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Solutions MCQs for NEET — Chemistry Questions with Answers

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The vapour pressure of a solvent decreases by 10 mm of mercury 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 if the decrease in vapour pressure is to be 20 mm of mercury then the mole fraction of the solvent is [CBSE PMT 1998]

0.8

0.6

0.4

None

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Explanation

$Δ P / P^{0} = X_{2}$

Hence $Δ P / P^{0} = X_{2} / X_{2}^{1}$ i.e. $10 / 20 = 0 .2 / X_{B}$ or $X_{B} = 0 .4$

Mole fraction of solvent = 1 – 0.4 = 0.6

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The vapour pressure of a solvent A is 0.80 atm. When a non-volatile substance B is added to this solvent its vapour pressure drops to 0.6 atm. the mole fraction of B in the solution is [MP PMT 2000]

0.25

0.50

0.75

0.90

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Explanation

$Δ P / P^{0} = X_{B}$ or $X_{B} = 0 .2 / 0 .8 = 0 .25$

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Osmotic pressure is 0.0821 atm at a temperature of 300 K. find concentration in mole/litre [Roorkee 1990]

0.033

0.066

0.33 × 10^–2

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Explanation

$C = \frac{P}{R T} = \frac{0 .0821}{0 .0821 \times 300} = \frac{1}{300} = 0 .33 \times 10^{- 2} m o l e / l i t r e$

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The osmotic pressure of 5% (mass-volume) solution of cane sugar at 150°C (mol. mass of sugar = 342) is [BHU 1995]

4 atm

5.07 atm

3.55 atm

2.45 atm

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Explanation

$C = \frac{5}{342} \times \frac{1}{100} \times 1000 = \frac{50}{342} M$ ;

$P = \frac{50}{342} \times 0 .082 \times 423 = 5 .07 a t m$

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The molal b.p. constant for water is $0 {.513}^{o} C k g m o l^{- 1}$ . When 0.1 mole of sugar is dissolved in 200 g of water, the solution boils under a pressure of 1 atm at [AIIMS 1991]

100.513°C

100.0513°C

100.256°C

101.025°C

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Explanation

$Δ T_{b} = K_{b} \times m = 0 .513 (\frac{0 .1}{200} \times 1000) = 0 .2565$ ;

$Δ T_{b} = 100 {.2565}^{o} C$

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An aqueous solution containing 1 g of urea boils at 100.25°C. The aqueous solution containing 3 g of glucose in the same volume will boil at [BHU 1994, CBSE 2000]

100.75°C

100.5°C

100°C

100.25°C

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Explanation

1 g urea $= \frac{1}{60} m o l$ , 3 g glucose $= \frac{3}{180} = \frac{1}{60} m o l$ .

Hence it will boil at the same temperature

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Solution of sucrose (Mol. Mass = 342) is prepared by dissolving 34.2 gm. of it in 1000 gm. of water. freezing point of the solution is (K_f for water is 1.86 K kg mol^–1) [AIEEE 2003]

272.814 K

278.1 K

273.15 K

270 K

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Explanation

Molality of the solution $= \frac{34 .2}{342} = 0 .1$

$Δ T_{f} = K_{f} \times m = 1 .86 \times 0 .1 = 0 .186 K$

Freezing point of solution $= 273 - 0 .186 = 272 .814 K$

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An aqueous solution of a weak monobasic acid containing 0.1 g in 21.7 g of water freezes at 272.817K. If the value of K_f for water is 1.86 K kg mol^–1, the molecular mass of the acid is [AMU 2002]

46.0

48.6

48.8

46.8

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Explanation

Mass of solvent $(W_{A}) = 21 .7 g$

Mass of solute $(W_{B}) = 0 .1 g$

Depression in freezing point, $(Δ T_{f}) = 273 - 272 .817 = 0 .183 K$

$Δ T_{f} = k_{f} \times m = k_{f} . \frac{W_{B}}{W_{A}} \times \frac{1000}{M_{B}}$

$M_{B} = \frac{k_{f} \times W_{B} \times 1000}{W_{A} \times Δ T_{f}} = \frac{1 .86 \times 0 .1 \times 1000}{21 .7 \times 0 .183} = 46 .8$

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What is the molality of solution of a certain solute in a solvent if there is a freezing point depression of 0.184° and if the freezing point constant is 18.4

0.01

0.001

100

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Explanation

$Δ T_{f} = K_{f} \times m$

or $m = \frac{Δ T_{f}}{K_{f}} = \frac{0 .184}{18 .4} = 0 .01$

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A solution containing 6.8 g of a nonionic solute in 100 g of water was found to freeze at $- 0 {.93}^{o} C$ . The freezing point depression constant of water is 1.86. Calculate the molecular weight of the solute

[Pb. PMT 1994; ISM Dhanbad 1994]

13.6

136

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Explanation

$M_{B} = \frac{1000 \times K_{f} \times W_{B}}{Δ T_{f} \times W_{A}} = \frac{1000 \times 1 .86 \times 6 .8}{100 \times 0 .93} = 136$

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