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Liquids in Liquids

PA = XAPA a...

Question

$P_{A} = X_{A} P_{A}$ and $P_{B} = X_{B} P_{B}$
$P_{T} = X_{A} P_{A} + X_{B} P_{B}$
Vapour pressure of mixtures of Benzene $(C_{6} H_{6})$ and toluene $(C_{7} H_{8})$ at $50^{\circ} C$ are given by $P_{M} = 179 X_{B} + 92$ where $X_{B}$ is mole fraction of $C_{6} H_{6}$ .
What is the vapour pressure of pure liquids?

A

P_{B} = 92 m m, P_{T} = 179 m m

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B

P_{B} = 271 m m, P_{T} = 92 m m

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C

P_{B} = 180 m m, P_{T} = 91 m m

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D

None of these

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Solution

The correct option is B $P_{B} = 271 m m, P_{T} = 92 m m$
as $x_{A} + x_{B} = 1$
$P T = x_{A} P_{A} + x_{B} P_{B} = (1 - x B) P_{A} + x_{B} P_{B}$
$P_{T} = P_{A} + x_{B} (P_{B} - P_{A}) ⟶ (1)$
Given, $P_{M} = 92 + 179 x_{B} ⟶ (2)$
on comparing equation $(1)$ & $(2)$ we have
$P_{A} = 92 m m, P_{B} - P_{A} = 179 m m$
$P_{B} = 179 + P_{A} = 179 + 92$
$P_{A} = 92 m m, P_{B} = 271 m m$

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