A solution has a $1 : 4$mole ratio of pentane  to hexane. The vapour pressures of the pure hydrocarbons at $20$ $°\mathrm{C}$ are $440 \mathrm{mm} \mathrm{Hg}$  for pentane and $120 \mathrm{mm} \mathrm{Hg}$ for hexane. The mole fraction of pentane in the vapour phase would be . 

Ratio of pentane ${\mathrm{C}}_5{\mathrm{H}}_{12}$ and hexane${\mathrm{C}}_6{\mathrm{H}}_{14}$ $\text{ i.e., }\frac{n_{{\mathrm{C}}_5{\mathrm{H}}_{12}}}{n_{{\mathrm{C}}_6{\mathrm{H}}_{14}}}=\frac{1}{4}$ 

Hence, mole fraction of $x_{{\mathrm{C}}_5{\mathrm{H}}_{12}}=\frac{1}{1+4}=\frac{1}{5}$ and mole fraction of $x_{{\mathrm{C}}_6{\mathrm{H}}_{14}}=\frac{4}{1+4}=\frac{4}{5}$

$P_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^o=440 \mathrm{mm} \mathrm{Hg}; P_{{\mathrm{C}}_6{\mathrm{H}}_{14}}^o=120 \mathrm{mm} \mathrm{Hg}$

According to Raoult's law :

$\begin{array}{l}{P}_T=P_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^0{x}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}+P_{{\mathrm{C}}_6{\mathrm{H}}_{14}}^0{x}_{{\mathrm{C}}_6{\mathrm{H}}_{14}}\\ =\left(440\times \frac{1}{5}\right)+\left(120\times \frac{4}{5}\right)\\ =88+96=184 \mathrm{mm} \mathrm{Hg}\end{array}$

Also, $P_{C_5{H}_{12}}=P_{C_5{H}_{12}}^o{x}_{C_5{H}_{12}}$                                      (1)

Here, $x_{{\mathrm{C}}_5{\mathrm{H}}_{12}}=$mole fraction of ${\mathrm{C}}_5{\mathrm{H}}_{12}$ in solution 

By Dalton's law:

$P_{{\mathrm{C}}_5{\mathrm{H}}_{12}}=P\times x_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^{\mathrm{\prime }}$                                (2)

where, $x{\mathit{\text{'}}}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}$ is the mole fraction of pentane in vapor phase.

From (1) and (2), we have : 

$P_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^o{x}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}=P\times x_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^{\mathrm{\prime }}$

$\therefore x_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^{\mathrm{\prime }}=\frac{P_{{\mathrm{C}}_5{\mathrm{H}}_{12}}^o{x}_{{\mathrm{C}}_5{\mathrm{H}}_{12}}}{P}=\frac{440}{184}\times \frac{1}{5}=0.478.$