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Solutions Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A solution containing 0.1 mole of a non-volatile solute in 1 kg of water has an osmotic pressure of 2.46 atm at 27°C. What is the van’t Hoff factor? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \), assume 1 L solution)

\( \Pi = i \cdot M \cdot RT \).

Molarity ≈ molality for dilute solution, so \( M = 0.1 \, \text{M} \).

\( 2.46 = i \times 0.1 \times 0.0821 \times 300 \).

\( i = \frac{2.46}{0.1 \times 0.0821 \times 300} \approx 1 \).

2
1.5
1
2.5
3

What is the boiling point elevation of a solution containing 10 g of fructose (molar mass = 180 g/mol) in 400 g of water? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))

Moles of fructose = \( \frac{10}{180} \approx 0.0556 \, \text{mol} \).

Molality = \( \frac{0.0556}{0.4} \approx 0.139 \, \text{mol/kg} \).

\( \Delta T_b = 0.52 \times 0.139 \approx 0.0723 \, \text{K} \).

0.06 K
0.07 K
0.0723 K
0.08 K
3

A 0.2 molal Na₂SO₄ solution has a van’t Hoff factor of 2.5. What is the freezing point depression? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))

\( \Delta T_f = i \cdot K_f \cdot m \).

\( \Delta T_f = 2.5 \times 1.86 \times 0.2 = 0.93 \, \text{K} \).

0.8 K
0.85 K
0.93 K
1.0 K
3

What is the molality of a solution containing 14 g of ethylene glycol (molar mass = 62 g/mol) in 350 g of water?

Moles of ethylene glycol = \( \frac{14}{62} \approx 0.2258 \, \text{mol} \).

Mass of solvent = 350 g = 0.35 kg.

Molality = \( \frac{0.2258}{0.35} \approx 0.645 \, \text{mol/kg} \).

0.6 mol/kg
0.645 mol/kg
0.7 mol/kg
0.5 mol/kg
2

What is the mass percentage of a solution prepared by dissolving 15 g of a solute in 85 g of water?

Total mass = \( 15 + 85 = 100 \, \text{g} \).

Mass % = \( \frac{15}{100} \times 100 = 15\% \).

10%
12%
15%
20%
3

What is the mole fraction of a solute if its solution lowers the vapor pressure of water from 25 mm Hg to 24 mm Hg?

\( \frac{p^0 - p}{p^0} = x_{\text{solute}} \).

\( \frac{25 - 24}{25} = 0.04 \).

0.02
0.03
0.05
0.04
4

A gas has a Henry’s law constant of 300 bar. If its partial pressure is 6 bar, what is the mole fraction in the solution?

\( p = K_H \cdot x \).

\( x = \frac{6}{300} = 0.02 \).

0.015
0.025
0.01
0.02
4

The vapor pressure of pure water is 25 mm Hg at a certain temperature. What is the vapor pressure of a solution containing 0.1 mole of a non-volatile solute in 0.9 mole of water?

Mole fraction of water = \( \frac{0.9}{0.9 + 0.1} = 0.9 \).

Vapor pressure = \( 25 \times 0.9 = 22.5 \, \text{mm Hg} \).

20 mm Hg
21 mm Hg
22 mm Hg
22.5 mm Hg
4

What is the mass of ethanol (molar mass = 46 g/mol) in 500 g of a solution if its mole fraction is 0.2 and water is the solvent?

Moles of water = \( \frac{500 - x}{18} \), moles of ethanol = \( \frac{x}{46} \).

Mole fraction = \( \frac{\frac{x}{46}}{\frac{x}{46} + \frac{500 - x}{18}} = 0.2 \).

Solving: \( \frac{x}{46} = 0.2 \left( \frac{x}{46} + \frac{500 - x}{18} \right) \).

Simplify: \( x \approx 138 \, \text{g} \) (after cross-multiplication and approximation).

92 g
138 g
46 g
100 g
2

A gas has a partial pressure of 5 bar and a Henry’s law constant of 500 bar. What is the mole fraction in the solution?

\( p = K_H \cdot x \).

\( x = \frac{5}{500} = 0.01 \).

0.005
0.015
0.02
0.01
4

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