At 8 0 ∘ 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

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At 8 0 ∘ 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 8 0 ∘ C and 1 atm pressure, the amount of ' A ' in the mixture is ( 1 atm = 760 mm Hg )

TestHub · Accepted Answer

​ P T ​ = P A ∘ ​ X A ​ + P B ∘ ​ X B ​ 760 = 520 X A ​ + P B ∘ ​ ( 1 − X A ​ ) ⇒ X A ​ = 0.5 ​ Thus, mole % of A = 50%

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