JEE Chemistry Solutions MCQs Set 18

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Multiple Choice Questions

Question 1: The degree of dissociation (α) of a weak electrolyte AxBy is related to the van't Hoff factor (i) by the expression:
(a) \( \alpha = \dfrac{i-1}{(x+y-1)} \)
(b) \( \alpha = \dfrac{i-1}{x+y+1} \)
(c) \( \alpha = \dfrac{x+y-1}{i-1} \)
(d) \( \alpha = \dfrac{x+y+1}{i-1} \)
Show Answer & Explanation

Answer: (a) \( \alpha = \dfrac{i-1}{(x+y-1)} \)

Explanation:
1. \( A_xB_y\rightleftharpoons xA+yB \) gives \( 1-\alpha+(x+y)\alpha = 1+(x+y-1)\alpha \) moles.
2. So \( i = 1+(x+y-1)\alpha \) and \( \alpha = \dfrac{i-1}{x+y-1} \).

Teacher's Note:
1. With x + y ions from one formula unit, each dissociating formula unit adds (x + y − 1) extra particles.

Question 2: One mole of a non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is:
(a) \( \dfrac13 \)
(b) \( \dfrac32 \)
(c) \( \dfrac23 \)
(d) \( \dfrac12 \)
Show Answer & Explanation

Answer: (c) \( \dfrac23 \)

Explanation:
1. \( \dfrac{P_s}{P^0} = x_{water} = \dfrac{2}{2+1} = \dfrac23 \).

Teacher's Note:
1. The ratio Ps/P0 is the mole fraction of the solvent.

Question 3: The correct relationship between the boiling points of very dilute solutions of AlCl3 (T1) and CaCl2 (T2), having the same molar concentration, is:
(a) \( T_1 = T_2 \)
(b) \( T_1 > T_2 \)
(c) \( T_2 = T_1 \)
(d) \( T_2 \geq T_1 \)
Show Answer & Explanation

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

Explanation:
1. AlCl3 gives 4 ions (i = 4) and CaCl2 gives 3 ions (i = 3).
2. The larger i gives a larger elevation in boiling point, so T1 > T2.

Teacher's Note:
1. In very dilute solutions, dissociation is nearly complete.

Question 4: The depression in freezing point of water containing ethylene glycol is 7.4 °C. Calculate the mass of ethylene glycol (molar mass 62) added to 500 g of water (Kf = 1.86 K kg mol⁻¹):
(a) 23.3 g
(b) 123.3 g
(c) 12.33 g
(d) 233.3 g
Show Answer & Explanation

Answer: (b) 123.3 g

Explanation:
1. \( m = \dfrac{7.4}{1.86} = 3.98 \) mol/kg, so moles in 0.5 kg \( = 1.99 \).
2. Mass \( = 1.99\times62 = 123.3 \) g.

Teacher's Note:
1. The depression gives the molality, then convert to moles and mass.

Question 5: The van't Hoff factor for a 0.1 M Ba(NO3)2 solution is 2.74. The degree of dissociation is:
(a) 100%
(b) 91.3%
(c) 87%
(d) 74%
Show Answer & Explanation

Answer: (c) 87%

Explanation:
1. Ba(NO3)2 gives 3 ions: \( \alpha = \dfrac{i-1}{n-1} = \dfrac{2.74-1}{2} = 0.87 \).

Teacher's Note:
1. The degree of dissociation is 87%.

Question 6: The vapour pressure of a dilute aqueous solution of glucose is 750 mm of mercury at 373 K. The mole fraction of the solute is:
(a) \( \dfrac1{7.6} \)
(b) \( \dfrac1{10} \)
(c) \( \dfrac1{76} \)
(d) \( \dfrac1{35} \)
Show Answer & Explanation

Answer: (c) \( \dfrac1{76} \)

Explanation:
1. At 373 K pure water has a vapour pressure of 760 mm Hg.
2. \( x_{solute} = \dfrac{760-750}{760} = \dfrac{1}{76} \).

Teacher's Note:
1. Water boils at 373 K, so its vapour pressure there is 760 mm Hg.

Question 7: Which of the following 0.1 M aqueous solutions will have the lowest freezing point?
(a) sodium chloride
(b) potassium sulphate
(c) glucose
(d) urea
Show Answer & Explanation

Answer: (b) potassium sulphate

Explanation:
1. K2SO4 gives 3 ions, NaCl gives 2 ions, and urea and glucose give 1 particle each.
2. The largest number of particles gives the largest depression and the lowest freezing point: K2SO4.

