The vapour pressure of pure liquids $A$ and $B$ are $450$ and $700$ mm Hg respectively. at $350$ K. Find out the composition of the liquid mixture if total vapour pressure is $600$ mm Hg. Also, find the composition of the vapour phase.

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Solution

Given: Vapour pressure of pure liquid A, $P o A = 450 m m o f H g$
Vapour pressure of pure liquid A, $P o A = 700 m m o f H g$
Total vapour pressure,a $P t o t a l = 600 m m o f H g$
Use the formula of Raoult’s law$
$600 = (450 - 700) X A + 700$
$250 X A = 100$
$X A =$ $\frac{100}{250}$ $= 0.4$
Use formula
$X B = 1 - X A$
Substitute the values
we get, $X B = 1 - 0.4 = 0.6$
use formula $P A = P o A \times X A = 450 \times 0.4 = 180 m m o f H g$
$P B = P o B \times X B = 700 \times 0.6 = 420 m m o f H g$
Now, in the vapour phase:
Mole fraction of liquid $A = \frac{180}{180 + 420} = 0.30$
Mole fraction of liquid $B, Y B = 1 - Y A = 1 - 0.30 = 0.70$

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