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The vapour pressure of pure benzene and toluence are 160 Torr and 60 Torr,respectively.The mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene would be

$\begin{array}{1 1}0.50\\0.84\\0.73\\0.27\end{array} $

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Answer : 0.73
$p_b=x_bp_b^*$=(0.5)(160 Torr)=80 Torr
$p_{total}=x_bp_b^*+x_tp_t^*$=(0.5)(160 Torr)+(0.5 Torr)(60 Torr)=110 Torr
$y_b=p_b/p_{total}=80/110=0.73$
answered Jul 22, 2014 by sreemathi.v
 

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