JEE Chemistry Solutions MCQs Set 30

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Chapter-wise Objective Questions: Solutions

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Multiple Choice Questions

Question 1: If 0.1 M glucose and 0.1 M urea solutions are placed on two sides of a semipermeable membrane to equal heights, then:
(a) There will be no net movement across the membrane
(b) Glucose will flow towards the urea solution
(c) Urea will flow towards the glucose solution
(d) Water will flow from the urea solution towards the glucose solution
Show Answer & Explanation

Answer: (a) There will be no net movement across the membrane

Explanation:
1. Equal molar concentrations of non-electrolytes are isotonic.
2. So there is no net solvent movement.
3. Answer: (a)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 2: If the mole fraction of the solvent in a solution decreases then:
(a) Vapour pressure increases
(b) Boiling point decreases
(c) Osmotic pressure increases
(d) All are correct
Show Answer & Explanation

Answer: (c) Osmotic pressure increases

Explanation:
1. A lower solvent mole fraction means more solute: vapour pressure falls, boiling point rises and osmotic pressure rises.
2. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 3: Which of the following solutions at the same temperature will be isotonic?
(a) 3.42 g cane sugar in one litre of water and 0.18 g glucose in one litre of water
(b) 3.42 g cane sugar in one litre of water and 0.18 g glucose in 0.1 litre of water
(c) 3.42 g cane sugar in one litre of water and 0.585 g NaCl in one litre of water
(d) 3.42 g cane sugar in one litre of water and 1.17 g NaCl in one litre of water
Show Answer & Explanation

Answer: (b) 3.42 g cane sugar in one litre of water and 0.18 g glucose in 0.1 litre of water

Explanation:
1. Sugar: 3.42/342 = 0.01 M.
2. (b) glucose: (0.18/180)/0.1 L = 0.01 M, equal.
3. (a) 0.001 M; (c) 0.01 M NaCl with i = 2 gives 0.02; (d) 0.02 M NaCl gives 0.04.
4. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 4: Equal volumes of 0.1 M urea and 0.1 M glucose are mixed. The mixture will have:
(a) Lower osmotic pressure
(b) The same osmotic pressure
(c) Higher osmotic pressure
(d) None of these
Show Answer & Explanation

Answer: (b) The same osmotic pressure

Explanation:
1. Mixing equal volumes of two solutions of the same molarity leaves 0.1 M total non-electrolyte solute.
2. The osmotic pressure is unchanged.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 5: A solution containing 4 g of a non-volatile organic solute per 100 mL was found to have an osmotic pressure equal to 500 cm of mercury at 27°C. The molecular weight of the solute is:
(a) 14.97
(b) 149.7
(c) 1697
(d) 1.497
Show Answer & Explanation

Answer: (b) 149.7

Explanation:
1. π = 500 cm Hg = 5000/760 = 6.58 atm.
2. C = π/RT = 6.58/(0.0821 × 300) = 0.267 mol L−1.
3. M = 40 g L−1/0.267 = 149.7.
4. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 6: If a 6.84% (w/V) solution of cane sugar (mol. wt. 342) is isotonic with a 1.52% (w/V) solution of thiocarbamide, the molecular weight of thiocarbamide is:
(a) 152
(b) 76
(c) 60
(d) 180
Show Answer & Explanation

Answer: (b) 76

Explanation:
1. Sugar: 68.4/342 = 0.2 M.
2. Thiocarbamide: 15.2/M = 0.2, so M = 76.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 7: The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose that will be isotonic with blood is (w/V):
(a) 5.41%
(b) 54.1%
(c) 3.5%
(d) 4.53%
Show Answer & Explanation

Answer: (a) 5.41%

Explanation:
1. C = π/RT = 7.65/(0.0821 × 310) = 0.3006 M.
2. Mass = 0.3006 × 180 = 54.1 g L−1 = 5.41% w/V.
3. Answer: (a)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 8: Equimolal solutions of A and B show depression in freezing point in the ratio 2 : 1. A remains in the normal state in solution. B will be in ____ state in solution.
(a) Normal
(b) Associated
(c) Hydrolysed
(d) Dissociated
Show Answer & Explanation

Answer: (b) Associated

Explanation:
1. A normal means iA = 1, so iB = 0.5.
2. i < 1 means B associates in solution.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 9: The van't Hoff factor (i) for a dilute solution of K3[Fe(CN)6] is:
(a) 10
(b) 4
(c) 5
(d) 0.25
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. K3[Fe(CN)6] gives 3K+ + [Fe(CN)6]3−, i.e. 4 ions, so i = 4.
2. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 10: The experimental molecular weight of an electrolyte will always be less than its calculated value because the van't Hoff factor i is:
(a) Less than 1
(b) Greater than 1
(c) One
(d) Zero
Show Answer & Explanation

