Solutions Part 2

Concept:The solution uses two colligative properties: boiling point elevation and relative lowering of vapour pressure. The van't Hoff factor $i$ is used to account for the electrolyte nature of the solute.Explanation:Boiling point of water is $373\,\text{K}$ and solution boils at $375\,\text{K}$.Elevation $\Delta T_b = 375 - 373 = 2\,\text{K}$.Mass of water: $100\,\text{mL} \times 1\,\text{g/mL} = 100\,\text{g} = 0.1\,\text{kg}$.Moles of solute = $\frac{W}{M}$, so molality $m = \frac{W/M}{0.1} = 10\frac{W}{M}$.For an electrolyte, $\Delta T_b = i K_b m$.$2 = i (0.52) \left(10\frac{W}{M}\right) = i\left(5.2\frac{W}{M}\right)$.Thus $i = \frac{2}{5.2}\cdot\frac{M}{W} = \frac{5}{13}\cdot\frac{M}{W}$.Relative lowering of vapour pressure: $\frac{\Delta p}{p^\circ} = \frac{640-600}{640} = \frac{40}{640} = \frac{1}{16}$.For electrolyte: $\frac{\Delta p}{p^\circ} = i x_2$.So $x_2 = \frac{1/16}{i} = \frac{1}{16} \cdot \frac{1}{\frac{5}{13}\cdot\frac{M}{W}} = \frac{1}{16} \cdot \frac{13}{5} \cdot \frac{W}{M} = \frac{13}{80} \cdot \frac{W}{M}$.Simplifying $\frac{13}{80} = \frac{1.3}{8}$, hence $x_2 = \frac{1.3}{8} \cdot \frac{W}{M}$.Answer:Option D: $\frac{1.3}{8} \times \frac{W}{M}$