JEE Main · 2009 · 12 Apr 2009 · Offline · Solutions

Two liquids X and Y form an ideal solution. At 300 K , vapour pressure of the solution containing 1 mol of

X

and 3 mol of

Y

is 550 mm Hg . At the same temperature, if 1 mol of

Y

is further added to this solution, vapour pressure of the solution increases by 10 mm Hg . Vapour pressure (in mmHg ) of

X

and Y in their pure states will be, respectively :

Answer: (C)

Solution

An ideal solution obeys

P_{T} = P_{X}^{\circ} x_{X} + P_{Y}^{\circ} x_{Y}

. With one mole of X and three of Y the first condition is

550 = \frac{1}{4} P_{X}^{\circ} + \frac{3}{4} P_{Y}^{\circ}

, or

P_{X}^{\circ} + 3 P_{Y}^{\circ} = 2200

. Adding another mole of Y makes the mixture one of X to four of Y and raises the pressure to 560, giving

P_{X}^{\circ} + 4 P_{Y}^{\circ} = 2800

. Subtracting the first relation from the second yields

P_{Y}^{\circ} = 600

mm Hg, and back-substitution gives

P_{X}^{\circ} = 400

mm Hg.

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option