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

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

• A. 0.180 °C
• B. 0.728 °C
• C. 1.144 °C
• D. 0.359 °C

Answer: (D)

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.