Question

 

The Kf for acetic acid is 3.90 oC/m and its freezing point is 16.6 oC. Calculate the freezing point of a 1.25 M solution of benzoic acid, C7H6O2 in acetic acid. The density of this solution is 1.067 g/mL. The density of pure acetic acid is 1.049 g/mL.

Answer 1

delta Tf = i*Kf*m

i = Vant hoff factor

Kf = freezing point constant

m = molality

delta Tf = depression in freezing point

Let us consider 1 L of solution

density of solution = 1.067 g /ml

1.067 = m / 1000

mass of solution = 1067 gms

density of acetic acid = 1.049 g/ml

1.049 = mass / 1000

mass = 1049 gms of acetic acid

Mass of benzoic acid = 1067-1049 = 18 gms

Molality = Moles of solute / mass of solvent in Kg

Molality = 18*1000 / 1049*122.123 = 0.14

Benzoic acid ---> C6H5COOH

C6H5COOH -----> C6H5COO- + H+

1.25 0 0

1.25-x x x

Ka = x^2 / 1.25-x = [C6H5COO-] [H+] / [C6H5COOH]

6.5 x 10-5. = x^2 / 1.25-x

8.125*10^-5 - 6.5*10^-5x - x^2 = 0

solve x by quadratic equation :

x = 8.98*10^-3

Vant hoff factor = sum of all species = 1.25+x = 1.25+8.98*10^-3 = 1.25898

delta Tf = 1.25898*0.14*3.9 = 0.689 C

Freezing point = 16.6 - 0.689 = 15.31 C