Question

assume that hexane and benzene mix to form an ideal solution. the vapor pressure of hexane and benzene at 25C are 151.3 torr and 100.8 torr respectively. The total vapor pressure of the solution is 130.1 torr at 25C. What is the mole fraction of benzene in the solution and in the vapor?

Answer 1

The probelem can be solved using Raoult's Law which sates that

Total pressure of the solution (p) = Sum of the partial pressures of the components (xp_a + x_b p_b)

Here p_a and p_b are partial pressures

x_a and x_b are mole fractions.

a) Mole fractions in the solution

Total pressure of the solution p= p_hexane x_hexane + p_benzene x_benzene

130.1= 151.3(x)+ (1-x) 100.8

130.1= 151.3x + 100.8 -100.8 x

130.1 = 50.5x +100.8

x= 0.5802 (mole fraction of hexane in solution)

mole fraction of benzene in solution = 1-x = 1-0.5802 =0.4198

b) Mole fractions in the vaapour

partial vapor pressure of benzene =p_benzene x_benzene = (100.8) (0.4198) =42.3158

Now mole fraction of benzene = 42.3158 /130.1 =0.3252

Simillarly we can calculatefor hexane also

partial vapor pressure of hexane = p_hexane x_hexane = (151.3) (0.5802) =87.7843

Now mole fraction of hexane = 87.7843 /130.1 =0.6747