Question

Two liquids $A$ and $B$ have, $P_A^0$ and $P_B\_{}^0=1:3$ at a certain temperature. Assume $A$ and $B$ form an ideal solution, and the ratio of mole of $A$ to $B$ in the liquid phase are $1:3,$ then mole fraction of $A$ in vapour phase in equilibrium with the solution is $y_A$ . Calculate $10y_A$ ?

Answer 1

Answer: 1

$Pt=P_A^0{x}_A+P_B^0{x}_B=\frac{1}{4}{P}_A^0+\frac{3}{4}\times 3P_A^0$
$=\frac{10}{4}{P}_A^0=\frac{5}{2}{P}_A^0$
$P_T.Y_A=P_A^0{x}_A$
$\Rightarrow y_A=\frac{P_A^{0}x\frac{1}{4}}{\frac{5}{2}{P}_A^0}=\frac{1}{10}10y_A=10\times 0.1=1$