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? Show all of the work needed to solve this problem.
Answers
DrBob222
mols glucose = grams/molar mass.
Sole for mols.
molality = m = mols/kg solvent.
Solve for m.
delta T = Kf*m
Solve for delta T. That's the freezing point depression.
Sole for mols.
molality = m = mols/kg solvent.
Solve for m.
delta T = Kf*m
Solve for delta T. That's the freezing point depression.