SolutionHard
Question
A mixture contains 1 mole of volatile liquid A ($P_{A}^{o}$= 100 mm Hg) and 3 moles of volatile liquid B ($P_{B}^{o}$= 80 mm Hg). If the solution behaves ideally, the total vapour pressure of the distillate is
Options
A.85 mm Hg
B.85.88 mm Hg
C.90 mm Hg
D.92 mm Hg
Solution
The mole fraction of A in distillate,
$X_{A}' = Y_{A} = \frac{X_{A}.P_{A}^{o}}{P_{\text{total}}} = \frac{\frac{1}{4} \times 100}{\frac{1}{4} \times 100 + \frac{3}{4} \times 80} = \frac{5}{17}$
Now, V.P. of distillate, $P = X_{A}'.P_{A}^{o} + X_{B}'.P_{B}^{o}$
$= \frac{5}{17} \times 100 + \frac{12}{17} \times 80 = 85.88\text{ mm Kg}$
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