    Question

# Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 2 mole of A and 3 mole of B is 500 mm Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increses by 20 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg);

A
pA=400 mm Hg and pB=600 mm Hg
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B
pA=300 mm Hg and pB=700 mm Hg
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C
pA=320 mm Hg and pB=620 mm Hg
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D
pA=430 mm Hg and pB=680 mm Hg
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Solution

## The correct option is C p∘A=320 mm Hg and p∘B=620 mm HgLet the vapour pressure of pure 'A' be p∘A and the vapour pressure of pure 'B' be p∘B Initially, Mole fraction of A, XA=25 Mole fraction of B, XB=35 Total vapour pressure of solution, =XA.p∘A+XB.p∘B500=25p∘A+35p∘Bor2500=2p∘A+3p∘B ....eq(i) After addition of 1 mol of B, Total vapour pressure of solution (2 mol of A + 4 mol B), =26p∘A+46p∘B520=26p∘A+46p∘Bor 3120=2p∘A+4p∘B ....eq(ii) Solving equations (i) and (ii), we get, p∘B=620 mm Hg⇒ vapour pressure of pure 'B' p∘A=320 mm Hg⇒ vapour pressure of pure 'A'  Suggest Corrections  0      Explore more