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A solution is made by dissolving 1.8 moles of sodium chloride (NaCl) in 0.198
kilograms 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 the steps taken to solve
this problem. (4 points)
[Hint: water boils at 100°C]

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Answer:

T(bp) = 104.64°C

Explanation:

molality = moles solute/Kg Solvent = 1.8 mol/0.198 Kg = 9.1 molal

ΔT(bp) = Kb·m = 0.51°C/m x 9.1 m = 4.64⁰C

Solution bp = 100°C + 4.64°C = 104.64°C

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