The vapour pressure of pure benzene and tolue | vedclass

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(D) For an equimolar solution,the mole fractions in the liquid phase are $\chi_b = \chi_t = 0.5$. The partial vapour pressure of benzene is $p_b = \ch

Vedclass · Accepted Answer

$0.73$

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(D) For an equimolar solution,the mole fractions in the liquid phase are $\chi_b = \chi_t = 0.5$. The partial vapour pressure of benzene is $p_b = \chi_b 	imes p_b^0 = 0.5 	imes 160 = 80 \ mm \ Hg$. The partial vapour pressure of toluene is $p_t = \chi_t 	imes p_t^0 = 0.5 	imes 60 = 30 \ mm \ Hg$. The total vapour pressure is $p_{	ext{total}} = p_b + p_t = 80 + 30 = 110 \ mm \ Hg$. The mole fraction of benzene in the vapour phase $(y_b)$ is given by $y_b = \frac{p_b}{p_{	ext{total}}} = \frac{80}{110} \approx 0.727 \approx 0.73$.

The vapour pressure of pure benzene and toluene are $160$ $mm \ Hg$ and $60$ $mm \ Hg$ respectively. The mole fraction of benzene in the vapour phase in contact with an equimolar solution of benzene and toluene is:

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