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Chemistry MCQs for NEET — Practice Questions with Answers

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34.2 g of cane sugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be

0.0099

1.1597

0.840

0.9901

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Explanation

$\frac{P_{A}^{0} - P_{A}}{P_{A}^{0}} = \frac{W_{B} / M_{A}}{W_{B} / M_{B} + W_{A} / M_{A}} = \frac{34 .2 / 342}{34 .2 / 342 + 180 / 18} = \frac{0 .1}{10 .1} = 0 .0099$

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Lowering in vapour pressure is the highest for [Roorkee 1989; BHU 1997]

0.2m urea

0.1 m glucose

0.1m MgSO₄

0.1m BaCl₂

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Explanation

$\frac{P_{A}^{0} - P_{A}}{P_{A}^{0}} = Molality \times (1 - α x + x α + γ x)$

The value of $P_{A}^{0} - P_{A}$ is maximum for BaCl₂.

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Vapour pressure of CCl₄ at 25°C is 143 mm Hg 0.5 g of a non-volatile solute (mol. wt. 65) is dissolved in 100 ml of CCl₄. Find the vapour pressure of the solution. (Density of $C C l_{4} = 1 .58 g / c m^{3})$ [CBSE PMT 1996]

141.93 mm

94.39 mm

199.34 mm

143.99 mm

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Explanation

$\frac{P^{0} - P_{s}}{P^{0}} = \frac{n_{2}}{n_{1}}$ ; $\frac{143 - P_{s}}{143} = \frac{0 .5 / 65}{158 / 154}$

or $P_{s} = 141 .93 m m$

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The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour pressure in contact with equimolar solution of benzene and toluene is [Pb. CET 1988]

0.50

0.6

0.27

0.73

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Explanation

For equimolar solutions, $X_{B} = X_{T} = 0 .5$

$P_{B} = X_{B} \times P_{B}^{0} = 0 .5 \times 160 = 80 m m$

$P_{T} = X_{T} \times P_{T}^{0} = 0 .5 \times 60 = 30 m m$

$P_{Total} = 80 + 30 = 110 m m$

Mole fraction of toluene in vapour phase $= \frac{30}{110} = 0 .27$

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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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