Question

A solution prepared by dissolving 90.0  mg of a sugar (a molecular compound and a nonelectrolyte) in 1.00 g of water froze at -0.930°C. What is the molar mass of this sugar (g/mol)? The value of K_f is 1.86°C/m.

Answer 1

Freezing point of solution = -0.930 ^oC

decrease in freezing point = (Freezing point of water) - (Freezing point of solution)

decrease in freezing point = (0.0 ^oC) - (-0.930 ^oC)

decrease in freezing point = 0.0 ^oC + 0.930 ^oC

decrease in freezing point = 0.930 ^oC

Molality of sugar = (decrease in freezing point) / (K_f)

Molality of sugar = (0.930 ^oC) / (1.86 ^oC/m)

Molality of sugar = 0.5 m

mass of water = 1.00 g = 1.00 x 10^-3 kg

moles of sugar = (molality of sugar) * (mass of water in kg)

moles of sugar = (0.5 m) * (1.00 x 10^-3 kg)

moles of sugar = 5 x 10^-4 mol

mass of sugar = 90.0 mg = 90.0 x 10^-3 g

molar mass of sugar = (mass of sugar) / (moles of sugar)

molar mass of sugar = (90.0 x 10^-3 g) / (5 x 10^-4 mol)

molar mass of sugar = 180 g/mol