Determine the mass of a nonvolatile, nonionizing compound that must be added to 4.79 kg of water to lower the freezing point to -1.3°C if the molar mass of the compound is 50.0 g/mol.

The mass we used from the solute is 175 g

Explanation:

ΔT = Kf . m . i → Formula for colligative property of freezing point depression

ΔT = T° freezing of pure solvent - T° freezing of solution

0°C - (-1.3°C) = 1.3 °C

Kf if the cryoscopic constant of solvent. In this case, for water and the value is 1.86 °C/m

m = molality → moles of solute / 1 kg of solvent

Let's replace → 1.3°C = 1.86°C . m

m = 1.3°C / 1.86 m  → 0.731 mol/kg

0.731 mol/kg is the molality, moles of solute in 1kg of solvent. But in this solution we do not have 1 kg of solvent, we have 4.79 kg. So:

0.731 mol/kg . 4.79 kg = 3.501 moles

This is the moles of the solute we used, so let's convert them to mass:

3.501 mol . 50 g / 1 mol = 175 g

Note: Freezing point depression formula is completed with the Van't Hoff factor (i); this number corresponds to the ions dissolved in the solution.

In this case the solute is non volatile and nonionizing, so i = 1

ΔT = Kf . m . i