Question

A solution contains 0.25 moles of non-volatile solute dissolved in one mole of solvent. Then calculate the % vapor pressure of solution relative to the vapor pressure of pure solvent.

• A. 20
• B. 40
• C. 60
• D. 80

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the vapor pressure of the solution (\(P_s\)) as a percentage of the vapor pressure of the pure solvent (\(P^\circ\)).
Step 2: Key Formula or Approach:
According to Raoult's Law for a non-volatile solute:
\[ P_s = P^\circ \cdot X_{\text{solvent}} \]
The mole fraction of solvent \(X_{\text{solvent}} = \frac{n_{\text{solvent}}}{n_{\text{solvent}} + n_{\text{solute}}}\).
Step 3: Detailed Explanation:
Given:
\(n_{\text{solute}} = 0.25\) mol
\(n_{\text{solvent}} = 1\) mol
Calculate mole fraction of solvent (\(X_{\text{solvent}}\)):
\[ X_{\text{solvent}} = \frac{1}{1 + 0.25} = \frac{1}{1.25} = 0.80 \]
Ratio of vapor pressures:
\[ \frac{P_s}{P^\circ} = X_{\text{solvent}} = 0.80 \]
Expressing as a percentage:
\[ % = \frac{P_s}{P^\circ} \times 100 = 0.80 \times 100 = 80%. \]
Step 4: Final Answer:
The vapor pressure of the solution relative to pure solvent is 80%.