For a 0.350 m solution, calculate boiling-point elevation ΔTb using Kb = 0.512 °C kg/mol and i ≈ 2.

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Multiple Choice

For a 0.350 m solution, calculate boiling-point elevation ΔTb using Kb = 0.512 °C kg/mol and i ≈ 2.

Explanation:
Boiling-point elevation is a colligative property, so it depends on how many solute particles are in solution, not on their identity. The formula is ΔTb = i Kb m, where m is the molality, Kb is the boiling-point elevation constant, and i is the van’t Hoff factor (how many particles the solute produces). Here, the solution has a molality of 0.350 mol/kg and the solute dissociates into two particles, so i ≈ 2. Compute ΔTb = (2)(0.512 °C·kg/mol)(0.350 mol/kg) = 0.3584 °C, which rounds to 0.359 °C. Therefore the boiling-point elevation is about 0.359 °C.

Boiling-point elevation is a colligative property, so it depends on how many solute particles are in solution, not on their identity. The formula is ΔTb = i Kb m, where m is the molality, Kb is the boiling-point elevation constant, and i is the van’t Hoff factor (how many particles the solute produces). Here, the solution has a molality of 0.350 mol/kg and the solute dissociates into two particles, so i ≈ 2. Compute ΔTb = (2)(0.512 °C·kg/mol)(0.350 mol/kg) = 0.3584 °C, which rounds to 0.359 °C. Therefore the boiling-point elevation is about 0.359 °C.

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