For a 0.10 m solution of a non-electrolyte in water, with Kf = 1.86°C·kg/mol, what is the freezing point depression ΔTf?

For a 0.10 m solution of a non-electrolyte in water, with Kf = 1.86°C·kg/mol, what is the freezing point depression ΔTf?

• A. 0.186°C
• B. 0.156°C
• C. 0.093°C
• D. 0.372°C

Answer: (A)

Freezing point depression is a colligative property, so it depends on how many solute particles are in solution, not their identity. For a non-electrolyte, the solute doesn’t dissociate, giving a van’t Hoff factor i = 1. The change in freezing point is ΔTf = i Kf m. With m = 0.10 m and Kf for water = 1.86 °C·kg/mol, you get ΔTf = 1 × 1.86 × 0.10 = 0.186 °C. So the solution’s freezing point is lowered by 0.186 °C (the actual freezing point would be Tf(solvent) − 0.186 °C, but the question asks for the depression itself).

Freezing point depression is a colligative property, so it depends on how many solute particles are in solution, not their identity. For a non-electrolyte, the solute doesn’t dissociate, giving a van’t Hoff factor i = 1. The change in freezing point is ΔTf = i Kf m. With m = 0.10 m and Kf for water = 1.86 °C·kg/mol, you get ΔTf = 1 × 1.86 × 0.10 = 0.186 °C. So the solution’s freezing point is lowered by 0.186 °C (the actual freezing point would be Tf(solvent) − 0.186 °C, but the question asks for the depression itself).