Question

At T(K), the vapour pressure of an aqueous solution of a non-volatile solute,
whose mole fraction is 0.02 is found to be 34.65 mm Hg.
What is the vapour pressure (in mm Hg) of pure water at the same temperature?

• A. 35.70
• B. 35.36
• C. 35.00
• D. 34.30

Hint:

In dilute solutions with non-volatile solutes, use Raoult’s law: \(P_{solution}} = \chi_{solvent}} \cdot P^0\), and remember \(\chi_{solvent}} = 1 - \chi_{solute}}\).

Answer 1

The Correct Option is B

Step 1: Use Raoult’s Law. Raoult's Law for a solution of non-volatile solute: \[ P_{\text{solution}} = \chi_{\text{solvent}} \cdot P^0 \] Where: 
- \(P_{\text{solution}} = 34.65 \, \text{mm Hg}\) 
- \(\chi_{\text{solute}} = 0.02 \Rightarrow \chi_{\text{solvent}} = 1 - 0.02 = 0.98\) 
- \(P^0 =\) vapour pressure of pure water 
Step 2: Solve for \(P^0\). \[ 34.65 = 0.98 \cdot P^0 \Rightarrow P^0 = \frac{34.65}{0.98} \approx 35.36 \, \text{mm Hg} \]