The vapour pressure of pure water at 25°C is 23.75 mmHg. The vapor pressure of the solution obtained by dissolving 34.2 g of sucrose ($C_{12}H_{22}O_{11}$) in 100 g of water will be:

23.33 mmHg

The correct answer is Option (1) → 23.33 mmHg

For a non-volatile solute like sucrose, vapour pressure lowering follows Raoult's law:

$P_{\text{solution}} = X_{\text{solvent}} \times P^0_{\text{solvent}}$

• $P_{\text{solution}} = \text{partial pressure of the solution}$
• $X_{\text{solvent}} = \text{mole fraction of the solvent}$
• $P^0_{\text{solvent}} = \text{vapour pressure of the pure solvent}$

Sucrose does not dissociate or evaporate, so only the mole fraction of water affects vapour pressure.

Moles of sucrose

Molar mass of sucrose = $342\text{ g/mol}$

$n_{\text{sucrose}} = \frac{34.2}{342} = 0.1\text{ mol}$

Moles of water

Molar mass of water = $18\text{ g/mol}$

$n_{\text{water}} = \frac{100}{18} = 5.56\text{ mol}$

Mole fraction of water

$X_{\text{water}} = \frac{5.56}{5.56 + 0.1} = \frac{5.56}{5.66} = 0.982$

Vapour pressure of solution

$P_{\text{solution}} = 0.982 \times 23.75 = 23.33\text{ mmHg (approx.)}$