Solutions Part 2

Concept:Boiling point elevation $\Delta T_b = K_b \cdot m$, where $m$ is molality (moles of solute per kg of solvent).Explanation:Given: For both compounds, mass of solute = 1 g, mass of solvent = 50 g = 0.05 kg, $K_b = 5 \; \text{K kg mol}^{-1}$.For any solute of molar mass $M$: $m = \frac{1/M}{0.05} = \frac{20}{M}$.Thus $\Delta T_b = K_b \cdot \frac{20}{M} = 5 \cdot \frac{20}{M} = \frac{100}{M}$.Hence $M = \frac{100}{\Delta T_b}$.1) For $PQ$: $\Delta T_b = 1.176 \; \text{K}$, so $M(PQ) = \frac{100}{1.176} \approx 85.03 \approx 85 \; \text{g mol}^{-1}$.Let molar masses of P = $x$ and Q = $y$. Then $x + y = 85 \quad \text{(1)}$.2) For $PQ_2$: $\Delta T_b = 0.689 \; \text{K}$, so $M(PQ_2) = \frac{100}{0.689} \approx 145.14 \approx 145 \; \text{g mol}^{-1}$.Thus $x + 2y = 145 \quad \text{(2)}$.Subtract (1) from (2): $(x + 2y) - (x + y) = 145 - 85 \Rightarrow y = 60$.Then from (1): $x + 60 = 85 \Rightarrow x = 25$.Therefore, molar masses: P = 25 g mol⁻¹, Q = 60 g mol⁻¹.Answer:Option A: 25, 60.