A solution is made by dissolving 0.608 mol of nonelectrolyte solute in 845 g of

Question

A solution is made by dissolving 0.608 mol of nonelectrolyte solute in 845 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. 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

Using the formula :

dTf = Tf(benzene) - Tf(solution) = mKf

where,

m = 0.608 mol

Kf for benzene = 5.12 oC/m

Feed values,

dTf = 0.608 x 5.12 = 3.113 oC

3.113 = 5.49 - Tf(solution)

Tf(solution) = 2.38 oC is the freezing point of solution

Now,

dTb = Tb(solution) - Tb(solvent) = Kb x m

Feed values for benzene and above data,

dTb = Tb(solution) - 80.1 = 2.53 x 0.608

Tb(solution) = 81.64 oC is the boiling point of solution

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