Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressure of pure…
Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at $T(\mathrm{~K})$ are $50 \mathrm{~mm} \mathrm{Hg}$ and $40 \mathrm{~mm} \mathrm{Hg}$ respectively. What is the mole fraction of toluene in vapour phase when $117 \mathrm{~g}$ of benzene is mixed with $46 \mathrm{~g}$ of toluene? (molar mass of benzene and toluene are 78 and $92 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively)
0.78
0.21
0.64
0.35
Solution
$p^{\circ}$ benzene $p_{(b)}^{\circ}=50 \mathrm{~mm} \mathrm{Hg}$ $p^{\circ}$ toluene $p_{(t)}^{\circ}=40 \mathrm{~mm} \mathrm{Hg}$
Mass of benzene $=117 \mathrm{~g}$
Mass of toluene $=46 \mathrm{~g}$
Molar mass of benzene $=78 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of toluene $=92 \mathrm{~g} \mathrm{~mol}^{-1}$
Number of moles of benzene $=\frac{117}{78}=1.5 \mathrm{~mol}$
Number of moles of toluene $=\frac{46}{92}=0.5 \mathrm{~mol}$
Mole fraction of toluene $\left(\chi_{(t)}\right)$
$
=\frac{\text { Number of moles of toluene }}{\text { Total number of moles }}
$
$=\frac{0.05}{(1.5+0.5)}=\frac{0.5}{2.0}=0.25$ and of benzene
$
\chi_{(b)}=0.75
$
$\because$ Total vapour pressure $\left(p_t\right)$
$
\begin{aligned}
& =p_{(t)}^{\circ} \times \chi_{(t)}+p_{(b)}^{\circ} \times \chi_{(b)} \\
& =p_{\text {(toluene })}+p_{\text {(benzene })}
\end{aligned}
$
(where, $p$ is the partial pressures of toluene and benzene respectively).
$
\begin{aligned}
p_T & =40 \times 0.25+50 \times 0.75 \\
& =10+37.5 \\
p_T & =47.5 \mathrm{~mm} \text { of } \mathrm{Hg}
\end{aligned}
$
Also,
$\because$ Mole fraction (in vapour phase)
$
=\frac{\text { Partial pressure }}{\text { Total pressure }}
$
$
\left.\therefore \chi_{(t)} \text { [in vapour phase }\right]=\frac{10}{47.5}=0.21
$
$\therefore$ Mole fraction (toluene) in vapour phase $=0.21$