Question

A solution is made by dissolving 0.531 mol of nonelectrolyte solute in 765 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution.

Answer 1

First calculate the molality of the solution as follows:

m = number of moles / mass of solvent in kg

given that 0.531 mol of nonelectrolyte solute and

765 g = benzene or 0.765 kg benzene

Therefore;

m = 0.531 m/ 0.765 kg

= 0.694 m

We know that T solvent or freezing point of benzene = 5.54 C

Kf fro benzene = 5.12 C/m

And

T solvent – T solution = iKf m

Here i = 1 for non electrolytes , Vant Hoff factor

Then;

5.5 C- T solution = 1 * 5.12 C/m * 0.694 m

5.5 C- T solution =3.55 C

T solution or the freezing point, Tf =1.95C

And T boiling point is calculated as follows:

T solution – T solvent = iKb m

T solution is the boiling point of the solution

T solvent is the boiling point of solvent , benzene 80.10 C

K b for benzene = 2.53 C/ m

Therefore;

T solution – 80.10 C = 1 * 2.53 C/ m * 0.694 m

T solution – 80.10 C = 1.76 C

T solution = 81.86 C