JEE Chemistry Solutions MCQs Set 25

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Multiple Choice Questions

Question 1: The mole fraction of glycerine \( \ce{C3H5(OH)3} \) in a solution of 36 g of water and 46 g of glycerine is:
(a) 0.46
(b) 0.36
(c) 0.20
(d) 0.40
Show Answer & Explanation

Answer: (c) 0.20

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

Teacher's Note:
1. The mole fractions of both components add to 1.
2. Here water is 0.80.

Question 2: For a weak monobasic acid HA with \( pK_a=4 \) at concentration 0.01 M, the van't Hoff factor is: (use \( K_a=C\alpha^2 \))
(a) 1.10
(b) 1.01
(c) 1.02
(d) 1.20
Show Answer & Explanation

Answer: (a) 1.10

Explanation:
1. \( K_a=10^{-4}=0.01\,\alpha^2 \), so \( \alpha=0.1 \).
2. \( i=1+(n-1)\alpha=1+0.1=1.10 \) for HA giving 2 particles.

Teacher's Note:
1. Weak electrolyte: i lies between 1 and the number of ions.
2. Use the small-alpha approximation, valid here as alpha = 0.1.

Question 3: The elevation in boiling point of an aqueous urea solution is 0.52 K (\( K_b=0.52 \) K kg mol\(^{-1}\)). The mole fraction of urea in the solution is:
(a) 0.982
(b) 0.567
(c) 0.943
(d) 0.018
Show Answer & Explanation

Answer: (d) 0.018

Explanation:
1. \( \Delta T_b=K_bm \), so m = 1 mol per kg of water.
2. Moles of water in 1 kg = 1000/18 = 55.56.
3. \( X_{urea}=\dfrac{1}{1+55.56}=0.0177\approx0.018 \).

Teacher's Note:
1. 0.982 is the mole fraction of water.
2. Molality 1 means 1 mol of solute in 55.56 mol of water.

Question 4: The density of a 1 M solution of the non-electrolyte glucose is 1.18 g/mL. If \( K_f \) of water is 1.86 K kg mol\(^{-1}\), the solution freezes at:
(a) -1.58 C
(b) -3.16 C
(c) -1.86 C
(d) +1.86 C
Show Answer & Explanation

Answer: (c) -1.86 C

Explanation:
1. 1 L of solution has mass 1180 g and contains 180 g of glucose.
2. Mass of water = 1180 - 180 = 1000 g, so the molality is 1.
3. \( \Delta T_f=1.86\times1=1.86 \) K; freezing point = -1.86 C.

Teacher's Note:
1. Here molality happens to equal molarity.
2. A 1 M solution of a heavier solute would differ.

Question 5: The total vapour pressure of a mixture of 1 mol of volatile A (\( p_A^0=100 \) mmHg) and 3 mol of volatile B (\( p_B^0=60 \) mmHg) is 75 mmHg. For this mixture:
(a) There is positive deviation from Raoult's law, the boiling point is lowered and A-B attraction is weaker than A-A and B-B
(b) There is negative deviation and the boiling point is raised
(c) The solution is ideal
(d) There is positive deviation, but the boiling point is raised
Show Answer & Explanation

Answer: (a) There is positive deviation from Raoult's law, the boiling point is lowered and A-B attraction is weaker than A-A and B-B

Explanation:
1. Ideal value = 0.25 x 100 + 0.75 x 60 = 25 + 45 = 70 mmHg.
2. Observed 75 mmHg is higher, so positive deviation.
3. Weaker A-B forces mean easier escape, higher vapour pressure and a lower boiling point.

Teacher's Note:
1. Positive deviation: delta H mix > 0, delta V mix > 0.
2. Examples: ethanol-water, benzene-methanol.

Question 6: Which solution has the highest freezing point? (i) 6 g urea in 100 g water, (ii) 6 g acetic acid in 100 g water, (iii) 6 g sodium chloride in 100 g water.
(a) 6 g acetic acid in 100 g water
(b) 6 g sodium chloride in 100 g water
(c) All have the same freezing point
(d) 6 g urea in 100 g water
Show Answer & Explanation

Answer: (d) 6 g urea in 100 g water

Explanation:
1. Moles: urea 0.1; acetic acid 0.1 (slight ionisation); NaCl 0.103 (completely ionised, i = 2).
2. Effective particles: urea 0.1, acetic acid a little above 0.1, NaCl about 0.21.
3. Smallest depression gives the highest freezing point: urea.

Teacher's Note:
1. Urea and acetic acid have the same molar mass (60 g/mol).
2. Electrolytes depress the freezing point more.

