At 80° C, the vapour pressure of pure liquid A is 520 mm of Hg and that of pure liquid B is 1000 mm of 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 of Hg )

Question

At 80° C, the vapour pressure of pure liquid A is 520 mm of Hg and that of pure liquid B is 1000 mm of 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 of Hg ) (A) 50 mol percent (B) 52 mol percent (C) 34 mol percent (D) 48 mol percent

Tardigrade Admin · Accepted Answer

We have, PA°=520 mm Hg and PB°=1000 mm Hg Let mole fraction of A in solution =xA and mole fraction of B in solution =xB Then, at 1 atm pressure i.e., at 760 mm Hg, PA° xA+PB° xB=PA° xA+PB°(1-xA)=760 mm Hg ⇒ 520 xA+1000-1000 xA=760 mm Hg ⇒ xA=(1/2) or 50 mol percent

Q. At 80∘C, the vapour pressure of pure liquid A is 520mm of Hg and that of pure liquid B is 1000mm of 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 (1atm=760mm of Hg)

50 mol percent

52 mol percent

34 mol percent

48 mol percent

Q. At $8 0^{\circ} C$ , the vapour pressure of pure liquid $A$ is $520 mm$ of $H g$ and that of pure liquid $B$ is $1000 mm$ of $H g$ . If a mixture solution of $A$ and $B$ boils at $8 0^{\circ} C$ and $1$ atm pressure, the amount of $A$ in the mixture is $(1 a t m = 760 mm$ of $H g)$