JEE Chemistry Solutions MCQs Set 16

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Multiple Choice Questions

Question 1: The ratio of the value of any colligative property for KCl solution to that for sugar solution (equimolal) is nearly:
(a) 1
(b) 2.5
(c) 0.5
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. KCl dissociates into two ions (i ≈ 2) and sugar does not (i = 1).
2. The ratio is \( \dfrac{2}{1} = 2 \).

Teacher's Note:
1. Colligative properties are proportional to i times the molality.

Question 2: Addition of 1 kg of a non-volatile solute to a volatile solvent increases the boiling point from 350 K to 355 K and decreases the freezing point from 250 K to 220 K. If Kb is 15.5 K kg mol⁻¹, then Kf will be:
(a) 10 K kg mol⁻¹
(b) 2.1 K kg mol⁻¹
(c) 93.0 K kg mol⁻¹
(d) 76 K kg mol⁻¹
Show Answer & Explanation

Answer: (c) 93.0 K kg mol⁻¹

Explanation:
1. \( \Delta T_b = 5 \) K and \( \Delta T_f = 30 \) K for the same molality.
2. \( \dfrac{K_f}{K_b} = \dfrac{\Delta T_f}{\Delta T_b} = 6 \Rightarrow K_f = 6\times15.5 = 93 \) K kg mol⁻¹.

Teacher's Note:
1. The same solution gives both changes, so the ratio of constants equals the ratio of the changes.

Question 3: An azeotropic mixture of two liquids boils at a lower temperature than either of them when:
(a) it shows a negative deviation from Raoult's law
(b) it does not deviate from Raoult's law
(c) it is saturated
(d) it shows a positive deviation from Raoult's law
Show Answer & Explanation

Answer: (d) it shows a positive deviation from Raoult's law

Explanation:
1. A positive deviation increases the vapour pressure above the ideal value.
2. A higher vapour pressure means the solution boils at a lower temperature, giving a minimum-boiling azeotrope.

Teacher's Note:
1. Positive deviation gives a minimum-boiling azeotrope.

Question 4: Two bottles A and B contain 1 M and 1 m aqueous solutions (density about 1 g/mL) of sulphuric acid respectively. Then:
(a) B is more concentrated than A
(b) it is not possible to compare the concentration
(c) A is more concentrated than B
(d) the concentration of A = concentration of B
Show Answer & Explanation

Answer: (c) A is more concentrated than B

Explanation:
1. A has 1 mol in 1 L of solution. B has 1 mol in 1000 g of water, so its total mass is 1098 g and the volume is about 1.1 L.
2. The same moles in a larger volume means B has a molarity below 1 M. A is more concentrated.

Teacher's Note:
1. Molality uses the mass of the solvent only, so the solution volume is larger than 1 L.

Question 5: The lowering of vapour pressure of 0.1 M aqueous solutions of NaCl, CuSO4 and K2SO4 are:
(a) in the ratio of 1.5 : 1 : 2.5
(b) in the ratio of 3 : 2 : 1
(c) in the ratio of 1 : 1 : 1.5
(d) all equal
Show Answer & Explanation

Answer: (c) in the ratio of 1 : 1 : 1.5

Explanation:
1. NaCl gives 2 ions, CuSO4 gives 2 ions and K2SO4 gives 3 ions.
2. The lowering of vapour pressure is proportional to i, so the ratio is 2 : 2 : 3 = 1 : 1 : 1.5.

Teacher's Note:
1. Equal molarity does not mean equal effect for electrolytes; count the ions.

Question 6: The boiling point of a 0.2 mol kg⁻¹ solution of X in water is greater than that of an equimolal solution of Y in water. Which one of the following statements is true in this case?
(a) The molecular mass of X is less than the molecular mass of Y
(b) The molecular mass of X is greater than the molecular mass of Y
(c) X is undergoing dissociation in water while Y undergoes no change
(d) Y is undergoing dissociation in water while X undergoes no change
Show Answer & Explanation

Answer: (c) X is undergoing dissociation in water while Y undergoes no change

Explanation:
1. Equal molality with a higher boiling point means more particles in the X solution.
2. This happens if X dissociates and Y does not. Molecular mass is irrelevant at equal molality.

