SolutionHard
Question
Water and chlorobenzene are immiscible liquids. Their mixture boils at 90°C under a reduced pressure of 9.031 × 104 Pa. The vapour pressure of pure water at 90°C is 7.031 × 104 Pa and the molecular mass of chlorobenzene is 112.5. On mass percent basis, chlorobenzene in the distillate is equal to
Options
A.50
B.22.2
C.64
D.36
Solution
$\frac{n_{\text{Chlorobenzene}}}{n_{\text{water}}} = \frac{P_{\text{Chlorobenzene}}^{o}}{P_{\text{Water}}^{o}}$
Or, $\frac{x/112.5}{(100 - x)/18} = \frac{9.031 \times 10^{4} - 7.031 \times 10^{4}}{7.031 \times 10^{4}} \Rightarrow x = 64$
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