The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Question

Philoid · Accepted Answer

Given,

P_A^o = 450 mm Hg

P_B^o = 700 mm Hg

p_total = 600 mm of Hg

By using Rault's law,

p_total = P_A + P_B

p_total = P_A^ox_A + P_B^ox_B

p_total = P_A^ox_A + P_B^o( 1 - x_A )

p_total = (P_A^o- P_B^o)x_A + P_B^o

600 = (450 - 700) x_A + 700

-100 = -250 x_A

x_A = 0.4

∴ x_B = 1 - x_A

x_B = 1 – 0.4

x_B = 0.6

Now,

P_A = P_A^ox_A

P_A = 450 × 0.4

P_A = 180 mm of Hg

and

P_B = P_B^ox

P_B = 700 × 0.6

P_B = 420 mm of Hg

Composition in vapour phase is calculated by

Mole fraction of liquid,

=

= 0.30

Mole fraction of liquid,

=

= 0.70

Note: Alternate method to find the Mole fraction of liquid B is

= 1 - Mole fraction of liquid A

= 1 – 0.30

= 0.70