The vapour pressure of pure benzene and toluene at a particular temperature are $100 \mathrm{~mm}$ and $50…

The vapour pressure of pure benzene and toluene at a particular temperature are $100 \mathrm{~mm}$ and $50 \mathrm{~mm}$ respectively. Then the mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is
  1. $0.67$
  2. $0.75$
  3. $0.33$
  4. $0.50$

Solution

Total vapour pressure $=$ vapour pressure of pure benzene $+$ vapour pressure of toluene
$=100+50=150 \mathrm{~mm}$
We know, $\mathrm{P}_{\mathrm{C}_{6} \mathrm{H}_{6}}^{\mathrm{o}}=\mathrm{P} \times \mathrm{X}_{\mathrm{C} 6 \mathrm{H} 6}$
$100=150 \times \mathrm{X}_{\mathrm{C} 6 \mathrm{H} 6}$
$X_{C 6 H 6}=\frac{100}{150}=0.67$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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