Assuming equal concentrations and complete dissociation, arrange these aqueous solutions by their freezing points. # Highest freezing point | Lowest freezing point | | | | | | | | | | | | | | | | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Answer Bank | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | K₂SO₄ | Li₃PO₄ | NaNO₃ | | | | | | | | | | | | | |
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Answer

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Step 1
: Understand the problem

The problem asks us to arrange aqueous solutions by their freezing points, assuming equal concentrations and complete dissociation. The solutions given are K₂SO₄, Li₃PO₄, and NaNO₃.

Step 2
: Know the concept

When an electrolyte (like the salts given) is added to water, it dissociates into ions. This increases the number of particles in the solution, which lowers the freezing point. The freezing point depression (ΔTf) is calculated using the formula ΔTf = Kf · m, where Kf is the cryoscopic constant for water (1.86 °C·kg/mol) and m is the molality of the solution. Since the concentrations are equal, the molality will be the same for all solutions, so the relative freezing point depression will depend only on the Kf values of the salts.

Final Answer

1. Li₃PO₄ 2. NaNO₃ 3. K₂SO₄ The freezing point decreases as the Kf value increases, so Li₃PO₄ has the highest freezing point, followed by NaNO₃ and K₂SO₄.