Question

A solution is made by dissolving 2.5 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

Calculate the mol of NaCl dissolved in 1.0kg water :


198 g / 1000 => 0.198 Kg

molality = moles solute / volume 

molality = 2.5 / 0.198

molality = 12.626 mol/kg

NaCl dissociates to form 2 ions 

increase in boiling point = molality * i * Kb

12.626 * 2 * 0.51  => 12.87 ºC

Water boils at 100.00 degrees Celsius , therefore:

12.87 + 100.00 => 112.87ºC