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:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 23.33 mmHg
Adding a non-volatile solute lowers the vapour pressure of the solvent. By Raoult’s law, the vapour pressure of the solution equals the vapour pressure of the pure solvent times the mole fraction of the solvent:
\( \displaystyle p_1 = x_1\,p_1^\circ \)
Moles of sucrose (\( M = 342\ \mathrm{g\,mol^{-1}} \)) and of water (\( M = 18\ \mathrm{g\,mol^{-1}} \)):
\( \displaystyle n_2 = \frac{34.2}{342} = 0.1\ \mathrm{mol} \)
\( \displaystyle n_1 = \frac{100}{18} = 5.556\ \mathrm{mol} \)
\( \displaystyle x_1 = \frac{5.556}{5.556 + 0.1} = \frac{5.556}{5.656} = 0.9823 \)
\( \displaystyle p_1 = 0.9823 \times 23.75\ \mathrm{mmHg} = 23.33\ \mathrm{mmHg} \)
The vapour pressure of the solution is 23.33 mmHg.
Shortcut: A non-volatile solute can only lower the vapour pressure, which rules out 24 mmHg and 23.75 mmHg at once. The lowering is \( 23.75 \times 0.1/5.656 \approx 0.42 \) mmHg, which is far smaller than the 1.25 mmHg drop needed for 22.5 mmHg.