At \(T(\mathrm{~K})\), the vapour pressures of pure liquids \(A\) and \(B\) are \(100 \mathrm{~mm}\) and…
At \(T(\mathrm{~K})\), the vapour pressures of pure liquids \(A\) and \(B\) are \(100 \mathrm{~mm}\) and \(160 \mathrm{~mm}\) respectively. An ideal solution is formed by mixing 2 moles of \(A\) and 3 moles of \(B\) at the same temperature. The mole fraction of \(A\) and \(B\) in the vapour state respectively are
\(0.706,0.294\)
\(0.294,0.706\)
\(0.40,0.60\)
\(0.60,0.40\)
Solution
Key Idea Vapour pressure of solution, \(p_{\text {total }}=p_A+p_{B^{\prime}}=\chi_A p_A^{\circ}+\chi_B p_B^{\circ} \quad\left[\because p_A=\chi_A p_A^{\circ}\right]\)
Also vapour pressure of component \(1, p_1=y_1 p_{\text {total }}\) where \(y\), is the mole fraction of component 1 in vapour phase.
Given,
Vapour pressure of pure liquid \(A, p_A^{\circ}=100 \mathrm{~mm}\)
Vapour pressure of pure liquid \(B, p_B^{\circ}=160 \mathrm{~mm}\)
\(\therefore\) Total vapour pressure of solution \(=p_A+p_B\)
\(\begin{aligned}
p_{\text {total }} & =\chi_A p_A^{\circ}+\chi_B p_B^{\circ}=\frac{2}{5} \times 100+\frac{3}{5} \times 160 \\
& =40+96=136 \mathrm{~mm}
\end{aligned}\)
Also \(\quad p_A=y_A p_{\text {total }}\) where, \(y_A\) is the mole fraction of \(A\).
Mole fraction of \(A, y_A=\frac{p_A}{p_{\text {total }}}=\frac{40}{136}=0.294\)
\(\therefore \quad y_B=1-y_A=1-0.294=0.706\)