  Question

# At 300 K, the vapour pressure of an ideal solution containing 1 mole of A and 3 moles of B is 500 mm Hg. At the same temperature, 2 moles of B are added to this solution. The vapour pressure of the solution increases by 10% of the original vapour pressure. The correct statement(s) about the vapour pressure are :The vapour pressure of A in the pure state is 1300 mm HgThe vapour pressure of B in the pure state is 650 mm HgThe ratio of the final vapour pressure to the initial vapour pressure is 2 : 1The ratio of the vapour pressure of pure B to the vapour pressure of pure A is 13 : 1

Solution

## The correct options are B The vapour pressure of B in the pure state is 650 mm Hg D The ratio of the vapour pressure of pure B to the vapour pressure of pure A is 13 : 1 nA = 1; nB = 3; PT = 500 mm Hg When 2 moles of B is added  nA = 1; nB = 5 As it is given that vapour pressure increases by 10%, P′T=500+500×10100=550 500 = P∘A(11+3)+P∘B(31+3) 2000 = P∘A+3P∘B−−−−−(1) 550 = P∘A(16)+P∘B(56) 3300 = P∘A+5P∘B−−−−−(2) Solving (1) and (2)  we get,  P∘A=50 mm Hg P∘B=650 mm Hg P∘AP∘B=65050=13:1  Suggest corrections   