Posted by Monique on .
A glucose solution contains 55.8g of glucose (C6H12O6) in 455g of water. Calculate the freezing poing and boiling point of the solution. (Assume a density of 1.00 g/ml for water.
This what I did, I got moles of C6H12O6, 0.89 moles C6H12O6
then I got the mass of the solution 455g/1000kg, 0.455 I then divided 0.89 solute mol/0.455 kg solvent. I was going to do the formula for freezing and boiling but what do I do with the density. I hope I havent confused you. I guess I don't know what to do with the density or if I have set this up right.
You ignore the density. Since it is 1.00 g/mL, that just means that you could calculate the volume of the solution but you don't need that.
moles glucose = 55.8/180.16 = not 0.89 but closer to 0.3 but that isn't exact.
m = moles/kg solvent = 0.3/0.455 = x
delta T = Kf*m and
delta T = Kb*m