At $298 \mathrm{~K}$, vapour pressures of two pure liquids $A$ and $B$ are 200 and $400 \mathrm{~mm} \mathrm{Hg}$ respectively, if mole fractions of $A$ and $B$ in solution are 0.7 and 0.3 respectively? What is the mole fraction of $B$ in vapour phase?

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Aicharya · Accepted Answer

Given, $ \begin{aligned} p_A^{\circ} & =200 \mathrm{~mm} \mathrm{Hg} \ p_B^{\circ} & =400 \mathrm{~mm} \mathrm{Hg} \end{aligned} $ Mole fraction of $A$ in solution $=\chi_A=0.7$ Mole fraction of $B$ in solution $=\chi_B=0.3$ $ \begin{aligned} p_{	ext {Total }} & =\chi_A p_A^{\circ}+\chi_B p_B^{\circ} \ & =(0.7 	imes 200)+(0.3 	imes 400)=140+120 \ & =260 \mathrm{~mm} \mathrm{Hg} \end{aligned} $ $\begin{gathered}	ext { Mole fraction of } B 	ext { in vapour phase }=\frac{p_B}{p_{	ext {Total }}} \ =\frac{p_B^{\circ} \chi_A}{p_{	ext {Total }}}=\frac{400 	imes 0.3}{260}=0.462\end{gathered}$

At $298 \mathrm{~K}$, vapour pressures of two pure liquids $A$ and $B$ are 200 and $400 \mathrm{~mm}…

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