Answer: (b) Greater than 1

Explanation:
1. Dissociation increases the particle count, so i > 1.
2. Observed molar mass = calculated molar mass/i, which is smaller.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 11: The substance A, when dissolved in solvent B, shows a molecular mass corresponding to A3. The van't Hoff factor will be:
(a) 1
(b) 2
(c) 3
(d) 1/3
Show Answer & Explanation

Answer: (d) 1/3

Explanation:
1. Three molecules associate into one particle.
2. i = observed particles/expected = 1/3.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 12: The molal elevation constant of water is 0.51. The boiling point of a 0.1 molal aqueous NaCl solution is nearly:
(a) 100.05°C
(b) 100.1°C
(c) 100.2°C
(d) 101.0°C
Show Answer & Explanation

Answer: (b) 100.1°C

Explanation:
1. ΔTb = i Kb m = 2 × 0.51 × 0.1 = 0.102.
2. Boiling point ≈ 100.1°C.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 13: Solute A is a ternary electrolyte and solute B is a non-electrolyte. If a 0.1 M solution of solute B produces an osmotic pressure of 2P, then a 0.05 M solution of A at the same temperature will produce an osmotic pressure equal to:
(a) P
(b) 1.5 P
(c) 2 P
(d) 3 P
Show Answer & Explanation

Answer: (d) 3 P

Explanation:
1. B: 0.1 RT = 2P, so RT = 20P.
2. A: i = 3, π = 3 × 0.05 × RT = 0.15 × 20P = 3P.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 14: Phenol associates in benzene as C6H5OH ⇌ ½(C6H5OH)2. If the degree of association of phenol is 40%, the van't Hoff factor i is:
(a) 1
(b) 0.8
(c) 1.4
(d) 0.6
Show Answer & Explanation

Answer: (b) 0.8

Explanation:
1. i = 1 − α(1 − 1/2) = 1 − α/2 = 1 − 0.2 = 0.8.
2. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 15: A 5.8% (w/vol) NaCl solution will exert an osmotic pressure closest to which one of the following?
(a) 5.8% (w/vol) sucrose solution
(b) 5.8% (w/vol) glucose solution
(c) 2 M sucrose solution
(d) 1 M glucose solution
Show Answer & Explanation

Answer: (c) 2 M sucrose solution

Explanation:
1. 5.8% NaCl = 58 g L−1 = 0.99 M and i ≈ 2, so effective concentration ≈ 2 M.
2. Closest is 2 M sucrose (the sucrose and glucose % solutions are only about 0.17 and 0.32 M).
3. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 16: Which one of the following solutions will have the highest osmotic pressure? (Assume all salts are completely dissociated.)
(a) 0.1 M Al2(SO4)3
(b) 0.1 M BaCl2
(c) 0.1 M Na2SO4
(d) The solution obtained by mixing equal volumes of (b) and (c)
Show Answer & Explanation

Answer: (a) 0.1 M Al2(SO4)3

Explanation:
1. Effective molarity = i × C: Al2(SO4)3 5 × 0.1 = 0.5; BaCl2 0.3; Na2SO4 0.3.
2. The mixture of (b) and (c) precipitates BaSO4 and has an even lower effective concentration.
3. Al2(SO4)3 is highest.
4. Answer: (a)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 17: Which one of the following pairs of solutions is expected to be isotonic at the same temperature?
(a) 0.1 M urea and 0.1 M NaCl
(b) 0.1 M urea and 0.2 M MgCl2
(c) 0.1 M NaCl and 0.1 M Na2SO4
(d) 0.1 M Ca(NO3)2 and 0.1 M Na2SO4
Show Answer & Explanation

Answer: (d) 0.1 M Ca(NO3)2 and 0.1 M Na2SO4

Explanation:
1. Compare i × C: (a) 0.1 and 0.2, (b) 0.1 and 0.6, (c) 0.2 and 0.3, (d) 0.3 and 0.3.
2. Only (d) matches.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 18: Two solutions of KNO3 and CH3COOH are prepared separately. The molarity of both is 0.1 M and their osmotic pressures are P1 and P2 respectively. The correct relationship is:
(a) P2 > P1
(b) P1 = P2
(c) P1 > P2
(d) P1/(P1 + P2) = P2/(P1 + P2)
Show Answer & Explanation

Answer: (c) P1 > P2

Explanation:
1. KNO3 is a strong electrolyte with i ≈ 2.
2. CH3COOH is a weak acid with i only slightly above 1.
3. So P1 > P2.
4. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 19: The correct relationship between the boiling points of very dilute solutions of AlCl3 (T1) and CaCl2 (T2), having the same molar concentration, is:
(a) T1 = T2
(b) T1 > T2
(c) T2 = T1
(d) T2 ≥ T1
Show Answer & Explanation