Teacher's Note:
1. At equal molality, the elevation depends on the number of particles, not on the molar mass.

Question 7: The azeotropic mixture of water (b.p. 100.15 °C) and HCl (b.p. −85 °C) boils at 108.5 °C. When this mixture is distilled, it is possible to obtain:
(a) pure water
(b) neither HCl nor H2O in their pure states
(c) pure HCl
(d) pure water as well as HCl
Show Answer & Explanation

Answer: (b) neither HCl nor H2O in their pure states

Explanation:
1. At the azeotropic composition, the vapour has the same composition as the liquid.
2. So distillation gives the azeotrope itself and neither component in pure form.

Teacher's Note:
1. A maximum-boiling azeotrope (negative deviation) distils unchanged at 108.5 °C.

Question 8: The molarity of 98% H2SO4 (d = 1.8 g/mL) by weight is:
(a) 18 M
(b) 4 M
(c) 6 M
(d) 10 M
Show Answer & Explanation

Answer: (a) 18 M

Explanation:
1. \( M = \dfrac{\%\times d\times10}{\text{molar mass}} = \dfrac{98\times1.8\times10}{98} = 18 \) M.

Teacher's Note:
1. Use M = (% by mass × density × 10) / molar mass.

Question 9: The freezing point of a 1 molal NaCl solution assuming NaCl to be 100% dissociated in water is (Kf = 1.86 K molality⁻¹):
(a) +3.72 °C
(b) +1.86 °C
(c) -3.72 °C
(d) -1.86 °C
Show Answer & Explanation

Answer: (c) -3.72 °C

Explanation:
1. \( \Delta T_f = iK_fm = 2\times1.86\times1 = 3.72 \).
2. The freezing point is lowered, so it is \( -3.72 \) °C.

Teacher's Note:
1. Remember that a freezing point is lowered, so the answer is negative.

Question 10: Which one is NOT equal to zero for an ideal solution?
(a) ΔH(mix)
(b) ΔV(mix)
(c) ΔS(mix)
(d) ΔP = P(observed) − P(Raoult)
Show Answer & Explanation

Answer: (c) ΔS(mix)

Explanation:
1. For an ideal solution the volume change, the enthalpy change and the deviation from Raoult's law are all zero.
2. The entropy of mixing is positive.

Teacher's Note:
1. Entropy always increases when two liquids mix.

Question 11: Colligative properties depend on:
(a) the relative number of solute molecules in solution and the nature of the solvent
(b) the relative number of solute molecules, irrespective of the nature of solvent and solute
(c) the relative number of solute molecules in solvent and the nature of solute
(d) the relative number of solute molecules and the nature of solute and solvent
Show Answer & Explanation

Answer: (b) the relative number of solute molecules, irrespective of the nature of solvent and solute

Explanation:
1. Colligative properties depend only on the number of solute particles relative to the solvent.
2. They do not depend on the chemical nature of the solute (the constants Kb, Kf do depend on the solvent, but the property itself is defined by the number of particles).

Teacher's Note:
1. Colligative means 'depending on the collection': it is the count of particles that matters.

Question 12: Henry's law constant for dissolution of CH4 in benzene at 298 K is \( 2\times10^5 \) mm of Hg. Then the solubility (mole fraction) of CH4 in benzene at 298 K under 760 mm of Hg is:
(a) \( 1\times10^{-2} \)
(b) \( 4\times10^{-7} \)
(c) \( 3.8\times10^{-3} \)
(d) \( 1.2\times10^{-5} \)
Show Answer & Explanation

Answer: (c) \( 3.8\times10^{-3} \)

Explanation:
1. \( p = K_Hx \Rightarrow x = \dfrac{760}{2\times10^5} = 3.8\times10^{-3} \).

Teacher's Note:
1. Henry's law: partial pressure = K_H × mole fraction of the dissolved gas.

