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Step-by-Step Solution
Step 1: Identify the concept
The freezing point of a solution is depressed relative to the pure solvent due to the presence of solute particles. For electrolytes, the solute dissociates into ions, increasing the effective number of particles in solution. The freezing point depression is directly proportional to the total number of particles (captured by the van’t Hoff factor, i ) and the molal concentration, m :
\Delta T_f = i \, K_f \, m
Step 2: Compare with the reference solution (0.10 M C₂H₅OH)
Ethanol ( \text{C₂H₅OH} ) is essentially non-dissociative in water (its van’t Hoff factor i \approx 1 ). Thus, for 0.10 M ethanol solution:
Effective concentration (in terms of particles) = 0.10 \times 1 = 0.10 \,\text{M}.
Step 3: Determine the van’t Hoff factor for each salt
Ba₃(PO₄)₂ dissociates into 3 Ba²⁺ and 2 PO₄³⁻ → i = 3 + 2 = 5
Na₂SO₄ dissociates into 2 Na⁺ and 1 SO₄²⁻ → i = 2 + 1 = 3
KCl dissociates into 1 K⁺ and 1 Cl⁻ → i = 1 + 1 = 2
Li₃PO₄ dissociates into 3 Li⁺ and 1 PO₄³⁻ → i = 3 + 1 = 4
Step 4: Calculate the effective concentration for each salt
Since each solution has the same initial molarity (0.10 M), multiplying by their respective i values gives the effective concentration (in terms of total particles):
Ba₃(PO₄)₂: 0.10 \times 5 = 0.50 \,\text{M (particles)}
Na₂SO₄: 0.10 \times 3 = 0.30 \,\text{M (particles)}
KCl: 0.10 \times 2 = 0.20 \,\text{M (particles)}
Li₃PO₄: 0.10 \times 4 = 0.40 \,\text{M (particles)}
Step 5: Compare each salt’s effective concentration with ethanol solution
The 0.10 M ethanol solution has an effective concentration of 0.10 M (since i=1 ). Each electrolyte listed has an effective concentration greater than 0.10 M, meaning each forms more particles in solution. Therefore, each salt solution will have a lower freezing point compared to the 0.10 M ethanol solution.
Final Answer
All four given salt solutions have freezing points lower than 0.10 M C₂H₅OH. Hence, the total number of solutions with lower freezing point is 4.
