Question

A 250 mL vial contains 100 mL of water with 1500 mg of benzene per liter. The vial is then filled with a 50/50 (by volume) toluene/benzene solution.

What will be the concentration of benzene in water after equilibrium is acheived? (mg/L)

Use a (mole fraction) partition coefficient, K_X for benzene in this system equal to 2476.

Answer 1

water volume = 100 ml

toulene/benzene solution = 250 - 100 = 150 ml

Let mol fraction of benzene in water = X , then mol fraction of benzene in toulene solution = 1-X

Kx = (mol fraction benzene in toulene solution / toule solution volume) / (Benzene in water / water volume)

2476 = ( 1-X)/250 / ( X) / 100

2476 = 100(1-X) / ( 250X)

619000 X = 100-100X

X = 0.00016155 = mol fraction of benzene in water

Benzene mentioned as 1500 mg per liter present . Initially we had 100 ml water , hence benzene mass = 1500 x 100/1000 = 150 mg = 0.15 g

Benzene moles = mass / Molar mass of Benzene = 0.15g / (78.11 g/mol) = 0.00192 mol

now 0.00016155 fraction of 0.00192 = 0.00016155 x 0.00192   = 3.1 x 10^-7

Thus moles of benzene in water = 3.1 x 10^-7

Benzene mass = moles x molar mass = 3.1 x 10^-7 mol x 78.11 g/mol = 2.423 x 10^-5 g = 2.423 x 10^-2 mg

             = 0.02423 mg

water volume = 100 ml = 0.1 L

Benzene concentration in mg/L = 0.02423 mg / 0.1 L = 0.2423 mg /L