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

Question

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, $0.706,0.294$, $0.294,0.706$, $0.40,0.60$, $0.60,0.40$

MARKS App · Accepted Answer

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 $A, y_A=\frac{p_A}{p_{\text {total }}}=\frac{40}{136}=0.294$ $\therefore \quad y_B=1-y_A=1-0.294=0.706$

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