Question

A solution made by dissolving aspirin, C₉H₈O₄, in chloroform, CH₃Cl had a boiling point of 61.84°C. What is the freezing point depression of this solution? Given the following information for the solvent: the boiling point and freezing points are 61.20°C and -63.50°C, respectively. K_b = 3.630 °C/m and K_f = 4.680°C/m .

Answer 1

Given that boiling point of solution is 61.84 °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

      = 3.630 °C/m

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

      = 4.680°C/m

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 =  boiling point of solution - boiling point of solvent

      = 61.84°C - 61.20°C = 0.64 °C

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

∆Tf = ∆Tb x Kf / Kb ---- Equation (3)

Then , ∆Tf = (0.64 °C ) x (4.68 ) / (3.63 ) = 0.825 °C

Therefore, freezing point depression of given solution = 0.825 °C