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: Understand the question.
We dissolve 2 mol of a non-volatile solute in 20 mol of water. Pure water's vapour pressure is $P_1^0 = 32$ mmHg. We want the solution's vapour pressure.
Step 2: Recall Raoult's law.
For a non-volatile solute, only the water gives vapour. The solution's vapour pressure equals the mole fraction of water times pure water's vapour pressure. \[ P = X_{water} \times P_1^0 \]
Step 3: Find the total moles.
\[ \text{Total} = 20 + 2 = 22 \text{ mol} \]
Step 4: Find the mole fraction of water.
\[ X_{water} = \frac{20}{22} = \frac{10}{11} \]
Step 5: Put values into Raoult's law.
\[ P = \frac{10}{11} \times 32 = \frac{320}{11} \approx 29.1 \text{ mmHg} \]
Step 6: Choose the answer.
The vapour pressure is about 29.1 mmHg, which is option 1. \[ \boxed{29.1\ \text{mmHg}} \]