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

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.

Solution

Given that

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

Use the formula 

 

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

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

Use formula 

Molar fraction of A,X_A = 0.714 /(0.714 + 5.556) = 0.114

Similarly

Molar fraction of B, X_B   = 1-  X_A

= 1 − 0.114 = 0.886

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

Use formula of Henry`s law

P_B  = P^o_B × X_B

Plug the values we get

P_B= 500 × 0.886 = 443 torr

Given that total vapour pressure of the solution, p_total = 475 torr

Use the formula

P_total   = p_A + p_B

p_A      = p_total − p_B

Plug the values we get

P_A      = 475 − 443

P_A      = 32 torr

Use formula of Henry`s law again we get

P_A  = P^o_A × X_A

Plug the values we get

32 = P^o_A × 0.114

P^o_A = 32/0.114  = 280.7 torr

So that the vapour pressure of pure liquid A = 280.7 torr.