Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?

Solution

Chemistry ma’am · 2 steps

Same molarity for all four, so the winner is whichever solute breaks into the most particles.

1. Count particles. Freezing point depression is a colligative property, so it depends on the number of dissolved particles. Since the molality is about the same for all four, the solute with the largest van't Hoff factor i wins.

   Δ T_f = i K_f m

   Solute	What happens in water	i
   hydrazine, N₂H₄	weak base, ionises only very slightly	about 1
   glucose	non-electrolyte, stays as whole molecules	1
   glycine	forms a zwitterion, which is still one particle	about 1
   KHSO₄	KHSO₄ → K⁺ + HSO₄⁻, then HSO₄⁻ ⇌ H⁺ + SO₄²⁻ in part	between 2 and 3

2. So. KHSO₄ is a strong electrolyte that gives at least 2 ions per formula unit, so it lowers the freezing point the most. The answer is (d).

The idea to keep. For colligative properties at equal concentration, rank solutes by i. Strong electrolytes beat weak ones, and non-electrolytes come last.

Answer from the NTA final answer key. Solution by PrepWiser.