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?
Answer & explanation
Correct answer: option 1
According to Raoult's law
P = PoAXA + PoBXB
p = 500 mm Hg, XA = \(\frac{1}{3}\), XB = \(\frac{2}{3}\)
500 = \(\frac{1}{3}\)PoA + \(\frac{2}{3}\) PoB
or 1500 = PoA + 2PoB ........ (i)
On adding 1 mole of B,
p = 525 mm Hg, XA = \(\frac{1}{4}\), XB = \(\frac{3}{4}\)
525 = \(\frac{1}{4}\)PoA + \(\frac{3}{4}\) PoB
or 2100 = PoA + 3PoB ........ (ii)
Subtracting (i) from (ii)
600 = PoB
1500 = PoA + 2 x 600
PoA = 300 mm Hg.