Solutions Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The vapor pressure of pure benzene is 100 mm Hg at a certain temperature. If a solution with a non-volatile solute has a vapor pressure of 95 mm Hg, what is the mole fraction of benzene in the solution?

Raoult’s law: \( p = p^0 \cdot x_{\text{benzene}} \).

\( 95 = 100 \cdot x_{\text{benzene}} \).

\( x_{\text{benzene}} = \frac{95}{100} = 0.95 \).

0.95
0.9
0.85
0.8
1

A 0.2 molal solution of NaBr in water has a freezing point depression of 0.744°C. What is the van’t Hoff factor? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))

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

\( 0.744 = i \times 1.86 \times 0.2 \).

\( i = \frac{0.744}{1.86 \times 0.2} = 2 \).

1.8
2
1.9
2.1
2

The vapor pressure of a pure solvent is 40 mm Hg. What is the vapor pressure of a solution with a mole fraction of solute 0.25?

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

Vapor pressure = \( 40 \times 0.75 = 30 \, \text{mm Hg} \).

30 mm Hg
10 mm Hg
35 mm Hg
25 mm Hg
1

A gas has a Henry’s law constant of 250 bar. If its mole fraction in a solution is 0.008, what is its partial pressure?

Henry’s law: \( p = K_H \cdot x \).

\( p = 250 \times 0.008 = 2 \, \text{bar} \).

2 bar
1.5 bar
2.5 bar
3 bar
1

What is the freezing point depression of a solution containing 12 g of a non-electrolyte solute (molar mass = 60 g/mol) in 200 g of water? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))

Moles of solute = \( \frac{12}{60} = 0.2 \, \text{mol} \).

Molality = \( \frac{0.2}{0.2} =appendicitis = 1 \, \text{mol/kg} \).

\( \Delta T_f = K_f \cdot m = 1.86 \times 1 = 1.86 \, \text{K} \).

1.0 K
1.5 K
1.86 K
2.0 K
3

What is the mole fraction of methanol (molar mass = 32 g/mol) in a solution containing 16 g of methanol and 72 g of water?

Moles of methanol = \( \frac{16}{32} = 0.5 \, \text{mol} \).

Moles of water = \( \frac{72}{18} = 4 \, \text{mol} \).

Total moles = \( 0.5 + 4 = 4.5 \).

Mole fraction = \( \frac{0.5}{4.5} \approx 0.111 \).

0.111
0.2
0.25
0.5
1

What is the vapor pressure of a solution made by mixing two volatile liquids with vapor pressures 300 mm Hg and 400 mm Hg, if their mole fractions in the solution are 0.4 and 0.6, respectively?

Total vapor pressure = \( p_1^0 \cdot x_1 + p_2^0 \cdot x_2 \).

\( 300 \times 0.4 + 400 \times 0.6 = 120 + 240 = 360 \, \text{mm Hg} \).

340 mm Hg
350 mm Hg
360 mm Hg
370 mm Hg
3

A solution is prepared by dissolving 15 g of a solute in water to make 250 mL of solution. If the osmotic pressure is 1.476 atm at 27°C, what is the molar mass of the solute? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

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

\( 1.476 = \frac{15}{M \times 0.25} \times 0.0821 \times 300 \).

\( M = \frac{15 \times 0.0821 \times 300}{1.476 \times 0.25} \approx 100 \, \text{g/mol} \).

90 g/mol
100 g/mol
110 g/mol
120 g/mol
2

A solution of 8 g of a solute (molar mass = 80 g/mol) in 500 g of water freezes at -0.372°C. What is the van’t Hoff factor? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))

Moles = \( \frac{8}{80} = 0.1 \).

Molality = \( \frac{0.1}{0.5} = 0.2 \, \text{mol/kg} \).

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

\( 0.372 = i \times 1.86 \times 0.2 \), \( i = \frac{0.372}{1.86 \times 0.2} = 1 \).

0.8
0.9
1
1.2
3

A solution is made by dissolving 20 g of a solute (molar mass = 40 g/mol) in water. If the freezing point depression is 0.465°C and \( K_f = 1.86 \, \text{K kg mol}^{-1} \), what is the total mass of the solution?

Moles = \( \frac{20}{40} = 0.5 \).

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

\( 0.465 = 1.86 \times \frac{0.5}{w} \), \( w = \frac{0.5 \times 1.86}{0.465} \approx 2 \, \text{kg} = 2000 \, \text{g} \).

Total mass = \( 20 + 2000 = 2020 \, \text{g} \).

2000 g
2010 g
2020 g
2030 g
3

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