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Q. Two liquids $A$ and $B$ have $p_{A}^{\circ}: p_{B}^{\circ}=1: 3$ at a certain temperature. If the mole fraction ratio of $x_{A}: x_{B}=1: 3$, the mole fraction of $A$ in vapor in equilibrium with the solution at a given temperature is

Solutions

Solution:

$ y_{A} =\frac{p_{A}^{0} x_{A}}{p_{A}^{\circ} x_{A}+p_{B}^{\circ} x_{B}} $

$=\frac{1}{1+\frac{p_{B}^{\circ} x_{B}}{p_{A}^{\circ} x_{A}}} $

$=\frac{1}{1+(3)(3)}=\frac{1}{10}=0.1$