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. 0.178
• B. 0.278
• C. 0.378
• D. 0.478

Answer 1

Answer: (D)

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