milk sample is found to be −0.2°C, while it should have been −0.5°C for pure milk. How much water has been added to pure milk to make the diluted sample?
(a) 2 cups of water to 3 cups of pure milk
(b) 1 cup of water to 3 cups of pure milk
(c) 3 cups of water to 2 cups of pure milk
(d) 1 cup of water to 2 cups of pure milk
Show Answer & Explanation

Answer: (c) 3 cups of water to 2 cups of pure milk

Explanation:
1. \( \DeltaT_f ∝ \) molality, so for the same amount of milk solute the water amounts are in inverse ratio: w(pure)/w(diluted) = \( 0.2/0.5 = 2/5 \).
2. So 2 parts of water in pure milk become 5 parts: 3 parts of water must be added to 2 parts of pure milk.

Teacher's Note:
1. Dilution lowers the molality and therefore the depression of freezing point.
2. The solute amount is the same in both samples.

Question 12: K₂HgI₄ is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is:
(a) 1.6
(b) 1.8
(c) 2.2
(d) 2.0
Show Answer & Explanation

Answer: (b) 1.8

Explanation:
1. K₂[HgI₄] → \( 2K^{+} \) + [HgI₄]²⁻, so \( n = 3 \) ions.
2. \( i = 1 + \alpha(n - 1) = 1 + 0.4 \times 2 = 1.8 \).

Teacher's Note:
1. Use \( i = 1 + (n - 1)\alpha \).
2. The complex ion does not dissociate further.

Question 13: 8 g of NaOH is dissolved in 18 g of H₂O. The mole fraction of NaOH in solution and the molality (in mol kg⁻¹) of the solution respectively are:
(a) 0.2, 11.11
(b) 0.167, 22.20
(c) 0.2, 22.20
(d) 0.167, 11.11
Show Answer & Explanation

Answer: (d) 0.167, 11.11

Explanation:
1. Moles of NaOH = \( 8/40 = 0.2 \); moles of water = \( 18/18 = 1 \).
2. x(NaOH) = \( 0.2/1.2 = 0.167 \).
3. Molality = 0.2/0.018 kg = 11.11 mol kg⁻¹.

Teacher's Note:
1. Mole fraction uses total moles in the denominator.
2. Molality uses kilograms of solvent.

Question 14: Molecules of benzoic acid (C₆H₅COOH) dimerise in benzene. w g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2 K. If the percentage association of the acid to form dimer in the solution is 80, then w is (given K_f = 5 K kg mol⁻¹, molar mass of benzoic acid = 122 g mol⁻¹):
(a) 1.8 g
(b) 1.0 g
(c) 2.4 g
(d) 1.5 g
Show Answer & Explanation

Answer: (c) 2.4 g

Explanation:
1. 2 C₆H₅COOH ⇌ (C₆H₅COOH)₂: for \( \alpha = 0.8 \), moles = \( 0.2 + 0.4 = 0.6 \) per mole, so \( i = 0.6 \).
2. \( \DeltaT_f = iK_f(w/122)(1000/30) \): \( 2 = 0.6 \times 5 \times w \times 1000/(122 \times 30) \).
3. \( w = 2.44 g \approx 2.4 g \).

Teacher's Note:
1. Association reduces the number of particles, so \( i \) < 1.
2. Use \( \DeltaT_f = iK_f m \).

Question 15: 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, the percentage association of acetic acid in benzene will be (K_f for benzene = 5.12 K kg mol⁻¹):
(a) 64.6%
(b) 80.4%
(c) 74.6%
(d) 94.6%
Show Answer & Explanation

Answer: (d) 94.6%

Explanation:
1. Molality = \( (0.2/60)/0.020 = 0.1667 \) mol kg⁻¹.
2. \( 0.45 = i \times 5.12 \times 0.1667 \), so \( i = 0.527 \).
3. For dimerisation \( i = 1 - \alpha/2 \), so \( \alpha = 2(1 - 0.527) = 0.946 \), i.e. 94.6%.

Teacher's Note:
1. Dimerisation reduces the particle count: \( i = 1 - \alpha/2 \).
2. Find \( i \) from the observed depression first.

Question 16: 18 g of glucose (C₆H₁₂O₆) is added to 178.2 g of water. The vapour pressure of water (in torr) for this aqueous solution is (p° of water = 760 torr):
(a) 76.0
(b) 752.4
(c) 759.0
(d) 7.6
Show Answer & Explanation

Answer: (b) 752.4

Explanation:
1. Moles of glucose = 0.1; moles of water = \( 178.2/18 = 9.9 \).
2. \( x_{glucose} = 0.1/10 = 0.01 \), so \( p = 760 \times (1 - 0.01) = 752.4 \) torr.

Teacher's Note:
1. Vapour pressure of solution = \( p^{\circ} \times x_{solvent} \).
2. Glucose is non-volatile.

Question 17: 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?
(a) They all have the same osmotic pressure
(b) 0.100 M Mg₃(PO₄)₂(aq) has the highest osmotic pressure
(c) 0.125 M Na₃PO₄(aq) has the highest osmotic pressure
(d) 0.500 M C₂H₅OH(aq) has the highest osmotic pressure
Show Answer & Explanation

Answer: (a) They all have the same osmotic pressure

Explanation:
1. π = iCRT. Ethanol: \( 1 \times 0.5 = 0.5 \). Mg₃(PO₄)₂: \( 5 \times 0.1 = 0.5 \).
2. KBr: \( 2 \times 0.25 = 0.5 \). Na₃PO₄: \( 4 \times 0.125 = 0.5 \).
3. All have iC = \( 0.5 M \), so they have the same osmotic pressure.

Teacher's Note:
1. Compute iC for each solution.
2. The osmotic pressure depends on the total particle concentration.

Question 18: K_f for water is 1.86 K kg mol⁻¹. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C₂H₆O₂) must you add to get the freezing point of the solution lowered to −2.8°C?
(a) 72 g
(b) 93 g
(c) 39 g
(d) 27 g
Show Answer & Explanation

Answer: (b) 93 g

Explanation:
1. \( \DeltaT_f = 2.8 = K_f \times (w/62)/1.0 \) kg, so \( w = 2.8 \times 62/1.86 \).
2. \( w = 93.3 g \approx 93 g \).

Teacher's Note:
1. Ethylene glycol is a non-electrolyte \( (i = 1) \).
2. Molar mass of \( C_{2}H_{6}O_{2} \) is 62 g/mol.

Question 19: The degree of dissociation (α) of a weak electrolyte, AₓB_y, is related to the van't Hoff factor (i) by the expression:
(a) α = (i − 1)/(x + y − 1)
(b) α = (i − 1)/(x + y + 1)
(c) α = (x + y − 1)/(i − 1)
(d) α = (x + y + 1)/(i − 1)
Show Answer & Explanation

Answer: (a) α = (i − 1)/(x + y − 1)

Explanation:
1. \( A_{x}B_y \) ⇌ \( xA^{y+} + yB^{x-} \) gives \( n = x + y \) particles.
2. \( i = 1 + \alpha(n - 1) = 1 + \alpha(x + y - 1) \), so \( \alpha = (i - 1)/(x + y - 1) \).

Teacher's Note:
1. \( n \) is the total number of ions produced per formula unit.
2. Rearrange \( i = 1 + (n - 1)\alpha \).

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