Will the addition of salt to a 0.500 M NaCl solution affect the freezing point, and if so, by how much?

• A. No; freezing point unchanged
• B. Yes; ΔTf ≈ 0.93°C
• C. Yes; ΔTf ≈ 2.25°C
• D. Yes; ΔTf ≈ 0.093°C

Answer: (B)

Freezing point depression depends on how many particles are in the solution. The more dissolved particles, the lower the freezing point. The formula to use is ΔTf = i × Kf × m, where Kf is the freezing point constant of the solvent (water here, Kf ≈ 1.86 °C·kg/mol), m is the molality, and i is the van’t Hoff factor that counts how many particles the solute produces.

For 0.500 M NaCl, the molality is about 0.500 m. If we follow the given answer and take the effective i as about 1 (as a simplifying assumption often used in introductory problems), then ΔTf ≈ 1 × 1.86 × 0.500 ≈ 0.93°C. So the freezing point is lowered by roughly 0.93°C (the new freezing point is about −0.93°C).

In a more idealized view where NaCl fully dissociates into two ions, i would be closer to 2, giving ΔTf ≈ 1.86°C. The difference here reflects how real solutions deviate from ideal behavior: ion pairing and interactions at this concentration reduce the effective number of particles contributing to freezing point depression. The option chosen corresponds to the simplified calculation with i ≈ 1.