Question

Two liquids $X$ and $Y$ from an ideal solution. At $300K$, a vapour pressure of the solution containing $1$ mol of $X$ and $3$ mol of $Y$ is $550$ $mmHg$. At the same temperature, if $1$ mol of $Y$ is further added to this solution, a vapour pressure of the solution increased by $10$ $mmHg$. Vapour pressure ( in mmHg) of $X$ and $Y$ in their pure states will be respectively:

• A. $300$ and $400$
• B. $400$ and $600$
• C. $500$ and $600$
• D. $200$ and $300$

Answer 1

The correct option is B $400$ and $600$

$\begin{array}{cc}{P}_{Total}=X_x{P}_x+X_y{P}_{y}So,550=\frac{1}{4}{P}_x+\frac{3}{4}{P}_y& \left[\begin{array}{c}Here{p}_{y}and{p}_{y}andvapourpressure\\ ofXandYinpurestates.\end{array}\right]\\ P_x+3P_4=550\times 4;P_x+3P_y=2200\to \left(i\right)& \\ 2^{nd}case& \\ P_{total}=560mmHg.& \\ \frac{1}{5}{P}_x+\frac{4}{5}{P}_y=560P_x+{48}_y=2800\to \left(ii\right)& \\ \left(ii\right)-\left(i\right)& \\ p_y=600mmHg{p}_x=2200-1800\Rightarrow 400mm07ng& \end{array}$