Question 15

Determine mole fractions in solution: Molality ($m$) = 5. Let mass of solvent A be $1000 \text{ g}$. Moles of B ($n_B$) = 5. Moles of A ($n_A$) = $\frac{1000 \text{ g}}{50 \text{ g/mol}} = 20$. Mole fraction of A ($x_A$) = $\frac{20}{20 + 5} = 0.8$. Mole fraction of B ($x_B$) = $\frac{5}{20 + 5} = 0.2$.

Find vapour pressure of pure B ($P_B^\circ$): Using Raoult's law: $P_T = P_A^\circ x_A + P_B^\circ x_B$ $100 = 105(0.8) + P_B^\circ(0.2)$ $100 = 84 + 0.2 P_B^\circ$ $16 = 0.2 P_B^\circ \Rightarrow P_B^\circ = 80 \text{ mm Hg}$.

Calculate molar volume of pure B in vapour phase ($V_{v,B}$): Using ideal gas law: $V_{v,B} = \frac{RT}{P_B^\circ}$ $P_B^\circ = \frac{80}{760} \text{ atm}$ $V_{v,B} = \frac{0.08 \times 300}{80/760} = \frac{24 \times 760}{80} = 228 \text{ L/mol}$.

Calculate molar volume of pure B in liquid phase ($V_{l,B}$): $V_{l,B} = \frac{\text{Molar mass}}{\text{Density}} = \frac{57 \text{ g/mol}}{0.5 \text{ g/mL}} = 114 \text{ mL/mol} = 0.114 \text{ L/mol}$.

Calculate the ratio: Ratio = $\frac{V_{v,B}}{V_{l,B}} = \frac{228}{0.114} = 2000$.