Answered

Can someone please explain how to do this?
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.

Answer :

MissPhiladelphia
We use one concept of colligative properties which is freezing point depression. We solve it as follows:

ΔTf = kf (m)      where kf is the freezing point depression constant and m is the molality of the solution

m = n solute / m solvent = 10.20 g (1 mol / 180.18 g ) / .355 kg = 0.16 mol/kg

ΔTf = kf (m) 
ΔTf = -1.86 (0.16) = - 0.2976 °C

Hope this answers the question.

Other Questions