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Azeotropic Mixtures

Volatile liqu...

Question

Volatile liquids A and B forming an azeotrope (at a particular temperature range) have vapour pressures of $180 m m$ and $100 m m$ respectively. If the mole fraction of A in the solution mixture is 0.5, and the mixture has a total vapour pressure of $200 m m$ .

Choose the correct option(s) for the azeotropic mixture.

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Solution

Given,
$Pure vapour pressure of A , p_{A}^{\circ} = 180 m m Pure vapour pressure of B , p_{B}^{\circ} = 100 m m$

$Mole fraction of A , XA=0.5∴Mole fraction of B, XB=1−0.5=0.5$

By Raoult's law
Total vapour pressure of solution,
$P_{T} = X_{A} . p_{A}^{\circ} + X_{B} . p_{B}^{\circ}$
$P_{T} = 180 \times 0.5 + 100 \times 0.5$
$= 140 m m .$

$P_{T} > X_{A} . p_{A}^{\circ} + X_{B} . p_{B}^{\circ}$
$200 m m > 140 m m$
Thus, the given mixture shows positive deviation from the Raoult's law and forms minimum boiling azeotrope

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Azeotropic Mixtures

Standard XII Chemistry

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