A solution is made by dissolving 0.559 mol of nonelectrolyte solute in 901 g of

Question

A solution is made by dissolving 0.559 mol of nonelectrolyte solute in 901 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution.Constants for freezing-point depression and boiling-point elevation calculations at 1 atm. The constants are as shown in the graph.

(°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

Based on all data given above,

The molality of the solution is 0.559 (mol) / 0.901(kg) = 0.6204 m

The normal freezing point of benzene 5.49 and the freezing point depression is 5.12° C per molal.

So the freezing point will be depressed by 0.6204 * 5.12° C = 3.18° C. So the resulting freezing point will be 5.49 - 3.18 = 2.31° C.

The normal boiling point is 80.1° C and the boiling point elevation is 2.53° C per molal. So the boiling point will be elevation by 0.6204 * 2.53° C = 1.57 ° C. So the boiling point will be 80.1 + 1.57 = 81.67 ° C.

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