At 300 K, the vapour pressure of an ideal solution containing 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg. The vapour pressure if the solution increases by 25 mm of Hg if one more mole of B is added to the above ideal solution at 300 K. What is the vapour pressure of A in its pure state?

300 mm of Hg

According to Raoult's law

P = P^o_AX_A + P^o_BX_B

p = 500 mm Hg, X_A = \(\frac{1}{3}\), X_B = \(\frac{2}{3}\)

500 = \(\frac{1}{3}\)P^o_A + \(\frac{2}{3}\) P^o_B

or 1500 = P^o_A + 2P^o_B    ........ (i)

On adding 1 mole of B,

p = 525 mm Hg, X_A = \(\frac{1}{4}\), X_B = \(\frac{3}{4}\)

525 = \(\frac{1}{4}\)P^o_A + \(\frac{3}{4}\) P^o_B

or 2100 = P^o_A + 3P^o_B    ........ (ii)

Subtracting (i) from (ii) 

600 = P^o_B

1500 = P^o_A + 2 x 600

P^o_A = 300 mm Hg.