Question 13: The molal elevation constant of water is 0.51. The boiling point of a 0.1 molal aqueous NaCl solution is nearly:
(a) 100.1 °C
(b) 100.05 °C
(c) 100.2 °C
(d) 101.0 °C
Show Answer & Explanation

Answer: (a) 100.1 °C

Explanation:
1. \( \Delta T_b = iK_bm = 2\times0.51\times0.1 = 0.102 \).
2. Boiling point \( = 100.1 \) °C.

Teacher's Note:
1. NaCl is an electrolyte with i = 2.

Question 14: Which one of the following electrolytes has the same value of van't Hoff's factor i as that of Al2(SO4)3 (if all are 100% ionised)?
(a) Al(NO3)3
(b) K3[Fe(CN)6]
(c) K2SO4
(d) K4[Fe(CN)6]
Show Answer & Explanation

Answer: (d) K4[Fe(CN)6]

Explanation:
1. \( Al_2(SO_4)_3 \to2Al^{3+}+3SO_4^{2-} \) gives 5 ions.
2. \( K_4[Fe(CN)_6]\to4K^++[Fe(CN)_6]^{4-} \) also gives 5 ions. The others give 4, 4 and 3.

Teacher's Note:
1. Count the ions each formula unit gives.

Question 15: The vapour pressure of a solvent decreases by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in vapour pressure is to be 20 mm of Hg?
(a) 0.4
(b) 0.8
(c) 0.2
(d) 0.6
Show Answer & Explanation

Answer: (d) 0.6

Explanation:
1. \( \dfrac{\Delta P}{P^0} = x_{solute} \Rightarrow \dfrac{10}{P^0} = 0.2 \), so \( P^0 = 50 \) mm.
2. For \( \Delta P = 20 \): \( x_{solute} = 0.4 \), so \( x_{solvent} = 0.6 \).

Teacher's Note:
1. The decrease in vapour pressure is proportional to the mole fraction of the solute.

Question 16: Which of the following gases does not obey Henry's law?
(a) He
(b) H2
(c) NH3
(d) O2
Show Answer & Explanation

Answer: (c) NH3

Explanation:
1. Henry's law holds for gases that do not react with the solvent.
2. NH3 reacts with water to form NH4+ and OH−, so it does not obey the law.

Teacher's Note:
1. Gases that ionize or react with the solvent (NH3, HCl) deviate from Henry's law.

Question 17: The molal elevation constant of water is 0.52 K molality⁻¹. The boiling point of a 1.0 molal aqueous KCl solution (assuming complete dissociation of KCl) should be:
(a) 100.52 °C
(b) 99.48 °C
(c) 98.96 °C
(d) 101.04 °C
Show Answer & Explanation

Answer: (d) 101.04 °C

Explanation:
1. \( \Delta T_b = 2\times0.52\times1.0 = 1.04 \).
2. Boiling point \( = 101.04 \) °C.

Teacher's Note:
1. The boiling point is raised, never lowered, by a non-volatile solute.

Question 18: The osmotic pressure of a solution containing 1 g of a protein in 100 mL of solution at 300 K is 1.66 bar. What is the molecular weight of the protein (R = 0.083 L bar mol⁻¹ K⁻¹)?
(a) 150
(b) 300
(c) 120
(d) 130
Show Answer & Explanation

Answer: (a) 150

Explanation:
1. \( M = \dfrac{wRT}{\pi V} = \dfrac{1\times0.083\times300}{1.66\times0.1} = 150 \) g/mol.

Teacher's Note:
1. Write V in litres (0.1 L) when using R in L bar.

Question 19: The vapour pressure of a solution of 5 g of a non-electrolyte in 100 g of water at a particular temperature is 2985 N m⁻². The vapour pressure of pure water at that temperature is 3000 N m⁻². The molecular weight of the solute is:
(a) 270
(b) 90
(c) 200
(d) 180
Show Answer & Explanation

Answer: (d) 180

Explanation:
1. \( \dfrac{3000-2985}{3000} = 0.005 = \dfrac{n}{N} \) (dilute), with \( N = \dfrac{100}{18} = 5.56 \).
2. \( n = 0.0278 \), so \( M = \dfrac{5}{0.0278} = 180 \) g/mol.

