The freezing point of a nonelectrolyte solution containing 30.0 g of a solute dissolved in 250.0 g of water is observed to be -2.50˚c. determine the molar mass of the solute if the freezing point elevation constant, kf, for water is 1.86 oc/m and the freezing point of pure water is 0.00 oc?

Molar mass of solute is 89.28 g/m

Explanation:

Colligative property of freezing point depression to solve this:

ΔT = Kf . m . i

i = number of particles, dissolved in solution. In this case, it is a nonelectrolyte, so i = 1.

m = molalilty (mol of solute/1kg of solvent

ΔT = T° freeze pure solvent - T° freeze solution

0°C - (-2.50°C) = 1.86 °C/m . m

2.50°C / 1.86 m/°C = m

1.34 mol solute/kg solvent = m

This means, that in 1000 g of solvent, we have 1.34 moles but we have 250 g of solvent, so let's make a rule of three.

1000 g 1.34 moles

250 g (2.50 . 1.34) / 1000 = 0.336 moles

To find the molar mass, we divide mass / moles

30 g/ 0.336 moles = 89.28 g/m

the answer is symmetrical : )