Solutions Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How many grams of CaCl₂ (molar mass = 111 g/mol) are needed to prepare 600 mL of a 0.2 M solution?

Moles of CaCl₂ = \( 0.2 \times 0.6 = 0.12 \, \text{mol} \).

Mass = \( 0.12 \times 111 = 13.32 \, \text{g} \).

13 g
13.32 g
14 g
12.5 g

A solute dissociates into 3 ions with a van’t Hoff factor of 2.2. What is the degree of dissociation?

\( i = 1 + \alpha (n - 1) \), where \( n = 3 \).

\( 2.2 = 1 + \alpha (3 - 1) \).

\( 2.2 = 1 + 2\alpha \), \( \alpha = \frac{1.2}{2} = 0.6 \).

0.5
0.7
0.8
0.6
4

A 0.1 molal solution of KCl has a van’t Hoff factor of 1.9. What is the freezing point depression? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))

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

\( \Delta T_f = 1.9 \times 1.86 \times 0.1 \approx 0.353 \, \text{K} \).

0.186 K
0.3 K
0.353 K
0.5 K
3

A solution of 0.1 mole of a solute in 1 kg of water freezes at -0.558°C. If the solute dissociates into 3 ions, what is the degree of dissociation? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))

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

\( 0.558 = i \times 1.86 \times 0.1 \).

\( i = \frac{0.558}{1.86 \times 0.1} = 3 \).

\( i = 1 + \alpha (n - 1) \), \( 3 = 1 + \alpha (3 - 1) \), \( 2\alpha = 2 \), \( \alpha = 1 \).

0.8
0.9
0.95
1
4

A 0.4 M solution of KNO₃ (molar mass = 101 g/mol) has a density of 1.1 g/mL. What is the mass percentage of KNO₃ in 250 mL of this solution?

Moles = \( 0.4 \times 0.25 = 0.1 \, \text{mol} \).

Mass of KNO₃ = \( 0.1 \times 101 = 10.1 \, \text{g} \).

Total mass = \( 250 \times 1.1 = 275 \, \text{g} \).

Mass % = \( \frac{10.1}{275} \times 100 \approx 3.673\% \).

3.5%
3.673%
4%
3%
2

A solution is prepared by dissolving 20 g of a solute in water to make 400 mL of solution with a density of 1.25 g/mL. If the molarity is 0.5 M, what is the molar mass of the solute?

Moles = \( M \times V = 0.5 \times 0.4 = 0.2 \, \text{mol} \).

Molar mass = \( \frac{\text{Mass}}{\text{Moles}} = \frac{20}{0.2} = 100 \, \text{g/mol} \).

(Density is extra info, not needed here.)

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

A gas has a Henry’s law constant of 80 bar. If its partial pressure above a solution is 4 bar, what is its mole fraction?

Henry’s law: \( p = K_H \cdot x \).

\( x = \frac{4}{80} = 0.05 \).

0.05
0.04
0.06
0.02
1

A solution of 0.03 mole of a non-volatile solute in 300 mL of water has an osmotic pressure of 1.231 atm at 27°C. What is the van’t Hoff factor? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

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

Molarity = \( \frac{0.03}{0.3} = 0.1 \, \text{M} \).

\( 1.231 = i \times 0.1 \times 0.0821 \times 300 \).

\( i = \frac{1.231}{0.1 \times 24.63} \approx 0.5 \), recalculate: \( i = 1 \) (non-electrolyte).

2
1.5
1
2.5
3

A solution of 0.05 mole of a non-volatile solute in 500 g of water has an osmotic pressure of 2.463 atm at 27°C. What is the van’t Hoff factor? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \), assume 0.5 L solution)

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

Molarity = \( \frac{0.05}{0.5} = 0.1 \, \text{M} \).

\( 2.463 = i \times 0.1 \times 0.0821 \times 300 \).

\( i = \frac{2.463}{0.1 \times 24.63} \approx 1 \).

2
1.5
1
2.5
3

What is the boiling point elevation of a 0.5 molal solution of a non-electrolyte in water? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))

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

\( \Delta T_b = 0.52 \times 0.5 = 0.26 \, \text{K} \).

0.2 K
0.24 K
0.26 K
0.3 K
3

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