Question

The boiling point of an aqueous solution is 101.42 °C. What is the freezing point?

Answer 1

Given that boiling point of an aqueous solution is 101.42 °C

We know that

Elevation of Boiling point ∆Tb = Kb.m ---- Equation (1)

Depression in Freezing point ∆Tf = Kf.m ---- Equation (2)


Kb = Ebullioscopic constant (boiling point elevation constant) of the solvent

      = 0.512 water

Kf = Cryoscopic constant (freezing point depression constant) of the solvent

      = 1.86 for water

m = molality of the solution

Divide the Equation (2) with Equation (1)

Then, ∆Tb / ∆Tf = Kb/ Kf

On rearranging

∆Tf = ∆Tb x Kf / Kb ---- Equation (3)

Since the normal boiling point of water = 100°C,

∆Tb = 101.42 °C - 100 °C = 1.42 °C

Substitute the values of ∆Tb, Kf, Kb in Equation (3)

∆Tf = ∆Tb x Kf / Kb ---- Equation (3)

Then , ∆Tf = (1.42 °C ) x (1.86 ) / (0.512 ) = 5.16 °C

Since the normal freezing point of water = 0°C and ∆Tf = 5.16 °C

Then, freezing point of the aqueous solution = 0°C -5.16 °C

                                                                       = -5.16 °C

Therefore, freezing point of the aqueous solution = -5.16 °C