Liquids A and B form an ideal solution in the entire composition range. At 350 K the vapour pressures of pure A and pure B are 7×103 pa and 12×103 pa respectively. The composition of vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is

P=xAP0A+xBP0B=0.4×7×103+0.6×12×103=10×103Pa=1×104PaX'A=PAP=XAP0AP=0.4×7×1031×104=0.28X'B=1−X'A=1−0.2=0.72

Liquids A and B form an ideal solution in the entire composition range. At 350 K the vapour pressures of pure A and pure B are 7×103 pa and 12×103 pa respectively. The composition of vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is

P=xAP0A+xBP0B=0.4×7×103+0.6×12×103=10×103Pa=1×104PaX'A=PAP=XAP0AP=0.4×7×1031×104=0.28X'B=1−X'A=1−0.2=0.72

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Liquids A and B form an ideal solution in the entire composition range. At 350 K the vapour pressures of pure A and pure B are 7×10³ pa and 12×10³ pa respectively. The composition of vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is

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$X' A = 0.76, X^{'} B = 0.24$

$X' A = 0.28, X^{'} B = 0.72$

$X' A = 0.4, X^{'} B = 0.6$

$X' A = 0.37, X^{'} B = 0.63$

answer is B.

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Detailed Solution

$P = x_{A} P^{0} A + x_{B} P^{0} B$
$= 0.4 \times 7 \times 10^{3} + 0.6 \times 12 \times 10^{3} = 10 \times 10^{3} Pa$
$= 1 \times 10^{4} Pa$
$X'_{A} = \frac{P_{A}}{P} = \frac{X_{A} {P^{0}}_{A}}{P} = \frac{0.4 \times 7 \times 103}{1 \times 10^{4}} = 0.28$
$X'_{B} = 1 - X'_{A} = 1 - 0.2 = 0.72$