Question 7: Which of the following statements is/are correct? (1) The fundamental cause of colligative properties is the higher entropy of the solution than of the pure solvent. (2) The freezing point of a hydrofluoric acid solution is higher than that of an equimolal hydrochloric acid solution. (3) 1 M glucose and 0.5 M NaCl solutions are isotonic at a given temperature.
(a) (1) and (2) only
(b) (2) and (3) only
(c) (1), (2) and (3) are all correct
(d) (1) and (3) only
Show Answer & Explanation

Answer: (c) (1), (2) and (3) are all correct

Explanation:
1. (1) Solution has higher entropy, so the solvent has less tendency to leave.
2. (2) HF is a weak acid and gives fewer particles than the strong acid HCl, so its freezing point is higher.
3. (3) NaCl (i = 2) at 0.5 M gives 1 M of particles, equal to 1 M glucose.

Teacher's Note:
1. Isotonic solutions have the same osmotic pressure.
2. Check i for each solute before comparing.

Question 8: The vapour pressure of a pure liquid A is 40 mmHg at 310 K. In a solution with liquid B the vapour pressure of A is 32 mmHg. If Raoult's law is obeyed, the mole fraction of A in the solution is:
(a) 0.8
(b) 0.5
(c) 0.2
(d) 0.4
Show Answer & Explanation

Answer: (a) 0.8

Explanation:
1. \( p_A=X_Ap_A^0 \), so \( X_A=32/40=0.8 \).

Teacher's Note:
1. Mole fraction cannot exceed 1.
2. X_B = 0.2.

Question 9: A solution of benzene (\( p^0=120 \) Torr) and toluene (\( p^0=80 \) Torr) contains 2.0 mol of each. The mole fraction of toluene in the vapour in equilibrium with it is:
(a) 0.50
(b) 0.25
(c) 0.60
(d) 0.40
Show Answer & Explanation

Answer: (d) 0.40

Explanation:
1. \( p_B=0.5\times120=60 \); \( p_T=0.5\times80=40 \).
2. \( y_T=\dfrac{40}{60+40}=0.40 \).

Teacher's Note:
1. Vapour is richer in the more volatile benzene (0.60).
2. Dalton's law gives the vapour composition.

Question 10: An azeotropic mixture of two liquids boils at a temperature lower than either of them when it:
(a) Shows negative deviation from Raoult's law
(b) Shows positive deviation from Raoult's law
(c) Shows ideal behaviour
(d) Is saturated
Show Answer & Explanation

Answer: (b) Shows positive deviation from Raoult's law

Explanation:
1. Positive deviation gives a maximum in the vapour pressure curve.
2. Maximum vapour pressure means minimum boiling point, so a minimum-boiling azeotrope.

Teacher's Note:
1. Negative deviation gives maximum-boiling azeotropes such as HNO3 + water.
2. An ideal solution has no azeotrope.

Question 11: The depression of freezing point of a 0.01 molal aqueous \( \ce{CH3COOH} \) solution is 0.02046 K (\( K_f=1.86 \)). Taking molality equal to molarity, the pH of the solution is:
(a) 2
(b) 3.2
(c) 3
(d) 4.2
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. \( i=\dfrac{0.02046}{1.86\times0.01}=1.10 \), so \( \alpha=0.10 \).
2. \( [H^+]=C\alpha=0.01\times0.10=10^{-3} \) M.
3. pH = 3.

Teacher's Note:
1. For a weak monobasic acid i = 1 + alpha.
2. Ka = C alpha^2 = 1 x 10^-4.

Question 12: Which of the following azeotropic solutions has a boiling point higher than the boiling points of its constituents?
(a) \( \ce{CH3CH2OH} \) and \( \ce{CH3COCH3} \)
(b) \( \ce{CHCl3} \) and \( \ce{CH3COCH3} \)
(c) \( \ce{CS2} \) and \( \ce{CH3COCH3} \)
(d) \( \ce{CH3CHO} \) and \( \ce{CS2} \)
Show Answer & Explanation

Answer: (b) \( \ce{CHCl3} \) and \( \ce{CH3COCH3} \)

Explanation:
1. A maximum-boiling azeotrope arises from negative deviation, with stronger A-B attraction.
2. Chloroform and acetone form a hydrogen bond (CCl3-H ... O=C), giving negative deviation.

Teacher's Note:
1. The other pairs show positive deviation (minimum-boiling azeotropes).
2. The azeotrope of acetone + chloroform boils at 64.7 C.

