Chemistry - Liquid Solution Question with Solution | TestHub

$\frac{{\mathrm{P}}_{{\mathrm{A}}^{\mathrm{o}}}}{{\mathrm{P}}_{{\mathrm{B}}^{\mathrm{o}}}}=\frac{1}{3}$
$\frac{{\text{n}}_{\text{A}}}{{\text{n}}_{\text{B}}}=\frac{1}{3}$
${\mathrm{X}}_{\mathrm{A}}=\frac{1}{4}$
${\mathrm{X}}_{\mathrm{B}}=\frac{3}{4}$
${\mathrm{P}}_{\mathrm{A}}={\mathrm{P}}_{\mathrm{A}}^0{\mathrm{X}}_{\mathrm{A}}={\mathrm{P}}_{\mathrm{A}}^0\times \frac{1}{4};\mathrm{}{\mathrm{P}}_{\mathrm{B}}={\mathrm{P}}_{\mathrm{B}}^0{\mathrm{X}}_{\mathrm{B}}={\mathrm{P}}_{\mathrm{B}}^0\times \frac{3}{4}$
$\frac{{\mathrm{P}}_{\mathrm{A}}^0}{{\mathrm{P}}_{\mathrm{B}}^0}=\frac{1}{3}\Rightarrow 3{\mathrm{P}}_{\mathrm{A}}^0={\mathrm{P}}_{\mathrm{B}}^0$
$\mathrm{P}={\mathrm{P}}_{\mathrm{A}}^0{\mathrm{X}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{X}}_{\mathrm{B}}={\mathrm{P}}_{\mathrm{A}}^0\times \frac{1}{4}+\frac{9}{4}{\mathrm{P}}_{\mathrm{A}}^0\mathrm{}\left[\because {\mathrm{P}}_{\mathrm{B}}^0\right]=3{\mathrm{P}}_{\mathrm{A}}^0$
$\mathrm{P}=\frac{10}{4}{\mathrm{P}}_{\mathrm{A}}^0\mathrm{}\mathrm{}{\mathrm{X}}_{\mathrm{A}}=\frac{\frac{1}{4{\mathrm{P}}_{\mathrm{A}}^0}}{\frac{10}{4}{\mathrm{P}}_{\mathrm{A}}^0}=\frac{1}{10}=0.1$