For an ideal mixture of A and B at fixed temperature, the total vapour pressure is $280 torr$ when $x_{A} = 0.30$ and $340 torr$ when $x_{A} = 0.60$ . Find the mole fraction of A in the vapour in equilibrium with a liquid having $x_{A} = 0.80$ (two decimal places).

Numerical answer type, tolerance 0.005.

Answer: 0.88

For an ideal mixture

P = p_{B}^{0} + x_{A} (p_{A}^{0} - p_{B}^{0})

. The slope is

(340 - 280) /0.30 = 200

, so

p_{A}^{0} - p_{B}^{0} = 200

; then

p_{B}^{0} = 280 - 0.30 \times 200 = 220

and

p_{A}^{0} = 420

torr. At

x_{A} = 0.80

P = 220 + 160 = 380

and

y_{A} = 0.80 \times 420/380 = 0.8842

, i.e.

0.88