A solution contains 50 g of a non-electrolytic solute per 100g of water at 25°C. The vapor pressure of water is 23.5 mmHg at this temperature. The molar mass of the solute is $250\, g\, mol^{-1}$. The vapor pressure of the solution at 25°C will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 22.67 mmHg
Given data:
- Mass of non-electrolyte solute = 50 g
- Mass of water (solvent) = 100 g
- Vapour pressure of pure water at 25°C = 23.5 mmHg
- Molar mass of solute = $250\, g\, mol^{-1}$
- Temperature = 25°C
Step 1: Write Raoult's law
For a non-volatile solute, vapour pressure of solution is:
$P_{\text{solution}}=X_{\text{solvent}}×X_{\text{solvent}}^ο$
So we need mole fraction of water.
Step 2: Calculate moles of each component
Moles of water $=\frac{100}{18}=5.56\text{mol}$
Moles of solute $=\frac{50}{100}=0.20\text{mol}$
Step 3: Calculate mole fraction of water
$X_{\text{water}}=\frac{\text{moles of water}}{\text{total moles}}$
$X_{\text{water}}=\frac{5.56}{5.56+0.20}=\frac{5.56}{5.76}=0.965$
Step 4: Calculate vapour pressure of solution
$P_{\text{solution}}=X_{\text{water}}×X_{\text{water}}^o$
$P_{\text{solution}}=0.965 × 23.5 = 22.67 \text{mmHg}$
Final Answer: Vapour pressure of the solution = 22.67 mmHg