Solutions Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A solution is made by dissolving 20 g of a solute in 80 g of water. If the mass percentage of the solute is doubled in a new solution with the same solute mass, what is the mass of water required?

Initial mass % = \( \frac{20}{20 + 80} \times 100 = 20\% \).

New mass % = \( 40\% = \frac{20}{20 + w} \times 100 \).

\( 0.4 (20 + w) = 20 \), \( 20 + w = 50 \), \( w = 30 \, \text{g} \).

30 g
40 g
50 g
60 g
1

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 260 mm Hg, what is the mole fraction of the first component?

\( P = P_1^0 \cdot x_1 + P_2^0 \cdot (1 - x_1) \).

\( 260 = 200 x_1 + 300 (1 - x_1) \).

\( 260 = 200 x_1 + 300 - 300 x_1 \).

\( 260 - 300 = -100 x_1 \), \( x_1 = 0.4 \).

0.3
0.4
0.6
0.5
2

A solution of two volatile liquids has vapor pressures of 300 mm Hg and 400 mm Hg for pure components. If the mole fraction of the second component in the vapor phase is 0.6, what is the total vapor pressure?

Vapor phase: \( y_2 = \frac{P_2^0 \cdot x_2}{P_{\text{total}}} \).

\( 0.6 = \frac{400 \cdot x_2}{P_{\text{total}}} \).

Liquid phase: \( P_{\text{total}} = 300 (1 - x_2) + 400 x_2 \).

Substitute: \( 0.6 P_{\text{total}} = 400 x_2 \), \( P_{\text{total}} = 300 + 100 x_2 \).

\( 0.6 (300 + 100 x_2) = 400 x_2 \), \( 180 + 60 x_2 = 400 x_2 \), \( 340 x_2 = 180 \), \( x_2 = \frac{180}{340} \approx 0.5294 \).

\( P_{\text{total}} = 300 (1 - 0.5294) + 400 \times 0.5294 \approx 352.96 \, \text{mm Hg} \).

350 mm Hg
352.96 mm Hg
360 mm Hg
340 mm Hg
2

The vapor pressure of pure water is 24 mm Hg at a certain temperature. What is the vapor pressure of a solution with a solute mole fraction of 0.25?

Mole fraction of water = \( 1 - 0.25 = 0.75 \).

Vapor pressure = \( 24 \times 0.75 = 18 \, \text{mm Hg} \).

18 mm Hg
20 mm Hg
16 mm Hg
22 mm Hg
1

A solution of two volatile liquids has vapor pressures of 150 mm Hg and 250 mm Hg for pure components. If the total vapor pressure is 190 mm Hg, what is the mole fraction of the first component?

\( P = P_1^0 \cdot x_1 + P_2^0 \cdot (1 - x_1) \).

\( 190 = 150 x_1 + 250 (1 - x_1) \).

\( 190 = 150 x_1 + 250 - 250 x_1 \).

\( 190 - 250 = -100 x_1 \), \( x_1 = 0.6 \).

0.5
0.6
0.4
0.7
2

How many moles of NaOH are present in 500 mL of a 0.2 M solution?

Molarity = \( \frac{\text{Moles of solute}}{\text{Volume in liters}} \).

Moles = Molarity × Volume = \( 0.2 \, \text{mol/L} \times 0.5 \, \text{L} = 0.1 \, \text{mol} \).

0.1 mol
0.2 mol
0.05 mol
1.0 mol
1

A 0.1 molal NaCl solution has a van’t Hoff factor of 1.8. What is the freezing point depression? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))

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

\( \Delta T_f = 1.8 \times 1.86 \times 0.1 \approx 0.335 \, \text{K} \).

0.186 K
0.3 K
0.335 K
0.4 K
3

The molar mass of a solute is 100 g/mol if 10 g of it in 1 L of solution produces an osmotic pressure of 0.246 atm at 27°C. (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

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

\( 0.246 = \frac{10}{M \times 1} \times 0.0821 \times 300 \).

\( M = \frac{10 \times 0.0821 \times 300}{0.246} \approx 100 \, \text{g/mol} \).

50 g/mol
75 g/mol
100 g/mol
150 g/mol
3

The partial pressure of a gas above a solution is 2 bar, and its Henry’s law constant is 100 bar. What is the mole fraction of the gas in the solution?

Henry’s law: \( p = K_H \cdot x \).

\( x = \frac{p}{K_H} = \frac{2}{100} = 0.02 \).

0.01
0.02
0.05
0.1
2

A solution of a non-volatile solute in water has a vapor pressure of 23.7 mm Hg at a temperature where pure water’s vapor pressure is 25 mm Hg. If the solution freezes at -0.372°C, what is the molality? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))

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

\( 0.372 = 1.86 \cdot m \).

\( m = \frac{0.372}{1.86} = 0.2 \, \text{mol/kg} \).

Cross-check: \( x_{\text{solute}} = \frac{25 - 23.7}{25} = 0.052 \), consistent for dilute solution.

0.18 mol/kg
0.19 mol/kg
0.21 mol/kg
0.2 mol/kg
4

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