Solutions Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the molality of a solution containing 10 g of glycerol (molar mass = 92 g/mol) in 200 g of water?

Moles of glycerol = \( \frac{10}{92} \approx 0.1087 \, \text{mol} \).

Mass of solvent = 200 g = 0.2 kg.

Molality = \( \frac{0.1087}{0.2} \approx 0.5435 \, \text{mol/kg} \).

0.5 mol/kg
0.5435 mol/kg
0.6 mol/kg
0.4 mol/kg
2

A solution boils at 100.26°C at 1 atm. What is the molality if \( K_b = 0.52 \, \text{K kg mol}^{-1} \)?

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

\( 100.26 - 100 = 0.52 \cdot m \).

\( m = \frac{0.26}{0.52} = 0.5 \, \text{mol/kg} \).

0.25 mol/kg
0.4 mol/kg
0.45 mol/kg
0.5 mol/kg
4

A solution of a gas in water has a mole fraction of 0.02 at 5 bar pressure. If the Henry’s law constant decreases by 20% due to a temperature change, what is the new partial pressure?

Initial: \( 5 = K_H \cdot 0.02 \), \( K_H = 250 \, \text{bar} \).

New \( K_H = 250 \times 0.8 = 200 \, \text{bar} \).

New \( p = 200 \times 0.02 = 4 \, \text{bar} \).

4.5 bar
5 bar
3.5 bar
4 bar
4

A solution of two volatile liquids has vapor pressures of 250 mm Hg and 350 mm Hg for pure components. If the mole fraction of the first is 0.6, what is the total vapor pressure?

Mole fraction of second = \( 1 - 0.6 = 0.4 \).

Total vapor pressure = \( 250 \times 0.6 + 350 \times 0.4 = 150 + 140 = 290 \, \text{mm Hg} \).

280 mm Hg
290 mm Hg
300 mm Hg
310 mm Hg
2

The osmotic pressure of a 0.01 M solution at 27°C is: (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

\( \Pi = MRT \).

\( \Pi = 0.01 \times 0.0821 \times 300 \approx 0.246 \, \text{atm} \).

0.2 atm
0.3 atm
0.246 atm
0.5 atm
3

What is the mole fraction of a solute if the vapor pressure of a solution is 27 mm Hg and that of pure solvent is 30 mm Hg?

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

\( \frac{30 - 27}{30} = \frac{3}{30} = 0.1 \).

0.08
0.09
0.11
0.1
4

A solution of a non-volatile solute reduces the vapor pressure of a solvent from 30 mm Hg to 27 mm Hg. What is the mole fraction of the solvent?

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

\( \frac{30 - 27}{30} = 0.1 \).

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

0.8
0.85
0.9
0.95
3

A solution of a non-volatile solute in water has a vapor pressure of 23.5 mm Hg at a temperature where pure water’s vapor pressure is 25 mm Hg. If the solution’s boiling point is 100.26°C at 1 atm, what is the molality? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))

\( \Delta T_b = 100.26 - 100 = 0.26 \, \text{K} \).

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

\( 0.26 = 0.52 \cdot m \), \( m = 0.5 \, \text{mol/kg} \).

Cross-check: \( x_{\text{solute}} = \frac{25 - 23.5}{25} = 0.06 \), but vapor pressure not needed here.

0.4 mol/kg
0.45 mol/kg
0.55 mol/kg
0.5 mol/kg
4

A 0.15 molal solution of CaCl₂ in water has a freezing point depression of 0.837°C. What is the van’t Hoff factor? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))

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

\( 0.837 = i \times 1.86 \times 0.15 \).

\( i = \frac{0.837}{1.86 \times 0.15} \approx 3 \).

2.5
3
2.8
3.2
2

The van’t Hoff factor of a 0.06 m K₂SO₄ solution is 2.5. What is the boiling point elevation? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))

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

\( \Delta T_b = 2.5 \times 0.52 \times 0.06 = 0.078 \, \text{K} \).

0.06 K
0.07 K
0.08 K
0.078 K
4

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0