Liquids A and B form an ideal solution with pure-liquid vapour pressures $p_{A}^{0} = 400 torr$ and $p_{B}^{0} = 120 torr$ at a given temperature. A liquid mixture is at equilibrium with vapour in which the mole fraction of A is $0.80$ . The total vapour pressure of this mixture, in torr, is

Answer: D

With

y_{A} = \frac{x _{A} p _{A}^{0}}{x _{A} p _{A}^{0} + x _{B} p _{B}^{0}}

\frac{x _{A}}{x _{B}} = \frac{y _{A} p _{B}^{0}}{y _{B} p _{A}^{0}} = \frac{0.80 \times 120}{0.20 \times 400} = 1.20

. So

x_{A} = 1.20/ (1 + 1.20) = 0.5455

and

x_{B} = 0.4545

. Then

P = 0.5455 (400) + 0.4545 (120) = 272.7 torr