Question

A solution is made by dissolving 1.00 moles of sodium chloride (NaCl) in 155 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

molality = 1.00 mol/ 0.155 Kg = 6.45
delta T = m x kb x i
i = Vant'Hoff factor for NaCl = 2

delta T = 6.45 x 0.51 x 2 = 6.6
boiling point = 100 + 6.1 = 106.1 °C

moles glucose = 15.5 g / 180 g/mol=0.0861
m = 0.0861 mol/ 0.245 Kg =0.351
delta T = 0.351 x 1.86 =0.65
freezing point = - 0.65 °C

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