Question

Vapour pressure of $A$ and $B$ mixture at $80^{\circ }C$ is given by $P(inmmHg)=100+200x_A,$ where $x_A$ is mole fraction of $A.$ Solution is prepared by mixing $2mol$ of $A$ and $3mol$ of $B$ and if the vapours are removed and condensed into liquid and again brought to the temperature of $-80^{\circ }C,$ If the ratio of mole fraction of $A$ in vapour state and mole fraction of $B$ in vapour state above the condensate is $x:1,$ find value of ${}^{\prime }{x}^{\prime }$ is -

Answer 1

Answer: 6

$P_T=100+200\times 0.4=180$
$P_A^0=300,P_B^0=100,y_A=\frac{300\times 0.4}{180}=\frac{2}{3}$
$y_A^{{}^{\prime }}=\frac{\frac{2}{3}\times 300}{\frac{2}{3}\times 300+\frac{1}{3}\times 100}=\frac{6}{7}$
$y_B^{{}^{\prime }}=\frac{1}{7},\frac{y_A^{\prime }}{y_B^{\prime }}=\frac{6}{1}=6$