Question

Solutions of two volatile liquids A and B obey Raoult’s law. At a certain temperature, it is found that when the total pressure above a given solution is 400 mm Hg, the mole fraction of A in the vapour is 0.45 and in the liquid it is 0.65. What are the vapour pressures of the two liquids in the pure state at the given temperature?

Answer 1

Given: The solution of volatile liquids A and B follows Raoult’s Law.

Total vapour pressure of the solution P_total = 400 mm Hg

Mole fraction of A in vapour phase y_A = 0.45

Mole fraction of A in liquid phase x_A = 0.65

Let the vapour pressures of A and B be `P_A^circ` and `P_B^circ` respectively.

By using Raoult’s law,

`P_A = x_AP_A^circ` and `P_B = x_BP_B^circ`

Total pressure `P_"total" = P_A + P_B`

Total pressure `P_"total"` = 400    ...(i)

The mole fraction of A in the vapour phase is given by

`y_A = P_A/P_"total"`

`y_A = (x_AP_A^circ)/400`

`0.45 = (0.65 * P_A^circ)/400`

⇒ `P_A^circ = (0.45 * 400)/0.65`

`P_A^circ` = 276.92 mm Hg

We know that,

`P_"total" = x_AP_A^circ + x_BP_B^circ`

⇒ `400 = 0.65 xx 276.92 + 0.35 xx P_B^circ`

⇒ `400 = 179.998 + 0.35 xx P_B^circ`

⇒ `0.35 xx P_B^circ = 220.002`

⇒ `P_B^circ = 220.002/0.35`

= 628.58 mm Hg

∴ The vapour pressures of the two liquids in the pure state at the given temperature are 276.92 mm Hg and 628.58 mm Hg.