SolutionMedium
Question
The vapour pressure of pure benzene at 88°C is 960 mm and that of toluene at the same temperature is 380 mm of benzene. At what mole fraction of benzene, the mixture will boil at 88°C?
Options
A.0.655
B.0.345
C.0.05
D.0.25
Solution
$P_{\text{total}} = X_{\text{Benzene}}.P_{\text{Benzene}}^{o} + X_{\text{Toulene}}.P_{\text{Toulene}}^{o}$
Or, $760 = x \times 960 + (1 - x) \times 380 \Rightarrow x = 0.655$
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