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

22.67 mmHg

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