An ideal solution is made from $60 g$ of A (molar mass $120 g mo l^{- 1}$ ) and $90 g$ of B (molar mass $60 g mo l^{- 1}$ ). The pure vapour pressures are $p_{A}^{0} = 360 torr$ and $p_{B}^{0} = 120 torr$ . Treating the vapour as ideal, find the mass percentage of A in the vapour phase at equilibrium (two decimal places).

Numerical answer type, answer in %, tolerance 0.01.

Answer: 66.67 %

Moles: A

= 0.50

, B

= 1.50

, so

x_{A} = 0.25

p_{A} = 0.25 \times 360 = 90

and

p_{B} = 0.75 \times 120 = 90

torr, so

y_{A} = y_{B} = 0.50

. Per mole of vapour: mass of A

= 0.5 \times 120 = 60 g

, mass of B

= 0.5 \times 60 = 30 g

. Mass percent of A

= 60/90 \times 100 = 66.67%