Question

The freezing point depression of a solution of nitrobenzene and a nonionic unknown was used to determine the molar mass of the unknown. Time-temperature data for the cooling of nitrobenzene and for the cooling of a solution containing 50.0 g of nitrobenzene and 5.00 mL of a nonionic liquid unknown. Density of the unknown is 0.714 g mL^-1. And the Kf of nitrobenzene is 6.87 degrees C Kg mol^-1.

What is the freezing point of the unknown solution? I know the answer is -0.9 celsius but I would to know how to solve it.

Answer 1

Mass of nitrobenzene = 50.0 g = 0.05 kg

Volume of unknown liquid = 5.00 mL

Density of unknown = 0.714 g/mL

Mass of unknown = 0.714 * 5.0

= 3.57 g

Kf of nitrobenzene = 6.87 oCKg /mol

Freezing point of solution = - 0.9 oC

Freezing point of nitrobenzene = 5.25 oC

Depression in freezing point, Tf = 5.25 + 0.9

= 6.15 oC

Tf = i * Kf * m

6.15 = 1 * 6.87 * m ......(i = 1 because unknown is non ionic)

m = 6.15 / 6.87

= 0.89

Molality = Moles of solute / Mass of solvent (kg)

0.89 = Moles of solute / 0.05

Moles of solute = 0.89*0.05

= 0.0445

Molar mass of unknown = 3.57 / 0.045

= 79.33 g/mol