Question

A solution is made by dissolving 0.577 mol0.577 mol of nonelectrolyte solute in 893 g893 g of benzene. Calculate the freezing point, ?f,Tf, and boiling point, ?b,Tb, of the solution. Constants can be found in the table of colligative constants.

Answer 1

Molality (m) = moles of solute / kg of solution

893 g × ( 1 kg / 1000 g ) = 0.893 kg

molality = 0.577 mol / 0.893 kg

molality = 0.646 mol/kg

Depression ion freezing point

∆Tf = i×m×Kf

for non electrolyte i = 1

m = 0.646 mol/kg

For benzene Kf = 5.12 °C/m

∆Tf = 1×0.646 m× 5.12 °C/m

∆Tf = 3.31 °C

∆Tf = Tf ( pure solvent) - Tf (solution)

3.31 °C = 5.5 °C = Tf solution

Tf solution = 5.5 -3.31

Tf solution = 2.19 °C

Here 5.5 °C freezing point of pure benzene solution.

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∆Tb = i× m × kb

∆Tb = 1× 0.646 m × 2.65 °C/m

∆Tb = 1.71 °C

∆Tb = Tb (solution) - Tb ( pure Solvent)

Tb solution =80.1 °C + 1.71 °C

Tb solution = 81.8 °C

Freezing point of the solution =