Question
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?
Answers
DrBob222
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.
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.