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At 80o C, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 80o C and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)

  • 52 mol percent

  • 34 mol percent

  • 48 mol percent

  • 50 mol percent 

Answer

D.

50 mol percent 

PT = PoXA + PoXB
760 = 520XA+ PoB(1-XA)
⇒ = 0.5
Thus, mole% of A = 50%

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