Question

Two liquids $X$ and $Y$ form an ideal solution at 300 K, vapour pressure of the solution containing 1 mole of $X$ and 3 mole of $Y$ is 550 mm Hg. At the same temperature, 1 mole of $Y$ is further added to this solution and the vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) 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

The correct option is C

400 and 600

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

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

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

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

On solving (i) and (ii), we get,

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