Question

What is the freezing point of a solution that contains 36.0 g of glucose in 500.0 g of water (Kf for water is 1.86C/m. The molar mass of glucose is 180.0 g per mole.)

I have been trying to solve this for 2 days, but I can't find a good enough explanation to understand it.

Answer 1

This freezing point business is usually based on molality, that is moles solute per kilogram of solvent.

molality = freezing point depression over Kf

In this case molality -10.3 degrees over -1.85 degrees Kf = 5.53 molal

This 5.53 molal solution is made up of l000 gms water and 5.53 moles glucose at 180 grams per mole for a total mass of 1997 grams

It volume would be l997 gms over 1.50 gms/ml or 1331 ml

We know that we have 5.53 moles of glucose dissolved in l331 ml of solution so now we can find how many moles of glucose in l000 ml or one liter of solution and this will be our Molarity

5.53 moles glucose over l331 ml = X moles glucose over l000 ml solution

cross multiply and solve for X moles glucose per liter solution

X = 4.15 moles glucose per liter = 4.15 Molar