Solutions

Concept:Use Raoult’s law for the solution and the ideal gas law to find molar volumes.Explanation:First, find the number of moles of solvent $\mathrm{A}$ in 1 kg:$n_A = \frac{1000 \text{ g}}{50 \text{ g mol}^{-1}} = 20 \text{ mol}$.The solution is 5 molal, so $n_B = 5$ mol.Mole fractions:$X_A = \frac{20}{20+5} = 0.8$, $X_B = \frac{5}{25} = 0.2$.Apply Raoult’s law: $P_T = X_A P_A^\circ + X_B P_B^\circ$.Given $P_T = 100$ mm Hg, $P_A^\circ = 105$ mm Hg.$100 = 0.8 \times 105 + 0.2 \times P_B^\circ$$100 = 84 + 0.2 P_B^\circ$$P_B^\circ = \frac{16}{0.2} = 80$ mm Hg.Molar volume of pure $\mathrm{B}$ in vapour phase (ideal gas):$V_{\text{vap}} = \frac{RT}{P}$.Convert pressure to atm: $P = \frac{80}{760}$ atm.$V_{\text{vap}} = \frac{0.08 \times 300}{80/760} = 24 \times \frac{760}{80} = 228$ L mol$^{-1}$.Molar volume of pure $\mathrm{B}$ in liquid phase:$V_{\text{liq}} = \frac{\text{Molar mass}}{\text{Density}} = \frac{57 \text{ g mol}^{-1}}{0.5 \text{ g mL}^{-1}} = 114$ mL mol$^{-1} = 0.114$ L mol$^{-1}$.Required ratio:$\frac{V_{\text{vap}}}{V_{\text{liq}}} = \frac{228 \times 1000}{114} = 2000$.Answer:2000