Question

Two liquids X and Y form are 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, if 1 mole of Y is further added to this solution, 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

Answer: (C)

400 and 600

Answer 2

Initially P_­m = P⁰_X.X_X + P⁰_Y.X_Y

550 = P⁰_X $\left( \frac { 1 } { 1 + 3 } \right)$ + P⁰_Y $\left( \frac { 3 } { 1 + 3 } \right)$

or P⁰_X + 3P⁰_Y = 2200 ...(i)

When 1 mole of Y is further added to it

P_m = P⁰_X.X_X + P⁰­_Y.X_Y

560 = P⁰_X $\left( \frac { 4 } { 1 + 4 } \right)$

\ P⁰_X + 4P⁰_Y = 2800 ...(ii)

By eqs. (i) and (ii)

P⁰_X = 400 mm and P⁰­_Y = 600 mm.