Question

A solution is made by dissolving 0.674 mol of nonelectrolyte solute in 775 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants may be found here.

Answer 1

Answer – We are given, moles of nonelectrolyte solute = 0.674 moles

Mass of solvent = 775 g

                          = 0.775 kg

Molality of nonelectrolyte solute = 0.674 moles / 0.775 kg

                                                      = 0.870 m

We know the , Kf value for benzene = 5.12 ^oC/m

We know,

∆Tf = Kf * m

       = 5.12 ^oC/m * 0.870 m

       = 4.45^oC

We know the, freezing point solution = Freezing point of pure solvent - ∆Tf

                                                         = 5.5^oC- 4.45^oC

                                                         = 1.05^oC

Now boiling point –

∆Tb = Kb * m

       = 2.53 ^oC/m * 0.870 m

       = 2.20^oC

Boiling point of solution = Boiling point of pur solvent + ∆Tb

                                       = 80.1^oC + 2.20^oC

                                       = 82.3^oC