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If x₁ and x₂ represent the mole fraction of a component A in the vapour phase and liquid mixture respectively and p_A⁰ and p_B⁰ represent vapours pressures of pure A and pure B, then total vapour pressure of the liquid mixture is

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$\frac{p_{B}^{0} x_{2}}{x_{1}}$

$\frac{p_{A}^{0} x_{2}}{x_{1}}$

$\frac{p_{A}^{0} x_{1}}{x_{2}}$

$\frac{p_{B}^{0} x_{1}}{x_{2}}$

answer is B.

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Detailed Solution

Mole fraction of component A in vapour phase is $x_{1}$

that means, $P_{A} = x_{1} P_{total} . . . (1)$

and mole fraction of compound A in liquid mixture is x₂

that means we have, $P_{A} = P_{A}^{o} x_{2} . . . (2)$

From (1) and (2), $P_{total} = \frac{P_{A}^{o} x_{2}}{x_{1}}$

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