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

Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 2.78 m Na2SO4(aq). Constants may be found here.

(°C/m)

Normal freezing

point (°C)

(°C/m)

Normal boiling

point (°C)

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

Explanation / Answer

Na2SO4   , i value = 3

molality = 2.78 m

freezing point of water = 0

To - Tf = Kf x i x m

0 - Tf = 1.86 x 3 x 2.78

freezinf point Tf = -15.5oC

Tb - To = Kb x i x m

Tb - 100 = 0.512 x 3 x 2.78

boiling point Tb = 104.27oC

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