The vapour pressure of water at 23°C is 19.8 mm Hg. 0.1 mol of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm Hg) of the resulting solution?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 19.602
Glucose is a non-volatile solute, so only water contributes to the vapour pressure. By Raoult’s law, the vapour pressure of the solution equals the mole fraction of the solvent times the vapour pressure of the pure solvent.
Moles of water: \( \displaystyle n_1=\frac{178.2\ \mathrm{g}}{18\ \mathrm{g\,mol^{-1}}}=9.9\ \mathrm{mol} \)
Mole fraction of water: \( \displaystyle x_1=\frac{n_1}{n_1+n_2}=\frac{9.9}{9.9+0.1}=0.99 \)
Vapour pressure of the solution:
\( \displaystyle p_1=x_1\,p_1^{\circ} \)
\( \displaystyle p_1=0.99\times19.8\ \mathrm{mm\,Hg} \)
\( \displaystyle p_1=19.602\ \mathrm{mm\,Hg} \)