Question

A solution is made by dissolving 1.8 moles of sodium chloride (NaCl) in 198 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution? Show all of the work needed to solve this problem.

Answer 1

First find the molality.\[m = \frac {moles_{solute}}{kilograms_{solvent}} = \frac {1.8 mol}{0.198 kg} = 9.1 m\]Now use the boiling point elevation formula. Your van't Hoff factor is 2 since NaCl dissociates into 2 ions. The boiling point elevation constant for water is 0.52 C/m.\[\Delta T = i K_b m = 2 \times 0.52 °C/m \times 9.1 m = 9.5 °C\]So the boiling point of this solution will be 9.5 degrees higher than the boiling point of water. So 100°C + 9.5°C = 109.5 °C.