Chemistry - Concentration Terms Question with Solution | TestHub

${\text{mole of CH}}_2\text{C}{\text{l}}_2\text{in liquid phase}=\frac{8.5}{85}=.1$

${\text{mole of CHCl}}_{\text{3}}\text{in liquid phase}=\frac{11.95}{119.5}=.1$

${\text{mole fraction of CH}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{in liquid phase}=\frac{.1}{.2}=\frac{1}{2}$

${\text{mole fraction of CHCl}}_{\text{3}}\text{in liquid phase}=\frac{.1}{.2}=\frac{1}{2}$

${\text{P}}_{\text{T}}={\text{X}}_{\text{C}{\text{H}}_2\text{C}{\text{l}}_2}\times {\text{vapour pressure}}_{\text{C}{\text{H}}_2\text{C}{\text{l}}_2}+{\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_3}\times {\text{vapour pressure}}_{\text{C}\text{H}\text{C}{\text{l}}_3}$

$=415\times \frac{1}{2}+200\times \frac{1}{2}=307.5$

${\text{P}}_{\text{T}}{\left({\text{X}}_{\text{A}}\right)}_{\text{V}\text{P}}={\left({\text{X}}_{\text{A}}\right)}_{\text{L}\text{P}}\times {\text{ Vapour pressure of CHCl}}_3$

$307.5\times {\left({\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_3}\right)}_{\text{V}\text{P}}=200\times \frac{1}{2}$

${\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_3}=\frac{100}{307.5}=.325$