Question

A solution has a $1: 4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $20^{\circ} C$ are $440 \,mm$ of $Hg$ for pentane and $120 \,mm$ of $Hg$ 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)

0.478

Answer 2

Total vapour pressure of mixture
$=$ (Mole fraction of pentane $\times$ VP of pentane) $+$ (Mole fraction of hexane $\times$ VP of hexane)
= VP of pentane in mixture $+$ VP of hexane in mixture.
$=\left(\frac{1}{5} \times 440+\frac{4}{5} \times 120\right)=184 \,mm$
$\because$ VP of Pentane in mixture.
$=$ VP of mixture $\times$ mole fraction of pentane in vapour phase
$88=184 \times$ mole fraction of pentane in vapour phase
$\therefore$ Mole fraction of pentane in vapour phase
$=\frac{88}{184}=0.478$