Teacher's Note:
1. Lowest freezing point means the largest van't Hoff factor at the same molarity.

Question 8: The molality of a solution containing 3 g of CH3COOH in 50 g of benzene is:
(a) 1 m
(b) 0.2 m
(c) 0.1 m
(d) 0.5 m
Show Answer & Explanation

Answer: (a) 1 m

Explanation:
1. Moles of acetic acid \( = 3/60 = 0.05 \).
2. \( m = \dfrac{0.05}{0.05\ \text{kg}} = 1 \) m.

Teacher's Note:
1. Molality uses the mass of the solvent in kilograms.

Question 9: Arrange the following aqueous solutions in the order of their increasing boiling points: (i) 10⁻⁴ M NaCl, (ii) 10⁻⁴ M urea, (iii) 10⁻³ M MgCl2, (iv) 10⁻² M NaCl.
(a) (i) < (ii) < (iv) < (iii)
(b) (iv) < (iii) = (i) = (ii)
(c) (ii) < (i) < (iii) < (iv)
(d) (ii) < (i) = (iii) < (iv)
Show Answer & Explanation

Answer: (c) (ii) < (i) < (iii) < (iv)

Explanation:
1. Particle concentrations: (i) \( 2\times10^{-4} \); (ii) \( 10^{-4} \); (iii) \( 3\times10^{-3} \); (iv) \( 2\times10^{-2} \).
2. The boiling point rises with the particle concentration: (ii) < (i) < (iii) < (iv).

Teacher's Note:
1. MgCl2 in (iii) is 10⁻³ M, ten times more concentrated than (i), so compare i × M carefully.

Question 10: The vapour pressure of water at room temperature is 23.8 mm of Hg. The vapour pressure of an aqueous solution of sucrose with mole fraction 0.1 is equal to:
(a) 23.9 mm Hg
(b) 31.44 mm Hg
(c) 21.42 mm Hg
(d) 24.2 mm Hg
Show Answer & Explanation

Answer: (c) 21.42 mm Hg

Explanation:
1. \( P_s = P^0x_{water} = 23.8\times0.9 = 21.42 \) mm Hg.

Teacher's Note:
1. Do not use the mole fraction of the solute directly. The solvent has a mole fraction of 0.9.

Question 11: Which aqueous solution has the minimum freezing point?
(a) 0.01 M NaCl
(b) 0.005 M MgSO4
(c) 0.005 M MgI2
(d) 0.005 M C2H5OH
Show Answer & Explanation

Answer: (a) 0.01 M NaCl

Explanation:
1. Particle concentrations: NaCl 0.01 × 2 = 0.02; C2H5OH 0.005 × 1 = 0.005; MgI2 0.005 × 3 = 0.015; MgSO4 0.005 × 2 = 0.010.
2. 0.01 M NaCl has the highest particle concentration, so the lowest freezing point.

Teacher's Note:
1. Compare i × M. Do not choose by the number of ions alone.

Question 12: 45 g of a non-volatile solute A is dissolved in 600 g of water. If the depression in freezing point is 2.2 K, what will be the molar mass of the solute (Kf = 1.86 K kg mol⁻¹)?
(a) 158.5
(b) 31.7
(c) 126.8
(d) 63.4
Show Answer & Explanation

Answer: (d) 63.4

Explanation:
1. \( M = \dfrac{1.86\times45\times1000}{2.2\times600} = 63.4 \) g/mol.

Teacher's Note:
1. Use the molar mass formula with the solvent mass in grams.

Question 13: The relationship between the values of osmotic pressure of solutions obtained by dissolving 6.00 g L⁻¹ of CH3COOH (π1) and 7.45 g L⁻¹ of KCl (π2) is:
(a) \( \pi_1 = \pi_2 \)
(b) \( \pi_1 < \pi_2 \)
(c) \( \pi_1 > \pi_2 \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \pi_1 < \pi_2 \)

Explanation:
1. Both solutions are 0.1 M: \( 6.00/60 = 0.1 \) and \( 7.45/74.5 = 0.1 \).
2. KCl is a strong electrolyte with i about 2, while acetic acid is a weak electrolyte with i just above 1. So \( \pi_1 < \pi_2 \).

