Question

A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

Answer 1

freezing point depression = freezing point constant * molal concentration of stuff dissolved

Answer 2

molal concentration of stuff dissolved in this case is:

(25.5 grams glucose) / (molecular weight of glucose) / (mass of solvent in kg)

Answer 3

molecular weight of glucose C6(H2O)6 is about 72 + 18*6 = 180 grams per mole

so molal concentration of stuff dissolved is approximately:

(25.5 grams / 180 grams-per-mole )/(.398 kg) = .356 m

*Remember that glucose doesn't dissociate like salt or strong acids. so the molal concentration of stuff dissolved is the same as the molality of glucose (and not twice like in NaCl)

so the freezing point depression is:

.356 m * -1.86 °C/m = -.662 deg Celsius (approximately)