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

Answer: (D)

Step 1:

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

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

Total pressure of solution,

$P_S=X_p{P}_p^0+X_H{P}_H^0$

$Where,X_p=molefractionofpentane,P_p^0=PureVapourpressureofpentane$

$X_H=molefractionofHexane,P_H^0=PureVapourpressureofHexane$

Step 2:

$P_S=\frac{1}{5}\times 440mmHg+\frac{4}{5}\times 120mmHg$

$=184mmHg$

Step 3:

Mole fraction of pentane in vapour phase =

$\frac{X_P{P}_p^0}{P_S}=\frac{88}{184}=0.478mmHg$

Hence, option (C) is correct.