Question

What is vapour pressure of a solution when 2 mol of a non-volatile solute are dissolved in 20 mol of water? \( P_1^0 = 32 \, \text{mmHg} \)

• A. 29.1 mmHg
• B. 12 mmHg
• C. 6 mmHg
• D. 9 mmHg

Hint:

Remember: Vapour pressure lowering is proportional to the mole fraction of solute. For dilute solutions, you can approximate, but here exact calculation is straightforward.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have a non-volatile solute dissolved in water. The vapour pressure of pure water at the given temperature is 32 mmHg. We need the vapour pressure of the solution.

Step 2: Key Formula or Approach:
Raoult’s law for a non-volatile solute: \( P_{\text{solution}} = X_{\text{solvent}} \times P_{\text{solvent}}^0 \), where \( X_{\text{solvent}} \) is the mole fraction of water.

Step 3: Detailed Explanation:
Moles of water = 20 mol, moles of solute = 2 mol. Total moles = 22 mol. Mole fraction of water = \( \frac{20}{22} = \frac{10}{11} \). \[ P_{\text{solution}} = \frac{10}{11} \times 32 \, \text{mmHg} = \frac{320}{11} \, \text{mmHg} \approx 29.09 \, \text{mmHg} \approx 29.1 \, \text{mmHg}. \]

Step 4: Final Answer:
Option (A) is correct.