Teacher's Note:
1. For dilute solutions the mole fraction of solute is approximately n/N.

Question 20: 1 mol of heptane (V.P. = 92 mm of Hg) was mixed with 4 mol of octane (V.P. = 31 mm of Hg). The vapour pressure of the resulting ideal solution is:
(a) 38.4 mm of Hg
(b) 43.2 mm of Hg
(c) 40.0 mm of Hg
(d) 46.2 mm of Hg
Show Answer & Explanation

Answer: (b) 43.2 mm of Hg

Explanation:
1. \( x_{heptane} = 0.2 \) and \( x_{octane} = 0.8 \).
2. \( P = 0.2\times92 + 0.8\times31 = 18.4+24.8 = 43.2 \) mm of Hg.

Teacher's Note:
1. Raoult's law for each component, then add the partial pressures.

Question 21: Solute A is a ternary electrolyte and solute B is a non-electrolyte. If a 0.1 M solution of 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) 2 P
(b) 3 P
(c) 1.5 P
(d) P
Show Answer & Explanation

Answer: (b) 3 P

Explanation:
1. For B: \( 2P = 0.1RT \).
2. For A (i = 3): \( \pi = 3\times0.05RT = 0.15RT = 1.5\times(0.1RT) = 1.5\times2P = 3P \).

Teacher's Note:
1. A ternary electrolyte gives three ions, so i = 3 for complete dissociation.

Question 22: On adding a non-volatile solute to a solvent, the vapour pressure of the solvent decreases and becomes z × (vapour pressure of the pure solvent), where z is the:
(a) mole fraction of solute
(b) mole fraction of solvent
(c) molality
(d) molarity
Show Answer & Explanation

Answer: (b) mole fraction of solvent

Explanation:
1. By Raoult's law, \( P_s = x_{solvent}P^0 \), so \( z = x_{solvent} \).

Teacher's Note:
1. The vapour pressure of the solution is the mole fraction of solvent times that of pure solvent.

Question 23: How many grams of a non-volatile solute of molecular weight 90 are to be dissolved in 97.5 g of water in order to decrease the vapour pressure of water by 2.5 percent?
(a) 18
(b) 25
(c) 12.5
(d) 9
Show Answer & Explanation

Answer: (c) 12.5

Explanation:
1. \( x_{solute} = 0.025 \). \( N = \dfrac{97.5}{18} = 5.417 \).
2. \( \dfrac{n}{n+5.417} = 0.025 \Rightarrow n = 0.1389 \) mol.
3. Mass \( = 0.1389\times90 = 12.5 \) g.

Teacher's Note:
1. A 2.5% lowering means the mole fraction of the solute is 0.025.

Question 24: At 88 °C benzene has a vapour pressure of 900 torr and toluene has a vapour pressure of 360 torr. What is the mole fraction of benzene in the mixture with toluene that will boil at 88 °C at 1 atm pressure (benzene-toluene form an ideal solution)?
(a) 0.740
(b) 0.416
(c) 0.688
(d) 0.588
Show Answer & Explanation

Answer: (a) 0.740

Explanation:
1. The solution boils when its vapour pressure is 760 torr: \( 760 = 900x + 360(1-x) \).
2. \( 400 = 540x \Rightarrow x = 0.74 \).

Teacher's Note:
1. Boiling occurs when the total vapour pressure equals the external pressure of 760 torr.

Question 25: Phenol associates in benzene as \( C_6H_5OH \rightleftharpoons \tfrac12(C_6H_5OH)_2 \). If the degree of association of phenol is 40%, the van't Hoff factor i is:
(a) 1.4
(b) 1
(c) 0.8
(d) 0.6
Show Answer & Explanation

Answer: (c) 0.8

Explanation:
1. Initially 1 mole. At equilibrium: \( (1-\alpha) + \dfrac\alpha2 = 1-\dfrac\alpha2 \).
2. \( i = 1-0.2 = 0.8 \).

Teacher's Note:
1. Each pair of molecules that associates reduces the number of particles by one half per molecule.

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