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Question

At 80oC, the vapour pressure of pure liquid A is 250 mm of Hg and that of pure liquid B is 1000 mm of Hg. If a solution of A and B boils at 80o and 1 atm pressure, the amount of A in the mixture is?

(1 atm =760 mm Hg).

A
50 mole percent
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B
52 mole percent
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C
32 mole percent
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D
48 mole percent
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Solution

The correct option is C 32 mole percent
1 atm = 760 mm Hg = PT

PT=P0AXA×P0BXB

760=250XA+1000(1XA)

240=750XA

XA=0.32

Hence mole % of A =32%

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