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A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at…
A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at $20^{\circ} \mathrm{C}$ are 440 $\mathrm{mm} \mathrm{Hg}$ for pentane and $120 \mathrm{~mm} \mathrm{Hg}$ for hexane. the mole fraction of pentane in the vapour phase would be:
0.200 0.549 0.786 0.478 Solution
$\begin{aligned}
\frac{n_{\mathrm{C}_5 \mathrm{H}_{12}}}{n_{\mathrm{C}_6 \mathrm{H}_{14}}} & =\frac{1}{4} \\
\Rightarrow \mathrm{X}_{\mathrm{C}_3 \mathrm{H}_{12}} & =\frac{1}{5} \text { and } \mathrm{X}_{\mathrm{C}_6 \mathrm{H}_{14}}=\frac{4}{5} \\
\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} & =440 \mathrm{~mm} \mathrm{Hg} ; \\
\mathrm{P}_{\mathrm{C}_6 \mathrm{H}_{14}}^{\circ} & =120 \mathrm{~mm} \mathrm{Hg} \\
\mathrm{P}_{\mathrm{T}} & =\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} \mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}} \\
& +\mathrm{P}_{\mathrm{C}_6 \mathrm{H}_{14}}^{\circ} \mathrm{X}_{\mathrm{C}_6 \mathrm{H}_{14}} \\
& =440 \times \frac{1}{5}+120 \times \frac{4}{5} \\
& =184 \mathrm{~mm} \text { of } \mathrm{Hg}
\end{aligned}$
By Raoult's Law,
$\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}=\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} \mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}$
By Dalton's Law,
$\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}=\mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} \mathrm{P}$
From (1) and (2),
$\begin{aligned}
\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}} & =440 \times \frac{1}{5}=88 \mathrm{~mm} \text { of } \mathrm{Hg} \\
\Rightarrow \quad 88 & =\mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} \times 184 \\
\mathrm{X}^{\prime} & =\frac{88}{184} ; \mathrm{X}^{\prime}=0.478 .
\end{aligned}$
Asked in: NEET 2005
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