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Question

A solution prepared by dissolving 360.0 mg of a sugar (a molecular compound and a nonelectrolyte) in 1.00 g of water froze at -3.72°C. What is the molar mass of this sugar? The value of K_f is 1.86°C/m.

The answer should be in g/mol.

Answer 1

∆T = imK

∆T = 3.72º (since water normally freezes at 0ºC)

i = 1 for a non electrolyte

m = molality = moles sugar / kg water = ?

K = freezing constant for water = 1.86º/m

Solving for m, we have...

m = ∆T / (i)(K) = 3.72 / (1)(1.86)

m = 2 moles/kg

We know we have 0.001 kg water (1.00 g) and we know we have 360 mg (0.360 g), so we can find molar mass which is grams/moles.

2 moles / kg x 0.001 kg = 0.002 moles

0.360 g / 0.002 moles = 180. g/mole = molar mass of the sugar