Water and chlorobenzene are immiscible liquids. Their mixture boils at under a reduced pressure of . — Solutions and Colligative Properties Chemistry Question
Question
Water and chlorobenzene are immiscible liquids. Their mixture boils at $90^\circ\text{C}$ under a reduced pressure of $9.031 \times 10^4 \text{ Pa}$. The vapour pressure of pure water at $90^\circ\text{C}$ is $7.031 \times 10^4 \text{ Pa}$ and the molecular mass of chlorobenzene is 112.5. On mass per cent basis, chlorobenzene in the distillate is equal to
💡 Solution & Explanation
For immiscible liquids boiling together, $P_{total} = P_{water}^\circ + P_{chlorobenzene}^\circ$. $P_{chlorobenzene}^\circ = 9.031 \times 10^4 - 7.031 \times 10^4 = 2.000 \times 10^4 \text{ Pa}$. Mass ratio in distillate $W_C / W_W = (P_C^\circ M_C) / (P_W^\circ M_W) = (2.000 \times 10^4 \times 112.5) / (7.031 \times 10^4 \times 18) \approx 225 / 126.558 = 1.777$. Mass % of chlorobenzene $= [1.777 / (1.777 + 1)] \times 100 \approx 64\%$. Therefore, correct answer is C.