Mole fraction of $A$ vapours above solution in mixture of $A$ and $B$ ( $X_{A} = 0.4$ ) will be:
[ $P_{A}^{\circ} = 100 m m, P_{B}^{\circ} = 200 m m$ ]

0.4

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0.8

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0.25

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None of the above

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Solution

The correct option is B $0.25$
Total pressure above the solution is

$P = P_{A}^{0} X_{A} + P_{B}^{0} X_{B}$

$P = 100 \times 0.4 + 200 (1 - 0.4) = 40 + 120 = 160 m m$

The vapor pressure of A is $P_{A} = 40 m m$

The mole fraction of A in the vapor is $\frac{P_{A}}{P} = \frac{40}{160} = 0.25$

Hence, the correct option is $C$

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Similar questions

(X_{A} = 0.4)

(P_{A}^{O} = 100 m m P_{B}^{O} = 200 m m)

Mole fraction of vapours of A above the solution in a mixture of A and B (X_A=0.4 ) is:

A

B

(X_{A} = 0.4)

P_{A}^{\circ} = 100 m m H g

P_{B}^{\circ} = 200 m m H g

Mole fraction of vapours of A above the solution in a mixture of A and B (X_A=0.4 ) is:

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