SolutionHard
Question
The immiscible liquid system containing aniline water boils at 98°C under a pressure of 760 mm. At this temperature, the vapour pressure of water is 700 mm. If aniline is distilled in steam at 98°C, then what percent of total weight of the distillate will be aniline?
Options
A.7.89
B.8.57
C.30.7
D.44.3
Solution
$\frac{n_{\text{aniline}}}{n_{\text{water}}} = \frac{P_{\text{aniline}}^{o}}{P_{\text{water}}^{o}} \Rightarrow \frac{m_{a}/93}{m_{w}/18} = \frac{760 - 700}{700} \Rightarrow \frac{m_{a}}{m_{w}} = 0.44$
∴ Mass percent of aniline in distillate
$= \frac{m_{a}}{m_{a} + m_{w}} \times 100 = 30.7\%$
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