A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m?

Answer 1

moles = grams/molar mass
Solve for moles.

molality = moles/kg solvent
Solve for molality

delta T = Kf*molality
Solve for delta T, then for the freezing point knowing the normal freezing point is zero C.