Of the following four aqueous solutions, total number of those solutions whose freezing point is lower than that of 0.10 M C₂H₅OH is __________ (Integer answer)
(i) 0.10 M Ba₃(PO₄)₂
(ii) 0.10 M Na₂SO₄
(iii) 0.10 M KCl
(iv) 0.10 M Li₃PO₄

Solution

Chemistry ma’am · 3 steps

Same concentration everywhere, so just compare how many particles each solute breaks into.

1. What lowers freezing point. Depression in freezing point is a colligative property: Δ T_f = i K_f m. Every solution here is 0.10 M in water, so the one with the larger van't Hoff factor i freezes lower.
2. Reference solution. C₂H₅OH is a non-electrolyte and stays as whole molecules, so i = 1. Any solute that gives more than one particle per formula unit beats it.

   Solute	Ions formed	i	Lower freezing point than ethanol?
   Ba₃(PO₄)₂	3Ba²⁺ + 2PO₄³⁻	5	Yes
   Na₂SO₄	2Na⁺ + SO₄²⁻	3	Yes
   KCl	K⁺ + Cl⁻	2	Yes
   Li₃PO₄	3Li⁺ + PO₄³⁻	4	Yes

3. So. All four electrolytes have i > 1, so all four freeze below 0.10 M ethanol. The answer is 4.

The common slip: 3. Leaving out Ba₃(PO₄)₂ because it is sparingly soluble gives 3. The question treats every solution as a full 0.10 M, so count it with i = 5.

The idea to keep. When concentrations are equal, rank colligative effects by i × m alone. A non-electrolyte has i = 1, so any dissolved electrolyte gives a bigger freezing point depression.

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