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.

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anonymous · Answer

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ΔT = kf x w/m x 1000/W

ΔT = depression of freezing point
kf = freezing pointconstant
w = mass of solute
m = molecular mass of solute
W = mass of solvent

Δ T = -1.86 x 25.5/180 x 1000/398
Δ T = - 0.662°C