Question

At ${80}^{\circ }C$ the vapour pressure of pure liquid A is 250 mm of Hg and that of pure liquid B is 1000 mm of Hg. If a solution of A and B boils at ${80}^{\circ }C$ and 1 atm pressure, the amount of A in the mixture is (1 atm = 760 mm Hg)

• A. 50 mole percent
• B. 52 mole percent
• C. 34 mole percent
• D. 48 mole percent

Answer 1

The correct option is C 34 mole percent

$P$= $P{°}_a$*$X_a$+ $P{°}_b$*($1$ - $X_a$)

$760$ = $250$*$X_a$ + $1000$*($1$ - $X_a$)

$X_a$ = $34$ mole percent