Question

A solution has a $1:4$ mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at ${20}^0\text{C}$ are $440\text{mm Hg}$ for pentane and $120\text{mm Hg}$ for hexane. The mole fraction of pentane in the vapour phase would be:

• A. $0.200$
• B. $0.549$
• C. $0.786$
• D. $0.478$

Answer 1

The correct option is D $0.478$

$\frac{n_{C_5}{H}_{12}}{n_{C_6}{H}_{14}}=\frac{1}{4}$
$\Rightarrow X_{C_5{H}_{12}}=\frac{1}{5}\text{and}{X}_{C_6{H}_{14}}=\frac{4}{5}$
$P_{C{}_5{H}_{12}}^0=40\text{mm Hg};P_{C_6{H}_{14}}^0=120\text{mm Hg}$
$P_T=P_{C_5{H}_{12}}^0{X}_{C_5{H}_{12}}+P_{C_6{H}_{14}}^0{X}_{C_6{H}_{14}}$
$=440\times \frac{1}{5}+120\times \frac{4}{5}=88+96=184\text{mm of Hg}$
By Raoult's law, $P_{C_5{H}_{12}}=P_{C_5{H}_{12}}^0{X}_{C_5{H}_{12}}....\left(1\right)$
$X_{C_5{H}_{12}}\to$ mole fraction of pentane in solution

By Dalton's Law, $P_{C_5{H}_{12}}=X_{C_5{H}_{12}}^{\prime }P.....\left(2\right)$
$X_{C_5{H}_{12}}^{\prime }\to$mole fraction of pentane above the solution.
From (1) and (2),
$P_{C_5{H}_{12}}=440\times \frac{1}{5}=88\text{mm of Hg}\Rightarrow 88=X_{C_5{H}_{12}}^{\prime }\times 184$
$X^{\prime }=\frac{88}{184};X^{\prime }=0.478$