Teacher's Note:
1. At equal molarity, the strong electrolyte gives more particles.

Question 14: The vapour pressure of pure A is 10 torr and at the same temperature, when 1 g of B is dissolved in 20 g of A, its vapour pressure is reduced to 9.0 torr. If the molecular mass of A is 200 amu, then the molecular mass of B is:
(a) 100 amu
(b) 90 amu
(c) 120 amu
(d) 75 amu
Show Answer & Explanation

Answer: (b) 90 amu

Explanation:
1. Relative lowering \( = \dfrac{10-9}{10} = 0.1 = x_B \).
2. \( n_A = 20/200 = 0.1 \). \( \dfrac{n_B}{n_B+0.1} = 0.1 \Rightarrow n_B = \dfrac{0.01}{0.9} = \dfrac{1}{90} \).
3. \( M_B = \dfrac{1}{1/90} = 90 \).

Teacher's Note:
1. Find the mole fraction of the solute from the relative lowering, then back-calculate the moles of B.

Question 15: If α is the degree of dissociation of K4[Fe(CN)6], then the abnormal mass of the complex in the solution will be:
(a) \( M_{normal}(1+2\alpha)^{-1} \)
(b) \( M_{normal}(1+3\alpha)^{-1} \)
(c) \( M_{normal}(1+4\alpha)^{-1} \)
(d) \( M_{normal}(1+\alpha)^{-1} \)
Show Answer & Explanation

Answer: (c) \( M_{normal}(1+4\alpha)^{-1} \)

Explanation:
1. \( K_4[Fe(CN)_6]\to4K^++[Fe(CN)_6]^{4-} \) gives 5 ions, so \( i = 1+(5-1)\alpha = 1+4\alpha \).
2. \( M_{abnormal} = \dfrac{M_{normal}}{i} = M_{normal}(1+4\alpha)^{-1} \).

Teacher's Note:
1. Observed (abnormal) molar mass is the normal value divided by i.

Question 16: 1 g of a protein polymer of molar mass \( 16\times10^4 \) g/mol is present in 800 mL of solution at 300 K. The osmotic pressure (in kilopascal) is [R = 8.314 J mol⁻¹ K⁻¹]:
(a) 0.2
(b) 0.04
(c) 0.02
(d) 0.4
Show Answer & Explanation

Answer: (c) 0.02

Explanation:
1. \( n = \dfrac{1}{1.6\times10^5} = 6.25\times10^{-6} \) mol, \( V = 8\times10^{-4} \) m³.
2. \( \pi = \dfrac{nRT}{V} = \dfrac{6.25\times10^{-6}\times8.314\times300}{8\times10^{-4}} \approx 19.5 \) Pa = 0.02 kPa.

Teacher's Note:
1. The very small osmotic pressure of a large molecule is still measurable with sensitive instruments.

Question 17: Among 0.1 M solutions of urea, Na3PO4 and Al2(SO4)3: (a) the vapour pressure and freezing point are the lowest for urea; (b) the vapour pressure and freezing point are the highest for urea; (c) the elevation in boiling point is the highest for Al2(SO4)3; (d) the depression in freezing point is the highest for Al2(SO4)3.
(a) (b), (c) and (d)
(b) only (a)
(c) (a), (b), (c) and (d)
(d) (b) and (c)
Show Answer & Explanation

Answer: (a) (b), (c) and (d)

Explanation:
1. Urea does not ionise (i = 1), Na3PO4 gives 4 ions and Al2(SO4)3 gives 5 ions.
2. Urea has the smallest effect, so it has the highest vapour pressure and freezing point: (b) is true and (a) is false.
3. Al2(SO4)3 gives the greatest boiling point elevation and freezing point depression: (c) and (d) are true.

Teacher's Note:
1. The more ions, the greater the colligative effect.

Question 18: The boiling points of C6H6, CH3OH, C6H5NH2 and C6H5NO2 are 80 °C, 65 °C, 184 °C and 212 °C respectively. Which will show the highest vapour pressure at room temperature?
(a) C6H5NH2
(b) C6H5NO2
(c) C6H6
(d) CH3OH
Show Answer & Explanation

Answer: (d) CH3OH

Explanation:
1. The liquid with the lowest boiling point is the most volatile and has the highest vapour pressure at a given temperature.
2. Methanol, with a boiling point of 65 °C, has the lowest boiling point in this list.

