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Standard XII

Chemistry

Raoult's Law

At 25∘C, ...

Question

At $25^{\circ} C$ , $1 m o l$ of $A$ having a vapour pressure of
$100 t o r r$ and $1 m o l$ of $B$ having a vapour pressure of $300 t o r r$ were mixed.
The vapour at equilibrium is removed, condensed and the condensate is heated back to $25^{\circ} C$ .
The vapour now formed are again removed, recondensed and analyzed.If the mole fraction of $A$ in this condensate is $1 \times 10^{- x}$ ,then what is the value of x?

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Solution

$P_{A}^{\circ} = 100, P_{B} = 300, X_{Z} = X_{B} = \frac{1}{2}$
$P_{T} = 200$
$Y_{A} = \frac{100 \frac{1}{2}}{200} = \frac{1}{4}$
On condensation $X_{Z} = \frac{1}{4}, X_{B} = \frac{3}{4}$
$P_{T} = 100 \times \frac{1}{4} + 300 \times \frac{3}{4} = 250$
$Y_{A} = \frac{25}{250} = 0.1$
on further condensation
$X_{A} = 0.1$

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P∘A=100,PB=300,XZ=XB=12PT=200YA=10012200=14On condensation XZ=14,XB=34PT=100×14+300×34=250YA=25250=0.1on further condensationXA=0.1