Chemistry - Liquid Solution Question with Solution | TestHub

Mole-fraction in vapour phase, Y for A

${\text{Y}}_{\text{A}}=\text{0.4}\text{,}{\text{Y}}_{\text{B}}=\text{0.6}$

$\begin{array}{c}{\text{P}}_{\text{A}}^{\circ }=\text{0.4}\\ {\text{P}}_{\text{B}}^{\circ }=\text{1.2}\end{array}\}\text{given}$

${\text{P}}_{\text{T}}={\text{P}}_{\text{A}}^{\circ }{\text{x}}_{\text{A}}+{\text{P}}_{\text{B}}^{\circ }{\text{x}}_{\text{B}}$...(i)

${\text{Y}}_{\text{A}}=\text{0.4}=\frac{{\text{P}}_{\text{A}}^{\circ }{\text{x}}_{\text{A}}}{{\text{P}}_{\text{T}}}\Rightarrow \frac{\text{0.4}{\text{x}}_{\text{A}}}{{\text{P}}_{\text{T}}}=\text{0.4}$

or${\text{P}}_{\text{T}}={\text{x}}_{\text{A}}$...(ii)

${\text{Y}}_{\text{B}}=\text{0.6}=\frac{{\text{P}}_{\text{B}}^{\circ }{\text{x}}_{\text{B}}}{{\text{P}}_{\text{T}}}\Rightarrow \frac{\text{1.2}{\text{x}}_{\text{B}}}{{\text{P}}_{\text{T}}}=\text{0.6}$

or${\text{P}}_{\text{T}}=2{\text{x}}_{\text{B}}$...(iii)

Comparing (ii) and (iii),

${\text{x}}_{\text{A}}=2{\text{x}}_{\text{B}}$

we know,${\text{x}}_{\text{A}}+{\text{x}}_{\text{B}}=1$

Hence,${\text{x}}_{\text{B}}=\frac{1}{3}$and${\text{x}}_{\text{A}}=\frac{2}{3}$

Putting in Equation (i)

${\text{P}}_{\text{T}}=\text{0.4}\times \frac{2}{3}+\text{1.2}\times \frac{1}{3}=\text{0.667 atm}$