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?
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?
Show Hint
- 35.70
- 35.36
- 35.00
- 34.30
The Correct Option is B
Solution and Explanation
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 \(P^0\). \[ 34.65 = 0.98 \cdot P^0 \Rightarrow P^0 = \frac{34.65}{0.98} \approx 35.36 \, \text{mm Hg} \]
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