Solutions Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A 0.1 M solution of K₃PO₄ (assuming complete dissociation) has an osmotic pressure of 0.984 atm at a certain temperature. What is the temperature? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

For K₃PO₄, \( i = 4 \) (3K⁺ + PO₄³⁻).

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

\( 0.984 = 4 \times 0.1 \times 0.0821 \times T \).

\( T = \frac{0.984}{0.4 \times 0.0821} \approx 30 \, \text{K} \).

25 K
27 K
30 K
32 K
3

A solute dissociates into 3 ions with a van’t Hoff factor of 2.4. What is the degree of dissociation?

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

\( 2.4 = 1 + \alpha (3 - 1) \).

\( 2.4 = 1 + 2\alpha \), \( \alpha = \frac{1.4}{2} = 0.7 \).

0.5
0.6
0.8
0.7
4

A 0.5 M solution of K₂SO₄ (molar mass = 174 g/mol) has a density of 1.15 g/mL. What is the mass of the solute in 300 mL of this solution?

Moles = \( 0.5 \times 0.3 = 0.15 \, \text{mol} \).

Mass of K₂SO₄ = \( 0.15 \times 174 = 26.1 \, \text{g} \).

(Density is extra info, not needed for molarity-based calculation.)

25 g
26.1 g
27 g
24.5 g
2

The solubility of a gas in a solvent follows Henry’s law with a constant of 180 bar. If the partial pressure is increased by 50% from 6 bar, what is the new mole fraction of the gas?

Initial \( p = 6 \, \text{bar} \), new \( p = 6 \times 1.5 = 9 \, \text{bar} \).

\( p = K_H \cdot x \).

New \( x = \frac{9}{180} = 0.05 \).

0.05
0.033
0.075
0.06
1

What is the molar mass of a solute if 3 g of it in 200 mL of solution produces an osmotic pressure of 0.738 atm at 27°C? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

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

\( 0.738 = \frac{3}{M \times 0.2} \times 0.0821 \times 300 \).

\( M = \frac{3 \times 0.0821 \times 300}{0.738 \times 0.2} \approx 50 \, \text{g/mol} \).

45 g/mol
48 g/mol
50 g/mol
55 g/mol
3

A solution of 18 g of glucose (molar mass = 180 g/mol) in 1 kg of water freezes at -0.186°C. What is the \( K_f \) of water?

Molality = \( \frac{18 / 180}{1} = 0.1 \, \text{mol/kg} \).

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

\( 0.186 = K_f \times 0.1 \).

\( K_f = \frac{0.186}{0.1} = 1.86 \, \text{K kg mol}^{-1} \).

1.0 K kg mol\(^{-1}\)
1.5 K kg mol\(^{-1}\)
1.8 K kg mol\(^{-1}\)
1.86 K kg mol\(^{-1}\)
4

A 0.3 M solution of Na₂SO₄ (assuming complete dissociation) has an osmotic pressure of 2.214 atm at a certain temperature. What is the temperature? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

For Na₂SO₄, \( i = 3 \) (2Na⁺ + SO₄²⁻).

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

\( 2.214 = 3 \times 0.3 \times 0.0821 \times T \).

\( T = \frac{2.214}{0.9 \times 0.0821} \approx 30 \, \text{K} \).

25 K
28 K
30 K
32 K
3

A solution reduces the vapor pressure of a solvent from 28 mm Hg to 26 mm Hg. What is the mole fraction of the solute?

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

\( \frac{28 - 26}{28} = \frac{2}{28} \approx 0.0714 \).

0.06
0.07
0.0714
0.08
3

The vapor pressure of pure water is 30 mm Hg at a certain temperature. A solution with a non-volatile solute has a vapor pressure of 28.5 mm Hg. If the solute’s molality is 0.5 mol/kg, what is the mass of water in the solution?

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

\( \frac{30 - 28.5}{30} = 0.05 = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}} \).

Molality = \( \frac{n_{\text{solute}}}{w_{\text{water}}} = 0.5 \).

Assume \( w_{\text{water}} = 1 \, \text{kg} \), then \( n_{\text{solute}} = 0.5 \).

\( n_{\text{water}} = \frac{1000}{18} \approx 55.56 \).

\( x_{\text{solute}} = \frac{0.5}{0.5 + 55.56} \approx 0.0089 \), adjust \( w_{\text{water}} = \frac{0.5 \times 18}{0.05} = 180 \, \text{g} \).

150 g
200 g
180 g
160 g
3

A solution lowers the vapor pressure of a solvent from 40 mm Hg to 38 mm Hg. What is the mole fraction of the solute?

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

\( \frac{40 - 38}{40} = \frac{2}{40} = 0.05 \).

0.02
0.04
0.05
0.06
3

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