Question 13: The depressions of freezing point of 0.05 molal aqueous solutions of (1) NaCl, (2) \( \ce{K2SO4} \), (3) glucose, (4) \( \ce{Al2(SO4)3} \) are measured. The maximum depression is shown by:
(a) NaCl
(b) \( \ce{K2SO4} \)
(c) Glucose
(d) \( \ce{Al2(SO4)3} \)
Show Answer & Explanation

Answer: (d) \( \ce{Al2(SO4)3} \)

Explanation:
1. van't Hoff factors for complete dissociation: NaCl 2, K2SO4 3, glucose 1, Al2(SO4)3 5.
2. Depression is proportional to i, so Al2(SO4)3 shows the largest.

Teacher's Note:
1. Same molality, so only i matters.
2. The ranking of i is the ranking of depression.

Question 14: The molecular weight of NaCl determined experimentally from elevation of boiling point or depression of freezing point is:
(a) Less than 58.5
(b) Greater than 58.5
(c) Equal to 58.5
(d) Cannot be predicted
Show Answer & Explanation

Answer: (a) Less than 58.5

Explanation:
1. NaCl dissociates, so i is about 2 and the observed colligative effect is larger than for undissociated molecules.
2. Normal molar mass / observed molar mass = i, so observed < 58.5 (about 29.3).

Teacher's Note:
1. Association (dimerisation) gives a higher observed molar mass.
2. Observed molar mass = calculated / i.

Question 15: On the basis of intermolecular forces, the correct order of decreasing boiling points of \( \ce{CH3OH} \), \( \ce{CH4} \) and \( \ce{H2} \) is:
(a) \( \ce{CH3OH} > \ce{H2} > \ce{CH4} \)
(b) \( \ce{CH4} > \ce{CH3OH} > \ce{H2} \)
(c) \( \ce{H2} > \ce{CH4} > \ce{CH3OH} \)
(d) \( \ce{CH3OH} > \ce{CH4} > \ce{H2} \)
Show Answer & Explanation

Answer: (d) \( \ce{CH3OH} > \ce{CH4} > \ce{H2} \)

Explanation:
1. Methanol has hydrogen bonding (b.p. 338 K).
2. Methane has stronger London forces than H2 because of more electrons (b.p. 112 K versus 20 K).

Teacher's Note:
1. Order of strength: hydrogen bonds > dipole forces > London forces.
2. London forces increase with polarisability.

Question 16: During depression of freezing point of a solution, the phases in equilibrium are:
(a) Liquid solvent and solid solute
(b) Liquid solvent and solid solvent
(c) Liquid solute and solid solute
(d) Liquid solute and solid solvent
Show Answer & Explanation

Answer: (b) Liquid solvent and solid solvent

Explanation:
1. The freezing point of the solution is the temperature at which pure solid solvent is in equilibrium with the solution.
2. The vapour pressure of the solid solvent equals that of the solvent in solution.

Teacher's Note:
1. Solute does not separate out; it stays in solution.
2. Eutectic point is where both solids separate together.

Question 17: Which of the following liquid pairs shows a positive deviation from Raoult's law?
(a) Acetone and chloroform
(b) Water and nitric acid
(c) Benzene and methanol
(d) Water and hydrochloric acid
Show Answer & Explanation

Answer: (c) Benzene and methanol

Explanation:
1. Methanol is hydrogen bonded in the pure liquid; adding benzene breaks these bonds and weakens attraction.
2. Hence the vapour pressure is higher than ideal.

Teacher's Note:
1. Acetone + chloroform, water + HNO3 and water + HCl all show negative deviation.
2. Other positive deviation pairs: ethanol + water, CCl4 + CHCl3.

Question 18: Consider 1 M solutions of the salts \( \ce{Na2SO4} \), \( \ce{BaCl2} \), NaCl and \( \ce{Al(NO3)3} \). Which has the lowest freezing point?
(a) \( \ce{Na2SO4} \)
(b) \( \ce{Al(NO3)3} \)
(c) \( \ce{BaCl2} \)
(d) NaCl
Show Answer & Explanation

Answer: (b) \( \ce{Al(NO3)3} \)

Explanation:
1. Number of ions: Na2SO4 3, BaCl2 3, NaCl 2, Al(NO3)3 4.
2. Maximum particles, maximum depression: Al(NO3)3.

Teacher's Note:
1. Same molarity, so compare i.
2. For more precise work activity effects reduce i, but not enough to reverse the order here.

Question 19: When 0.6 g of urea is dissolved in 100 g of water, the water will boil at (\( K_b \) of water = 0.52 K kg mol\(^{-1}\), normal boiling point 100 C):
(a) 373.052 K
(b) 273.52 K
(c) 372.48 K
(d) 273.052 K
Show Answer & Explanation

Answer: (a) 373.052 K

Explanation:
1. Moles of urea = 0.6/60 = 0.01; molality = 0.01/0.1 = 0.1.
2. \( \Delta T_b=0.52\times0.1=0.052 \) K.
3. Boiling point = 373 + 0.052 = 373.052 K.

