Solutions Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the freezing point depression of a solution containing 18 g of sucrose (molar mass = 342 g/mol) in 400 g of water? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))

Moles of sucrose = \( \frac{18}{342} \approx 0.0526 \, \text{mol} \).

Molality = \( \frac{0.0526}{0.4} \approx 0.1315 \, \text{mol/kg} \).

\( \Delta T_f = 1.86 \times 0.1315 \approx 0.2446 \, \text{K} \).

0.2 K
0.2446 K
0.3 K
0.186 K
2

The solubility of a gas in a liquid increases with an increase in:

According to Henry’s law, solubility of a gas is directly proportional to its partial pressure.

Pressure
Temperature
Volume
Density
1

What is the osmotic pressure of a 0.025 M solution of a non-electrolyte at 27°C? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

\( \Pi = MRT \).

\( \Pi = 0.025 \times 0.0821 \times 300 \approx 0.6158 \, \text{atm} \).

0.5 atm
0.6 atm
0.6158 atm
0.7 atm
3

The vapor pressure of pure water at a certain temperature is 50 mm Hg. What is the vapor pressure of a solution with a solute mole fraction of 0.1?

Mole fraction of water = \( 1 - 0.1 = 0.9 \).

Vapor pressure = \( 50 \times 0.9 = 45 \, \text{mm Hg} \).

45 mm Hg
40 mm Hg
50 mm Hg
55 mm Hg
1

The van’t Hoff factor of a 0.1 m K₂SO₄ solution is 2.4. What is the boiling point elevation? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))

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

\( \Delta T_b = 2.4 \times 0.52 \times 0.1 = 0.1248 \, \text{K} \).

0.1 K
0.11 K
0.12 K
0.1248 K
4

What is the boiling point elevation of a solution containing 12 g of glucose (molar mass = 180 g/mol) in 600 g of water? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))

Moles of glucose = \( \frac{12}{180} \approx 0.0667 \, \text{mol} \).

Molality = \( \frac{0.0667}{0.6} \approx 0.111 \, \text{mol/kg} \).

\( \Delta T_b = 0.52 \times 0.111 \approx 0.0577 \, \text{K} \).

0.05 K
0.06 K
0.0577 K
0.07 K
3

A solution contains 23 g of methanol (molar mass = 32 g/mol) and 72 g of water. If 64 g of water is added, what is the new mole fraction of methanol?

Moles of methanol = \( \frac{23}{32} \approx 0.7188 \).

Initial moles of water = \( \frac{72}{18} = 4 \).

New moles of water = \( \frac{72 + 64}{18} = \frac{136}{18} \approx 7.5556 \).

Total moles = \( 0.7188 + 7.5556 \approx 8.2744 \).

Mole fraction = \( \frac{0.7188}{8.2744} \approx 0.0869 \).

0.0869
0.1
0.12
0.08
1

The vapor pressure of pure water is 25 mm Hg at a certain temperature. A solution with a non-volatile solute has a vapor pressure of 23 mm Hg. If the solute’s molar mass is 60 g/mol, what is the mass of solute in 180 g of water?

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

\( \frac{25 - 23}{25} = 0.08 \).

Moles of water = \( \frac{180}{18} = 10 \).

\( x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + 10} = 0.08 \).

\( n_{\text{solute}} = 0.08 (n_{\text{solute}} + 10) \), \( n_{\text{solute}} - 0.08 n_{\text{solute}} = 0.8 \), \( 0.92 n_{\text{solute}} = 0.8 \), \( n_{\text{solute}} \approx 0.8696 \).

Mass = \( 0.8696 \times 60 \approx 52.18 \, \text{g} \).

50 g
51 g
52.18 g
53 g
3

A 0.25 M solution of a solute in 400 mL of water has an osmotic pressure of 1.845 atm at 27°C. If the solute dissociates into 2 ions, what is the degree of dissociation? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

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

\( 1.845 = i \times 0.25 \times 0.0821 \times 300 \).

\( i = \frac{1.845}{0.25 \times 24.63} \approx 0.3 \times 6 = 1.8 \).

\( i = 1 + \alpha \), \( 1.8 = 1 + \alpha \), \( \alpha = 0.8 \).

0.7
0.9
1
0.8
4

A 0.25 molal solution of a solute in water has a boiling point elevation of 0.26°C. If the solute dissociates into 2 ions, what is the degree of dissociation? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))

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

\( 0.26 = i \times 0.52 \times 0.25 \).

\( i = \frac{0.26}{0.52 \times 0.25} = 2 \).

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

0.8
1
0.9
0.7
2

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