Calculate the freezing‑point depression (ΔT_f) of a solution prepared by dissolving 6.00 g of aluminium sulfate octahydrate (Al₂(SO₄)₃·18H₂O, M = 666.4 g·mol⁻¹) in 250.0 g of water. Assume complete dissociation into five ions and use K_f = 1.86 K·kg·mol⁻¹ for water.
Answer
ΔT_f = 1.66 K
The correct answer is ΔT_f = 1.66 K — Use the freezing‑point depression formula: ΔT_f = i·K_f·m = 5 × 1.86 K·kg·mol⁻¹ × 0.0360 mol·kg⁻¹ = 0.3348 K × 5 = 1.67 K (rounded to 1.66 K).
Step-by-step solution
- Determine the number of moles of Al₂(SO₄)₃·18H₂O: n = 6.00 g / 666.4 g·mol⁻¹ = 9.01×10⁻³ mol.
- Calculate the molality of the solute: m = n / mass of solvent (kg) = 9.01×10⁻³ mol / 0.2500 kg = 0.0360 mol·kg⁻¹.
- Identify the van 't Hoff factor i for Al₂(SO₄)₃, which dissociates into 2 Al³⁺ and 3 SO₄²⁻, giving i = 5.
- Use the freezing‑point depression formula: ΔT_f = i·K_f·m = 5 × 1.86 K·kg·mol⁻¹ × 0.0360 mol·kg⁻¹ = 0.3348 K × 5 = 1.67 K (rounded to 1.66 K).
Common mistake: Using the molarity instead of molality or forgetting to multiply by the van 't Hoff factor i
Concept tested: Freezing‑point depression with van 't Hoff factor
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