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Liquids in Liquids

Vapour pressu...

Question

Vapour pressure of liquids $A$ and $B$ at $298 K$ is $300 m m H g$ and $450 m m H g$ . Calculate the mole fraction of $A$ in the mixture.

[Given total vapour pressure of solution $= 405 m m H g$ .]

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Solution

It is given that,
$P_{A}^{0} = 300$ mm of $H g$
$P_{B}^{0} = 450$ mm of $H g$
$P_{T o t a l} = 405$ mm of $H g$
From Raoults law, we have
$P_{t o t a l} = P_{A} + P_{B}$
$P_{t o t a l} = P_{A}^{0} X_{A} + P_{B}^{0} (1 - X_{A})$
$P_{t o t a l} = P_{A}^{0} X_{A} + P_{B}^{0} - P_{B}^{0} X_{A}$
$P_{t o t a l} = (P_{A}^{0} - P_{B}^{0}) X_{A} + P_{B}^{0}$
$405 = (300 - 450) X_{A} + 450$
$- 45 = - 150 X_{A}$
$X_{A} = 0.3$
Therefore,
$X_{B} = 1 - X_{A}$
$= 1 - 0.3$
$= 0.7$
$P_{A} = P_{A}^{0} X_{A}$
$= 300 \times 0.3$
$= 90 m m o f H g$
$P_{B} = P_{B}^{0} \times X_{B}$
$= 450 \times 0.7$
$= 315 m m o f H g$
Now in the vapour phase:
Mole fraction of liquid A,
$A = \frac{P_{A}}{(P_{A} + P_{B})} = \frac{90}{90 + 315} = 0.22$
Thus mole fraction of liquid A is 0.22

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