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The vapour pressure of pure benzene and toluene are 160 and $60 \mathrm{~mm} \mathrm{Hg}$ respectively. The mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is
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The correct answer is:
$0.73$
For equimolar solution,
$\begin{aligned} & \chi_b=\chi_t=0.5 \\ & \mathrm{p}_b=\chi_b \times \mathrm{p}_b^0=0.5 \times 160 \\ & =80 \mathrm{~mm} \\ & \mathrm{p}_b=\chi_t \times \mathrm{p}_b^0=0.5 \times 60 \\ & =30 \mathrm{~mm}\end{aligned}$
$\mathrm{p}_{\text {total }}=80+30=110 \mathrm{~mm}$
Mole fraction of benzene in vapour phase
$=\frac{\mathrm{p}_b}{\mathrm{p}_{\text {total }}}=\frac{80}{110}=0.727=0.73$
$\begin{aligned} & \chi_b=\chi_t=0.5 \\ & \mathrm{p}_b=\chi_b \times \mathrm{p}_b^0=0.5 \times 160 \\ & =80 \mathrm{~mm} \\ & \mathrm{p}_b=\chi_t \times \mathrm{p}_b^0=0.5 \times 60 \\ & =30 \mathrm{~mm}\end{aligned}$
$\mathrm{p}_{\text {total }}=80+30=110 \mathrm{~mm}$
Mole fraction of benzene in vapour phase
$=\frac{\mathrm{p}_b}{\mathrm{p}_{\text {total }}}=\frac{80}{110}=0.727=0.73$
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