Question

# Two liquids 'A' and 'B' form an ideal solution. At 300K, the vapour pressure of a solution containing 1 mol of A and 3 mol of B is 550 mm Hg. At the same temperature if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of 'A' and 'B' in their pure states.

A
pA=400 mm Hg and pB=600 mm Hg
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B
pA=600 mm Hg and pB=400 mm Hg
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C
pA=300 mm Hg and pB=700 mm Hg
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D
pA=700 mm Hg and pB=300 mm Hg
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Solution

## The correct option is A p∘A=400 mm Hg and p∘B=600 mm Hg Let 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=14 Mole fraction of B, XB=34 Total vapour pressure of solution, =XA.p∘A+XB.p∘B550=14p∘A+34p∘Bor2200=p∘A+3p∘B ....eq(i) After addition of 1 mol of B, Total vapour pressure of solution (1 mol of A + 4 mol B), =15p∘A+45p∘B560=15p∘A+45p∘Bor 2800=p∘A+4p∘B ....eq(ii) Solving equations (i) and (ii), we get, p∘B=600 mm Hg⇒ vapour pressure of pure 'B' p∘A=400 mm Hg⇒ vapour pressure of pure 'A'

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