A freezing point depression problem: A solute lowers the freezing point by 6.36 °C in 1.00 kg of water. Using Kf for water = 1.86 °C/m and assuming the solute dissociates into two particles, which compound is the unknown solute? (CsCl, KCl, LiF, NaCl)

• A. NaCl
• B. CsCl
• C. KCl
• D. LiF

Answer: (A)

Freezing point depression scales with the number of solute particles in solution. For an electrolyte that dissociates into two ions, the depression is ΔTf = i × Kf × m, and here i is 2.

Plugging in the numbers: ΔTf = 6.36 °C, Kf = 1.86 °C·kg/mol, and the solvent is 1.00 kg, so m = ΔTf / (i × Kf) = 6.36 / (2 × 1.86) ≈ 1.71 m. This means 1.71 moles of solute per kilogram of water, or about 1.71 mol of solute for 1.00 kg of solvent.

Since the problem specifies the solute dissociates into two particles, it must behave as a 1:1 electrolyte in solution, producing two ions per formula unit. All the listed salts—NaCl, KCl, CsCl, LiF—fit that description, so any of them could be the solute under the given assumption. NaCl is a common, representative example of a two-particle electrolyte, which is why it’s the standard pick in this context. The key takeaway is understanding how to use the freezing point depression equation with the appropriate i to find the required molality.