Answers
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) / (Kf)
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
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