Answer: (b) T1 > T2

Explanation:
1. AlCl3 gives 4 ions (i = 4) and CaCl2 gives 3 ions (i = 3).
2. Greater i gives greater elevation, so T1 > T2.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 20: If α is the degree of dissociation of K4[Fe(CN)6], then the abnormal molar mass of the complex in the solution is:
(a) Mnormal/(1 + 2α)
(b) Mnormal/(1 + 3α)
(c) Mnormal/(1 + α)
(d) Mnormal/(1 + 4α)
Show Answer & Explanation

Answer: (d) Mnormal/(1 + 4α)

Explanation:
1. K4[Fe(CN)6] → 4K+ + [Fe(CN)6]4−, n = 5 ions.
2. i = 1 + (n − 1)α = 1 + 4α.
3. Abnormal mass = Mnormal/i.
4. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 21: Which solution will have the least vapour pressure?
(a) 0.1 M BaCl2
(b) 0.1 M urea
(c) 0.1 M Na2SO4
(d) 0.1 M Na3PO4
Show Answer & Explanation

Answer: (d) 0.1 M Na3PO4

Explanation:
1. The greatest number of particles gives the largest lowering.
2. Na3PO4 gives 4 ions (i = 4), the most among the four.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 22: The freezing point of a 1% aqueous solution of calcium nitrate will be:
(a) 0°C
(b) Above 0°C
(c) 1°C
(d) Below 0°C
Show Answer & Explanation

Answer: (d) Below 0°C

Explanation:
1. A dissolved solute depresses the freezing point below that of pure water (0°C).
2. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 23: Which of the following solutions will have the highest boiling point?
(a) 1% glucose in water
(b) 1% sucrose in water
(c) 1% NaCl in water
(d) 1% urea in water
Show Answer & Explanation

Answer: (c) 1% NaCl in water

Explanation:
1. For 1 g per 100 g, moles: glucose 0.0056, sucrose 0.0029, urea 0.0167, NaCl 0.0171 × i(2) = 0.034.
2. NaCl gives the largest number of particles, so the highest boiling point.
3. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 24: The freezing point of an equimolal aqueous solution will be highest for:
(a) C6H5NH3Cl
(b) Ca(NO3)2
(c) La(NO3)3
(d) C6H12O6 (glucose)
Show Answer & Explanation

Answer: (d) C6H12O6 (glucose)

Explanation:
1. Depression ∝ i: C6H5NH3Cl 2, Ca(NO3)2 3, La(NO3)3 4, glucose 1.
2. The smallest depression (glucose) means the highest freezing point.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 25: When mercuric iodide is added to the aqueous solution of potassium iodide:
(a) The boiling point does not change
(b) The freezing point is raised
(c) The freezing point is lowered
(d) The freezing point does not change
Show Answer & Explanation

Answer: (b) The freezing point is raised

Explanation:
1. HgI2 + 2KI → K2[HgI4].
2. Four ions (2K+ + 2I−) become three (2K+ + HgI42−), so the particle number falls and the freezing point rises.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 26: The molecular weight of benzoic acid in benzene as determined by the depression in freezing point method corresponds to:
(a) Ionisation of benzoic acid
(b) Dimerization of benzoic acid
(c) Trimerization of benzoic acid
(d) Solvation of benzoic acid
Show Answer & Explanation

Answer: (b) Dimerization of benzoic acid

Explanation:
1. In benzene benzoic acid forms hydrogen-bonded dimers, so the observed molar mass is about twice the normal value (i about 0.5).
2. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 27: A 0.1 M solution of K2SO4 is dissolved to the extent of 90%. Its osmotic pressure at 27°C is:
(a) 6.89 atm
(b) 0.689 atm
(c) 0.344 atm
(d) 3.4 atm
Show Answer & Explanation

Answer: (a) 6.89 atm

Explanation:
1. i = 1 + 2α = 1 + 2 × 0.9 = 2.8.
2. π = iCRT = 2.8 × 0.1 × 0.0821 × 300 = 6.9 atm.
3. Answer: (a)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 28: Which one has the same van't Hoff factor i as that of Hg2Cl2 (treated as dissociating into Hg22+ and 2Cl−)?
(a) NaCl
(b) Na2SO4
(c) Al(NO3)3
(d) Al2(SO4)3
Show Answer & Explanation

Answer: (b) Na2SO4

Explanation:
1. Hg2Cl2 gives 3 ions, so i = 3.
2. Na2SO4 also gives 3 ions (2Na+ + SO42−).
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

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