A solution has a 1:4 mole ratio of pentane to hexane. The vapour pressure of the

Question

A solution has a 1:4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at

20^{\circ}

is 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be

Options:

A . 0.549

B . 0.200

C . 0.786

D . 0.478

Answer: Option D
:
D
Mole fractionof pentane invapourphase =

Y_{1} = \frac{P_{1}}{P_{1} + P_{2}} = \frac{P_{1}^{0} X_{1}}{P_{1}^{0} X_{1} + P_{2}^{0} X_{2}}

= \frac{440 \times \frac{1}{5}}{440 \times \frac{1}{5} + 120 \times \frac{1}{5}} = 0.478

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