Chemistry - Liquid Solution Question with Solution | TestHub

$$\begin{array}{rl}\frac{n_{{\mathrm{C}}_5{\mathrm{H}}_{12}}}{n_{{\mathrm{C}}_6{\mathrm{H}}_{14}}}& =\frac{1}{4}\\ \Rightarrow {\mathrm{X}}_{{\mathrm{C}}_3{\mathrm{H}}_{12}}& =\frac{1}{5}\text{ and }{\mathrm{X}}_{{\mathrm{C}}_6{\mathrm{H}}_{14}}=\frac{4}{5}\\ {\mathrm{P}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^{\circ }& =440\mathrm{mm}\mathrm{Hg};\\ {\mathrm{P}}_{{\mathrm{C}}_6{\mathrm{H}}_{14}}^{\circ }& =120\mathrm{mm}\mathrm{Hg}\\ {\mathrm{P}}_{\mathrm{T}}& ={\mathrm{P}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^{\circ }{\mathrm{X}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}\\ & +{\mathrm{P}}_{{\mathrm{C}}_6{\mathrm{H}}_{14}}^{\circ }{\mathrm{X}}_{{\mathrm{C}}_6{\mathrm{H}}_{14}}\\ & =440\times \frac{1}{5}+120\times \frac{4}{5}\\ & =184\mathrm{mm}\text{ of }\mathrm{Hg}\end{array}$$ By Raoult's Law, $${\mathrm{P}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}={\mathrm{P}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^{\circ }{\mathrm{X}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}$$ By Dalton's Law, $${\mathrm{P}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}={\mathrm{X}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^{\prime }\mathrm{P}$$ From (1) and (2), $$\begin{array}{rl}{\mathrm{P}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}& =440\times \frac{1}{5}=88\mathrm{mm}\text{ of }\mathrm{Hg}\\ \Rightarrow 88& ={\mathrm{X}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^{\prime }\times 184\\ {\mathrm{X}}^{\prime }& =\frac{88}{184};{\mathrm{X}}^{\prime }=\mathrm{0.478.}\end{array}$$