Mole fraction of pentane solution =$$\frac{1}{1 + 4} = \frac{1}{5}$$

Mole fraction of hexane solution =$$\frac{4}{1 + 4} = \frac{1}{5}$$

The total pressure of the solution,

$$P_{s} = X_{P}P_{P}^{O} + X_{H}P_{P}^{O}$$ $$\Rightarrow \frac{1}{5} * 440 mm Hg + \frac{4}{5} * 120 mm Hg$$ $$\Rightarrow 184 mm Hg$$

Therefore, the fraction of pentane in the vapour phase.

$$\Rightarrow \frac{X_{P}P_{P}^{O}}{P_{s}}$$ $$\Rightarrow\frac{88}{184}$$ $$\Rightarrow 0.478$$

Therefore, the mole fraction of pentane in the vapour phase would be 0.478

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