At 80 ∘ 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 80 ∘ C and 1 atm pressure, the amount of 'A' in the mixture is: ( 1 atm =760 mm Hg )

Question

At 80 ∘ 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 80 ∘ C and 1 atm pressure, the amount of 'A' in the mixture is: ( 1 atm =760 mm Hg ), 52 mole percent, 34 mole percent, 48 mole percent, 50 mole percent

MARKS App · Accepted Answer

P T = P A ∘ X A + P B ∘ X B Mixture solution boil at 1 atm = 760 mm Hg = total pressure of solution. 7 6 0 = 520X A + 1 0 00 1 - X A ⇒ 7 6 0 = 5 2 0 X A + 1 0 0 0 - 1 0 0 0 X A ⇒ 4 8 0 X A = 2 4 0 ​ X A = 0.5 , mole percentage of A=50%

Search any question & find its solution

Looking for more such questions to practice?