Teacher's Note:
1. The boiling point of water is raised, never lowered.
2. 100 C = 373 K.

Question 20: A 2% (w/v) solution of glucose has the same elevation in boiling point as a 5% solution of a non-volatile solute. The molar mass of the solute is:
(a) 180 g/mol
(b) 72 g/mol
(c) 18 g/mol
(d) 450 g/mol
Show Answer & Explanation

Answer: (d) 450 g/mol

Explanation:
1. Equal elevations mean equal molalities (mole per unit volume): 2/180 = 5/M.
2. \( M=\dfrac{5\times180}{2}=450 \).

Teacher's Note:
1. The same method applies to equal osmotic pressures or equal depressions.
2. Compare mass percent divided by molar mass.

Question 21: A 1.0 molal solution with the lowest freezing point is that of:
(a) HCl
(b) KCl
(c) \( \ce{FeCl3} \)
(d) \( \ce{MgCl2} \)
Show Answer & Explanation

Answer: (c) \( \ce{FeCl3} \)

Explanation:
1. \( i \): FeCl3 4, MgCl2 3, HCl 2, KCl 2.
2. The highest i gives the lowest freezing point.

Teacher's Note:
1. For equal molality only i differs.
2. FeCl3 gives Fe3+ and 3 Cl-.

Question 22: Which of the following has the lowest freezing point and the highest boiling point?
(a) 1.0 m sodium chloride
(b) 1.5 m aluminium nitrate
(c) 1.5 m calcium chloride
(d) 1.5 m magnesium phosphate \( \ce{Mg3(PO4)2} \)
Show Answer & Explanation

Answer: (d) 1.5 m magnesium phosphate \( \ce{Mg3(PO4)2} \)

Explanation:
1. Effective molality: Mg3(PO4)2 1.5 x 5 = 7.5; Al(NO3)3 1.5 x 4 = 6.0; CaCl2 1.5 x 3 = 4.5; NaCl 1.0 x 2 = 2.0.
2. Largest value gives lowest freezing point and highest boiling point.

Teacher's Note:
1. Both colligative effects follow the same ranking.
2. Ignoring ion-pair formation here.

Question 23: 20 g of compound X is dissolved in 500 mL of water. The osmotic pressure of the solution is 500 mm Hg at 10 C. The molar mass of X is:
(a) 1115.42 g/mol
(b) 1150.70 g/mol
(c) 1412.84 g/mol
(d) 1163.88 g/mol
Show Answer & Explanation

Answer: (c) 1412.84 g/mol

Explanation:
1. \( \pi=500/760=0.6579 \) atm; T = 283 K.
2. \( M=\dfrac{wRT}{\pi V}=\dfrac{20\times0.0821\times283}{0.6579\times0.5}=1412.8 \) g/mol.

Teacher's Note:
1. Convert mm Hg to atm and degrees C to kelvin.
2. Use V in litres.

Question 24: The aqueous solution having an osmotic pressure nearest to that of an equimolar solution of \( \ce{K4[Fe(CN)6]} \) is:
(a) \( \ce{K2SO4} \)
(b) \( \ce{Al2(SO4)3} \)
(c) \( \ce{Na3PO4} \)
(d) \( \ce{C6H12O6} \)
Show Answer & Explanation

Answer: (b) \( \ce{Al2(SO4)3} \)

Explanation:
1. K4[Fe(CN)6] gives 4 K+ and 1 [Fe(CN)6]4-: 5 ions.
2. Al2(SO4)3 gives 2 Al3+ and 3 SO4(2-): 5 ions.
3. Equal i means nearly equal osmotic pressure.

Teacher's Note:
1. Na3PO4 gives 4 ions, K2SO4 3, glucose 1.
2. Osmotic pressure depends on the number of particles.

Question 25: The elevation in boiling point of a solution of 13.44 g of \( \ce{CuCl2} \) (M = 134.4) in 1 kg of water is (\( K_b=0.52 \) K kg mol\(^{-1}\)):
(a) 0.16 K
(b) 0.05 K
(c) 0.10 K
(d) 0.21 K
Show Answer & Explanation

Answer: (a) 0.16 K

Explanation:
1. Moles = 13.44/134.4 = 0.1; m = 0.1 mol/kg.
2. CuCl2 gives 3 ions, i = 3.
3. \( \Delta T_b=3\times0.52\times0.1=0.156\approx0.16 \) K.

Teacher's Note:
1. Without dissociation the answer would be 0.052 K.
2. Assume complete dissociation.

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