A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20° C are 440 mm Hg for pentane and 120 ~mm…

$$\begin{aligned} \frac{n_{\mathrm{C}_5 \mathrm{H}_{12}}}{n_{\mathrm{C}_6 \mathrm{H}_{14}}} & =\frac{1}{4} \\ \Rightarrow \mathrm{X}_{\mathrm{C}_3 \mathrm{H}_{12}} & =\frac{1}{5} \text { and } \mathrm{X}_{\mathrm{C}_6 \mathrm{H}_{14}}=\frac{4}{5} \\ \mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} & =440 \mathrm{~mm} \mathrm{Hg} ; \\ \mathrm{P}_{\mathrm{C}_6 \mathrm{H}_{14}}^{\circ} & =120 \mathrm{~mm} \mathrm{Hg} \\ \mathrm{P}_{\mathrm{T}} & =\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} \mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}} \\ & +\mathrm{P}_{\mathrm{C}_6 \mathrm{H}_{14}}^{\circ} \mathrm{X}_{\mathrm{C}_6 \mathrm{H}_{14}} \\ & =440 \times \frac{1}{5}+120 \times \frac{4}{5} \\ & =184 \mathrm{~mm} \text { of } \mathrm{Hg} \end{aligned}$$nC6​H14​​nC5​H12​​​⇒XC3​H12​​PC5​H12​∘​PC6​H14​∘​PT​​=41​=51​ and XC6​H14​​=54​=440 mmHg;=120 mmHg=PC5​H12​∘​XC5​H12​​+PC6​H14​∘​XC6​H14​​=440×51​+120×54​=184 mm of Hg​ By Raoult's Law, $$\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}=\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} \mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}$$PC5​H12​​=PC5​H12​∘​XC5​H12​​ By Dalton's Law, $$\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}=\mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} \mathrm{P}$$PC5​H12​​=XC5​H12​′​P From (1) and (2), $$\begin{aligned} \mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}} & =440 \times \frac{1}{5}=88 \mathrm{~mm} \text { of } \mathrm{Hg} \\ \Rightarrow \quad 88 & =\mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} \times 184 \\ \mathrm{X}^{\prime} & =\frac{88}{184} ; \mathrm{X}^{\prime}=0.478 . \end{aligned}$$PC5​H12​​⇒88X′​=440×51​=88 mm of Hg=XC5​H12​′​×184=18488​;X′=0.478.​