Question

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.

Answer 1

Number of moles of liquid A,

= 0.714 mol

Number of moles of liquid B,

= 5.556 mol

Then, mole fraction of A,

= 0.114

And, mole fraction of B, x_B = 1 − 0.114

= 0.886

Vapour pressure of pure liquid B, = 500 torr

Therefore, vapour pressure of liquid B in the solution,

= 500 × 0.886

= 443 torr

Total vapour pressure of the solution, p_total = 475 torr

Vapour pressure of liquid A in the solution,

p_A = p_total − p_B

= 475 − 443

= 32 torr

Now,

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.