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)

Options

A. 50 mol percent
B. 52 mol percent
C. 34 mol percent
D. 48 mol percent

Answer

A

Explanation

We have, `P_A^@ = 520 mm Hg and P_B^@=1000 mm Hg`
Let mole fraction of A in solution `= x_A`
and mole fraction of B in solution `=x_B`
Then, at 1 atm pressure i.e., at 760 mm Hg,
`P_A^@ x_A + P_B^@x_B =P_A^@ x_A + P_B^@(1-x_A) = 760 mm Hg`
`implies 520 x_A + 1000 - 1000 x_A = 760 mm Hg`
`implies x_A = 1/2` or 50 mol percent