Teacher's Note:
1. Lower boiling point means higher vapour pressure at room temperature. Here, 65 °C is the lowest of the four.

Question 19: Which solution will have the least vapour pressure?
(a) 0.1 M BaCl2
(b) 0.1 M Na2SO4
(c) 0.1 M Na3PO4
(d) 0.1 M urea
Show Answer & Explanation

Answer: (c) 0.1 M Na3PO4

Explanation:
1. The lowest vapour pressure corresponds to the greatest number of particles in solution.
2. Na3PO4 gives 4 ions (i = 4), the most.

Teacher's Note:
1. The vapour pressure falls as the particle concentration rises.

Question 20: The normality of a 1% H2SO4 solution (w/V) is approximately:
(a) 0.2
(b) 0.02
(c) 0.4
(d) 0.1
Show Answer & Explanation

Answer: (a) 0.2

Explanation:
1. 1% w/V means 10 g per litre. Equivalent mass \( = 49 \).
2. \( N = \dfrac{10}{49} = 0.204 \approx 0.2 \).

Teacher's Note:
1. Normality = grams per litre / equivalent mass.

Question 21: When equimolal aqueous solutions of glucose, sodium chloride and barium nitrate are compared, the vapour pressures of the solutions will be in the following order:
(a) glucose > NaCl > Ba(NO3)2
(b) glucose = NaCl = Ba(NO3)2
(c) NaCl > Ba(NO3)2 > glucose
(d) Ba(NO3)2 > NaCl > glucose
Show Answer & Explanation

Answer: (a) glucose > NaCl > Ba(NO3)2

Explanation:
1. Van't Hoff factors: glucose 1, NaCl 2, Ba(NO3)2 3.
2. More particles lower the vapour pressure more, so glucose has the highest and Ba(NO3)2 the lowest.

Teacher's Note:
1. The vapour pressure falls as the particle concentration rises.

Question 22: If Raoult's law is obeyed, the vapour pressure of the solvent in a solution is directly proportional to:
(a) the mole fraction of the solvent and solute
(b) the mole fraction of the solute
(c) the mole fraction of the solvent
(d) the volume of the solution
Show Answer & Explanation

Answer: (c) the mole fraction of the solvent

Explanation:
1. \( P_{solvent} = x_{solvent}P^0 \).

Teacher's Note:
1. Raoult's law relates the partial pressure to the mole fraction of the same component.

Question 23: Which has the maximum freezing point?
(a) 1 molar CaCl2 solution
(b) 1 molar NaCl solution
(c) 1 molar urea solution
(d) 1 molar KCl solution
Show Answer & Explanation

Answer: (c) 1 molar urea solution

Explanation:
1. The highest freezing point means the smallest depression, i.e. the fewest particles.
2. Urea does not ionise (i = 1), whereas the salts give 2 or 3 ions.

Teacher's Note:
1. Non-electrolytes lower the freezing point the least at equal molarity.

Question 24: A 0.04 molal aqueous solution of a monoprotic acid has a depression in freezing point equal to 0.1 K. Find the van't Hoff factor for the acid (Kf = 1.86 K kg mol⁻¹):
(a) 2
(b) 2.5
(c) 1
(d) 1.34
Show Answer & Explanation

Answer: (d) 1.34

Explanation:
1. \( i = \dfrac{\Delta T_f}{K_fm} = \dfrac{0.1}{1.86\times0.04} = 1.34 \).

Teacher's Note:
1. A weak acid is only partly ionised, so i lies between 1 and 2.

Question 25: What is the freezing point of a solution containing 8.1 g of HBr in 100 g of water, assuming the acid to be 90% ionised? (Kf for water = 1.86 K molality⁻¹)
(a) -0.35 °C
(b) -3.53 °C
(c) 0 °C
(d) 0.85 °C
Show Answer & Explanation

Answer: (b) -3.53 °C

Explanation:
1. Moles of HBr \( = 8.1/81 = 0.1 \), so \( m = 1 \). \( i = 1+0.9 = 1.9 \).
2. \( \Delta T_f = 1.9\times1.86\times1 = 3.53 \), so the freezing point is \( -3.53 \) °C.

Teacher's Note:
1. HBr gives 2 ions, so i = 1 + α.

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