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

0.4

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0.8

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0.625

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None of these

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Solution

The correct option is D $0.625$
Mole fraction of A is $X_{A} = 0.4$

The mole fraction of B is $X_{B} = 1 - 0.4 = 0.6$

The expression for the total pressure of the solution is

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

Substitute values in the above expression

$P = 100 \times 0.4 + 200 \times 0.6 = 160 m m H g$

But $P_{A}^{o} = P Y_{A}$

$100 = 160 Y_{A}$

$Y_{A} = 0.625$

Thus the mole fraction of A in vapours is 0.625

Hence, the correct option is $C$

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Similar questions

A

A

B

X_{A} = 0.4

P_{A}^{\circ} = 100 m m, P_{B}^{\circ} = 200 m m

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

(X_{A} = 0.4)

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

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

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

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