3.56 grams of an unknown non-electrolytic solute were dissolved in 40.0 g of water (Kfp = 1.86 K/m). If the freezing point of the solution was -0.820 oC, determine the molar mass of the unknown solute.

201.8 g/mol is the molar mass for the compound

Explanation:

To solve this, we apply the formula for freezing point depression:

Freezing point for solution = - (Kf . m - Freezing point por pure solvent)

Freezing point for solution =  - Kf . m + Freezing point por pure solvent

Freezing point for pure solvent = 0°C, therefore  - 0.820°C = - Kf . m

Kf = 1.86 °C/m ; m = molality (mol/kg), moles of solute in 1kg of solvent

It is more common to say 1.86 °C/m instead of K/m

0.820°C / 1.86 m°C = m → m = 0.441 molal (These moles are in 1kg of solvent, but our mass is 40 g, 1kg = 1000g)

We determine the moles of solute → Molal . mass of solvent

0.441 mol/1000g . 40 g = 0.0176 moles

Molar mass → mass of compound (g) in 1 mol → 3.56 g / 0.0176 moles = 201.8 g/mol

the answer is 303.15 or 303