An ideal binary liquid of A and B ( $p_{A}^{0} = 200 torr$ , $p_{B}^{0} = 50 torr$ ) with $x_{A} = 0.50$ is partly vaporised. The vapour is completely condensed to a liquid, which is then partly vaporised again at the same temperature. Find the mole fraction of A in the vapour from this second vaporisation (two decimal places).

Numerical answer type, tolerance 0.005.

Answer: 0.94

First stage:

y_{1} = \frac{0.5 \times 200}{0.5 \times 200 + 0.5 \times 50} = \frac{100}{125} = 0.80

. The condensate has

x_{A} = 0.80

. Second stage:

y_{2} = \frac{0.8 \times 200}{0.8 \times 200 + 0.2 \times 50} = \frac{160}{170} = 0.9412

, i.e.

0.94