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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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Verified Answer
The correct answer is:
0.462
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 $=\chi_B=0.3$
$$
\begin{aligned}
p_{\text {Total }} & =\chi_A p_A^{\circ}+\chi_B p_B^{\circ} \\
& =(0.7 \times 200)+(0.3 \times 400)=140+120 \\
& =260 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
$\begin{gathered}\text { Mole fraction of } B \text { in vapour phase }=\frac{p_B}{p_{\text {Total }}} \\ =\frac{p_B^{\circ} \chi_A}{p_{\text {Total }}}=\frac{400 \times 0.3}{260}=0.462\end{gathered}$
$$
\begin{aligned}
p_A^{\circ} & =200 \mathrm{~mm} \mathrm{Hg} \\
p_B^{\circ} & =400 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
Mole fraction of $A$ in solution $=\chi_A=0.7$
Mole fraction of $B$ in solution $=\chi_B=0.3$
$$
\begin{aligned}
p_{\text {Total }} & =\chi_A p_A^{\circ}+\chi_B p_B^{\circ} \\
& =(0.7 \times 200)+(0.3 \times 400)=140+120 \\
& =260 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
$\begin{gathered}\text { Mole fraction of } B \text { in vapour phase }=\frac{p_B}{p_{\text {Total }}} \\ =\frac{p_B^{\circ} \chi_A}{p_{\text {Total }}}=\frac{400 \times 0.3}{260}=0.462\end{gathered}$
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