A solution of a non-volatile solute in water freezes at −0.30°C. The vapour pressure of pure water is 23.51 mm Hg and Kf for water is 1.86 K kg mol−1. What is the vapour pressure of the solution at 298 K?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 23.442 mm Hg
The freezing point gives the molality of the solution; the molality gives the mole fraction of the solute, and Raoult’s law then gives the vapour pressure. The amount of solute in the water does not change with temperature, so the molality found at the freezing point also holds at 298 K.
Depression of freezing point: pure water freezes at 0°C, so ΔTf = 0 − (−0.30) = 0.30°C = 0.30 K.
Molality:
\( \displaystyle \Delta T_f=K_f\,m \)
\( \displaystyle m=\frac{0.30\,\mathrm{K}}{1.86\,\mathrm{K\,kg\,mol^{-1}}}=0.161\,\mathrm{mol\,kg^{-1}} \)
Moles in 1 kg of water: n2 = 0.161 mol of solute; n1 = 1000/18 = 55.56 mol of water.
Mole fraction of the solute:
\( \displaystyle x_2=\frac{n_2}{n_1+n_2}=\frac{0.161}{55.56+0.161}=0.00290 \)
Vapour pressure of the solution (non-volatile solute, so only water contributes):
\( \displaystyle p_1=x_1\,p_1^0=(1-x_2)\,p_1^0 \)
\( \displaystyle p_1=(1-0.00290)\times23.51\,\mathrm{mm\,Hg} \)
\( \displaystyle p_1=23.442\,\mathrm{mm\,Hg} \)
The vapour pressure of the solution is 23.442 mm Hg, a little below that of pure water, as a non-volatile solute always lowers it.