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)

Options

• 52 mol per cent

• 34 mol per cent

• 48 mol per cent

• 50 mol per cent

Answer 1

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 50 mol per cent.

Explanation:

Given: `P_A^circ = 520` mm Hg (vapour pressure of pure A)

`P_B^circ = 1000` mm Hg (vapour pressure of pure B)

Total pressure `P_"total"` = 760 mm Hg (boiling point condition)

By using Raoult’s law

`P_"total" = P_A^circ * x_A + P_B^circ * x_B`

Since x_B = 1 − x_A​, substitute

760 = 520 x_A​ + 1000(1 − x_A​)

760 = 520 x_A ​+ 1000 − 1000 x_A​

760 = 1000 − 480 x_A

480 x_A ​= 240

`x_A = 1/2`

x_A ​= 0.5

x_A ​= 50 per cent