Solutions Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A solution contains 6 g of a solute (molar mass = 120 g/mol) in 300 mL of solution. What is the molarity?

Moles of solute = \( \frac{6}{120} = 0.05 \, \text{mol} \).

Volume = 300 mL = 0.3 L.

Molarity = \( \frac{0.05}{0.3} \approx 0.1667 \, \text{M} \).

0.15 M
0.1667 M
0.2 M
0.1 M
2

What is the volume of water required to prepare 200 mL of a 0.5 M solution using 4 g of NaOH (molar mass = 40 g/mol)?

Moles of NaOH = \( \frac{4}{40} = 0.1 \, \text{mol} \).

Molarity = \( \frac{\text{Moles}}{\text{Volume in L}} \), so \( 0.5 = \frac{0.1}{V} \).

Volume = \( \frac{0.1}{0.5} = 0.2 \, \text{L} = 200 \, \text{mL} \).

Since total volume is 200 mL, water volume = 200 mL (assuming solute volume is negligible).

200 mL
100 mL
250 mL
150 mL
1

What is the osmotic pressure of a 0.015 M solution of a non-electrolyte at 27°C? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

\( \Pi = MRT \).

\( \Pi = 0.015 \times 0.0821 \times 300 \approx 0.3695 \, \text{atm} \).

0.3 atm
0.35 atm
0.3695 atm
0.4 atm
3

What is the mole fraction of a solute if the vapor pressure of a solution is 22 mm Hg and that of pure solvent is 25 mm Hg?

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

\( \frac{25 - 22}{25} = \frac{3}{25} = 0.12 \).

0.1
0.11
0.13
0.12
4

A solution contains 30 g of a solute in 120 g of water. If the mass percentage of the solute is increased to 25% by evaporating water, what mass of water remains?

Initial mass % = \( \frac{30}{30 + 120} \times 100 = 20\% \).

New mass % = \( 25\% = \frac{30}{30 + w} \times 100 \).

\( 0.25 (30 + w) = 30 \), \( 30 + w = 120 \), \( w = 90 \, \text{g} \).

90 g
100 g
80 g
110 g
1

A gas has a Henry’s law constant of 400 bar. If its partial pressure is 8 bar, what is the mole fraction in the solution?

\( p = K_H \cdot x \).

\( x = \frac{8}{400} = 0.02 \).

0.01
0.015
0.025
0.02
4

What is the mass of water required to prepare 300 g of a 5% by mass solution of a solute?

Mass of solute = \( \frac{5}{100} \times 300 = 15 \, \text{g} \).

Mass of water = \( 300 - 15 = 285 \, \text{g} \).

280 g
290 g
285 g
295 g
3

A 0.2 M solution of Na₂CO₃ (assuming complete dissociation) has an osmotic pressure of 1.968 atm at a certain temperature. What is the temperature? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

For Na₂CO₃, \( i = 3 \) (2Na⁺ + CO₃²⁻).

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

\( 1.968 = 3 \times 0.2 \times 0.0821 \times T \).

\( T = \frac{1.968}{0.6 \times 0.0821} \approx 40 \, \text{K} \).

35 K
38 K
40 K
42 K
3

What is the mass of urea (molar mass = 60 g/mol) required to prepare 2 kg of a 0.5 molal aqueous solution?

Molality = \( \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \).

Moles = \( 0.5 \times 2 = 1 \, \text{mol} \).

Mass = \( 1 \times 60 = 60 \, \text{g} \).

30 g
50 g
55 g
60 g
4

A gas has a Henry’s law constant of 120 bar. If its mole fraction in a solution is 0.025, what is its partial pressure?

Henry’s law: \( p = K_H \cdot x \).

\( p = 120 \times 0.025 = 3 \, \text{bar} \).

3 bar
2.5 bar
3.5 bar
4 bar
1

Performance Summary

Score:

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