Question

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

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

Answer 1

Answer: (C)

$P_T=P_X^{\circ }{x}_X+P_Y^{\circ }{x}_Y$

where, $P_T=$ Total pressure

$P^{\circ }x=$ Vapour pressure of $X$ in pure state

$P^{\circ }{}_Y=$ Vapour pressure of $Y$ in pure state

$x_X=$ Mole fraction of $X=1/4$

$x_Y=$ Mole fraction of $Y=3/4$

(i) When $T=300K,P_T=550mmHg$

$\therefore 550=P_X^{\circ }\left(\frac{1}{4}\right)+P_Y^{\circ }\left(\frac{3}{4}\right)$

$\Rightarrow P_X^{\circ }+3P_Y^{\circ }=2200$ ...(1)

(ii) When at $T=300K,1$ mole of $Y$ is added,

$P_T=(550+10)mmHg$

$\therefore x_X=1/5$ and $x_Y=4/5$

$\Rightarrow 560=P_X^{\circ }\left(\frac{1}{5}\right)+P_Y^{\circ }\left(\frac{4}{5}\right)$

or $P_X^{\circ }+4P_Y^{\circ }=2800$ ...(2)

On solving equations (1) and (2), we get

$P_Y^{\circ }=600mmHg$ and $P_X^{\circ }=400mmHg$