JEE Main 2012 — Solutions Question with Solution

Let vapour pressure of $A=P_A^0$ Vapour pressure of $B=P_B^0$ In first solution, Mole fraction of $A\left(x_A\right)=\frac{1}{1+2}=\frac{1}{3}$ Mole fraction of $B\left(x_B\right)=\frac{2}{1+2}=\frac{2}{3}$ According to Raoult's law, Total vapour pressure $$\begin{array}{rl}& =250=P_A^0{x}_A+P_B^0{x}_B\\ & 250=\frac{1}{3}{P}_A^0+\frac{2}{3}{P}_B^0\end{array}$$ In second solution Mole fraction of $A\left(x_A\right)=\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}$ Mole fraction of $B\left(x_B\right)=\frac{2}{4}=\frac{1}{2}$ $\therefore$ Total vapour pressure $$\begin{array}{rl}& =300=P_A^0{x}_A+P_B^0{x}_B\\ & 300=\frac{1}{2}{P}_A^0+\frac{1}{2}{P}_B^0\end{array}$$ Multiplying equation (i) by $\frac{1}{2}$ and equation (ii) by $\frac{1}{3}$ $$\begin{array}{rl}& \frac{1}{6}{P}_A^0+\frac{2}{6}{P}_B^0=125\\ & \frac{1}{6}{P}_A^0+\frac{1}{6}{P}_B^0=100\\ & \frac{1}{6}{P}_B^0=25\\ & P_B^0=25\times 6=150\mathrm{mm}\mathrm{Hg}\end{array}$$ On substituting value of $P_B^0$ in equation (ii) we get $$\begin{array}{rl}& 300=P_A^0\times \frac{1}{2}+150\times \frac{1}{2}\\ & P_A^0=450\mathrm{mm}\mathrm{Hg}\end{array}$$