Question

A solution is made by dissolving 0.717 mol of nonelectrolyte solute in 817 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution.

Answer 1

Ans:   Tf =1.0 deg C
           Tb = 82.3 deg C

Explanation:

We know the molal freezing and boiling point depression constants Kf and Kb values for benzene as below.

Kf= 5.12
Kb= 2.53

0.717 mol of nonelectrolyte solute dissolved in 817 g of benzene

Hence the molaliity(m) of the solute = 0.717 mol / 0.817 kg = 0.8776 m (mol/kg)

m = 0.8776 m (mol/kg)

Depression in frzing point = molality x Kf
                                          = 0.8776 x 5.12 = 4.4933

Elevation in Boiling point = molality x Kb
                                         = 0.8776 x 2.53 = 2.2203

As we know the Boiling and freezing points of benzene B.P =80.1deg C, F.P= 5.5 deg C

1) The freezing point, Tf of the solution.
     Tf = 5.5 - 4.4933 = 1.0067 deg C (1.0 deg C)

2) The boiling point, Tb of the solution.   

      Tb = 80.1 + 2.2203 = 82.3203 deg C (82.3 deg C)