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

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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 )

Quantrex Academy · Accepted Answer

Correct answer: D. 50 mol percent

aligned & P_T=P_A^ X_A+P_B^ X_B & 760=520 X_A+P_B^ (1-X_A ) & X_A=0.5 aligned Thus, mole % of A=50 %

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 )

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