As the ratio of pentane to hexane = 1 : 4

$\therefore$ Mole fraction of pentane = 1/5

Mole fraction of hexane = 4/5

Total vapour pressure

= $\left( {{1 \over 5} \times 440 + {4 \over 5} \times 120} \right)$

= 184 mm of Hg

$\therefore$ Vapour pressure of pentane in mixture

= (Vapour pressure of mixture pentane $\times$ Mole fraction of in vapour phase)

$\Rightarrow$ 88 = 184 × mole fraction of pentane in vapour phase

$\Rightarrow$ Mole fraction of pentane in vapour phase

= ${{88} \over {184}}$ = 0.478