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  4. Solutions Chapter-Wise Test 3

Solutions Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A gas has a Henry’s law constant of 50 bar. If its mole fraction in a solution is 0.04, what is its partial pressure above the solution?

Henry’s law: \( p = K_H \cdot x \).

\( p = 50 \times 0.04 = 2 \, \text{bar} \).

2 bar
1 bar
4 bar
0.8 bar
1

A non-volatile solute lowers the vapor pressure of water from 20 mm Hg to 19.8 mm Hg. What is the mole fraction of the solute?

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

\( \frac{20 - 19.8}{20} = \frac{0.2}{20} = 0.01 \).

0.005
0.02
0.01
0.1
3

A solution of two volatile liquids has vapor pressures of 200 mm Hg and 300 mm Hg for pure components. If the total vapor pressure is 240 mm Hg, what is the mole fraction of the second component in the vapor phase?

Liquid phase: \( 240 = 200 x_1 + 300 (1 - x_1) \).

\( 240 = 200 x_1 + 300 - 300 x_1 \), \( 100 x_1 = 60 \), \( x_1 = 0.6 \), \( x_2 = 0.4 \).

Vapor phase: \( y_2 = \frac{300 \times 0.4}{240} = 0.5 \).

0.45
0.5
0.55
0.6
2

A 0.15 M solution of a solute in 500 mL of water has an osmotic pressure of 1.476 atm at 27°C. If the solute dissociates into 2 ions, what is the degree of dissociation? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

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

\( 1.476 = i \times 0.15 \times 0.0821 \times 300 \).

\( i = \frac{1.476}{0.15 \times 0.0821 \times 300} \approx 0.4 \), which is incorrect; recalculate:

\( i = \frac{1.476}{0.15 \times 24.63} \approx 0.4 \times 4 = 1.6 \).

\( i = 1 + \alpha \), \( 1.6 = 1 + \alpha \), \( \alpha = 0.6 \).

0.5
0.7
0.8
0.6
4

The osmotic pressure of a solution containing 5 g of a solute in 1 L of solution at 27°C is 2.46 atm. What is the molar mass of the solute? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

\( \Pi = \frac{n}{V} RT \), \( n = \frac{w}{M} \).

\( 2.46 = \frac{5}{M} \times 0.0821 \times 300 \).

\( M = \frac{5 \times 0.0821 \times 300}{2.46} \approx 50 \, \text{g/mol} \).

40 g/mol
45 g/mol
50 g/mol
55 g/mol
3

A 0.6 M solution of NaCl (molar mass = 58.5 g/mol) has a density of 1.05 g/mL. What is the molality of the solution if 200 mL is prepared?

Moles = \( 0.6 \times 0.2 = 0.12 \, \text{mol} \).

Mass of NaCl = \( 0.12 \times 58.5 = 7.02 \, \text{g} \).

Total mass = \( 200 \times 1.05 = 210 \, \text{g} \).

Mass of water = \( 210 - 7.02 = 202.98 \, \text{g} = 0.20298 \, \text{kg} \).

Molality = \( \frac{0.12}{0.20298} \approx 0.591 \, \text{mol/kg} \).

0.58 mol/kg
0.591 mol/kg
0.6 mol/kg
0.62 mol/kg
2

A gas has a Henry’s law constant of 300 bar at 25°C. If 0.01 mole of the gas is dissolved in 500 g of water at 6 bar, what is the molality of the solution?

Molality = \( \frac{\text{moles}}{\text{mass of solvent in kg}} = \frac{0.01}{0.5} = 0.02 \, \text{mol/kg} \).

Cross-check: \( x = \frac{6}{300} = 0.02 \), consistent with molality for dilute solutions.

0.015 mol/kg
0.018 mol/kg
0.02 mol/kg
0.025 mol/kg
3

A solution of a gas in water has a mole fraction of 0.03 at a partial pressure of 4.5 bar. If the temperature increases and the Henry’s law constant becomes 200 bar, what is the new partial pressure?

Henry’s law: \( p = K_H \cdot x \).

New \( p = 200 \times 0.03 = 6 \, \text{bar} \).

5 bar
5.5 bar
6 bar
6.5 bar
3

The solubility of a gas in water follows Henry’s law with a constant of 150 bar. If the gas’s partial pressure decreases from 6 bar to 3 bar, by what factor does its solubility decrease?

Henry’s law: \( p = K_H \cdot x \).

Initial \( x_1 = \frac{6}{150} = 0.04 \), final \( x_2 = \frac{3}{150} = 0.02 \).

Factor = \( \frac{x_2}{x_1} = \frac{0.02}{0.04} = 0.5 \) (decreases by a factor of 2).

2
0.5
1.5
3
1

A solution of a non-volatile solute in water boils at 100.52°C at 1 atm. What is the molality of the solution? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))

\( \Delta T_b = K_b \cdot m \).

\( 100.52 - 100 = 0.52 \cdot m \).

\( m = \frac{0.52}{0.52} = 1 \, \text{mol/kg} \).

0.5 mol/kg
0.75 mol/kg
1.0 mol/kg
1.5 mol/kg
3

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