Question

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 10mm Hg.So, vapour pressure (in mm Hg) of X and Y in their pure states will be, respectively,

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

Answer 1

The correct option is A

400 and 600

Mole fraction of X = $\frac{1}{1+3}=\frac{1}{4}$
Mole fraction of Y = $\frac{3}{1+3}=\frac{3}{4}$

Total pressure = Partial pressure of X * Mole fraction of X + partial pressure of Y * mole fraction of Y

$550=\frac{P_x}{4}+\frac{3P_y}{4}$ ------------1

$2200=P_x^{\circ }+3P_y^{\circ }$

When one mole of Y is added then mole fraction of X becomes $\frac{1}{1+4}=\frac{1}{5}$ and mole fraction of Y becomes $\frac{4}{1+4}=\frac{4}{5}$ . Now the total pressure given as

$560=\frac{P_x}{5}+\frac{4P_y}{5}$ ------------2

$2800=P_x^{\circ }+4P_y^{\circ }$

Solving eqns 1 and 2 , we get $P_x^{\circ }$ = 400 mm Hg and $P_y^{\circ }$ = 600 mm Hg.