Assuming 100% dissociation, calculate the freezing point and boiling point of 2.

Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 2.39 m Na2SO4(aq). Constants may be found here.

Colligative Constants

(°C/m)

Normal freezing

point (°C)

(°C/m)

Normal boiling

point (°C)

Constants for freezing-point depression and boiling-point elevation calculations at 1 atm:
Solvent Formula Kf value*

(°C/m)

Normal freezing

point (°C)

Kb value

(°C/m)

Normal boiling

point (°C)

water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol C2H6O 1.99 –117.3 1.22 78.4 carbon
tetrachloride   CCl4 29.8 –22.9 5.03 76.8 camphor   C10H16O 37.8 176
Assuming 100% dissociation, calculate the freezing point and boiling point of 2.39 m Na2SO4(aq). Constants may be found here. Number T. Number Previous & Give Up & View Solution 0 Check Answer Nex Hint e addition of solute causes a depression in freezing point and an elevation in boiling point according to tne equations where AT is the change in freezing or boiling point, K is a constant specific to the solvent, m is the molal concentration, and i is the number of dissociated ions per formula unit. Note, molality is sometimes represented as b.

Explanation / Answer

Answer – We are given, molality of Na2SO4 (aq) = 2.39 m

The Van’t off factor, i = 3 , Kf = 1.86 oC/m , Kb = 0.512 oC/m

So we know the formula

Tf = i*Kf*m

       = 3*1.86 oC/m *2.39 m

       = 13.3oC

So the Freezing point = Freezing point of pure solvent - Tf

                                   = 0.0oC - 13.3oC

                                   = -13.3oC

Tb = i*Kb*m

       = 3*0.512 oC/m *2.39 m

       = 3.67oC

So the Boiling point = Boiling point of pure solvent + Tb

                                = 100.0oC +3.67oC

                                = 103.67oC

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