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

• A. 0.549
• B. 0.2
• C. 0.786
• D. 0.478

Answer 1

The correct option is D 0.478

Mole fraction of pentane in vapour phase = $Y_1=\frac{P_1}{P_1+P_2}=\frac{P_1^0{X}_1}{P_1^0{X}_1+P_2^0{X}_2}$
$=\frac{440\times \frac{1}{5}}{440\times \frac{1}{5}+120\times \frac{1}{5}}=0.478$