100 g of liquid A (molar mass 140 g mol^–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Given-

Mass of liquid A, W_A = 100g, Molar mass, M_A = 140 g mol^-1

Mass of liquid B, W_B = 1000 g, Molar mass, M_B = 180 g mol^-1

Using the formula below calculate the no. of moles in liquid A and B.

→

Number of moles of liquid A, M_A = 100/140 = 0.714 mol^-1

Number of moles of liquid B, M_B = 1000/ 180 = 5.556 mol^-1

Using the formula,

,

we calculate the mole fraction of liquids A and B.

→ Mole fraction of A,

∴

→ Mole fraction of B,

Vapour pressure of pure liquid B, P^o_B = 500 torr (given)

According to Henry`s law,

→

Having given the total vapour pressure of the solution, P_total = 475 torr,

→

Using Henry`s law again to get the vapour pressure of pure liquid A,

Thus, the vapour pressure of pure liquid A = 280.7 torr and

vapour pressure of liquid A in the solution = 32 torr