Liquids A and B ( $p_{A}^{0} = 200 torr$ , $p_{B}^{0} = 80 torr$ at the working temperature) form an ideal solution. A mixture of $0.60 mol$ of A and $0.40 mol$ of B is prepared, and then $0.50 mol$ of a non-volatile, non-dissociating solute is dissolved in it. Find the total vapour pressure in torr, to one decimal place.

Numerical answer type, answer in torr, tolerance 0.1.

Answer: 101.3 torr

Total moles

= 0.60 + 0.40 + 0.50 = 1.50

. The mole fractions are

x_{A} = 0.60/1.50 = 0.4000

and

x_{B} = 0.40/1.50 = 0.2667

; the solute contributes nothing to the vapour.

P = 0.4 (200) + 0.2667 (80) = 101.333 \approx 101.3 torr