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Liquids A and B form an ideal solution, with vapour pressures of the pure liquids 200 torr and 100 torr respectively at a given temperature. An…

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Liquids A and B form an ideal solution, with vapour pressures of the pure liquids

200

torr and

100

torr respectively at a given temperature. An equimolar liquid mixture is allowed to equilibrate with its vapour. The vapour is completely condensed to a new liquid, which is again equilibrated with its own vapour at the same temperature. The mole percent of A in this second vapour is

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Answer: 80

Step 1: liquid

x_A=0.5

, vapour

y_A=\dfrac{200(0.5)}{200(0.5)+100(0.5)}=\dfrac23

. Step 2: the condensed liquid has

x_A=\frac23

x_B=\frac13

;

P=200\cdot\frac23+100\cdot\frac13=166.7

torr and

y_A=\dfrac{133.3}{166.7}=0.80

. So

80

mole percent (equivalent to one more stage of distillation).

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