For boiling-point elevation with i ≈ 2 and Kb = 0.512 °C kg/mol, what is ΔTb for a 0.350 m solution?

For boiling-point elevation with i ≈ 2 and Kb = 0.512 °C kg/mol, what is ΔTb for a 0.350 m solution?

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

Answer: (C)

Boiling-point elevation is a colligative property, depending on the number of dissolved particles. Use ΔTb = i × Kb × m. With a dissociating solute into about two particles, i ≈ 2. The given molality is 0.350 m, and Kb is 0.512 °C·kg/mol. Multiply: 2 × 0.512 × 0.350 = 0.3584 °C, which rounds to 0.359 °C. So the rise in boiling point is about 0.359 °C.

Boiling-point elevation is a colligative property, depending on the number of dissolved particles. Use ΔTb = i × Kb × m. With a dissociating solute into about two particles, i ≈ 2. The given molality is 0.350 m, and Kb is 0.512 °C·kg/mol. Multiply: 2 × 0.512 × 0.350 = 0.3584 °C, which rounds to 0.359 °C. So the rise in boiling point is about 0.359 °C.