Question

Please help, I've been stuck on this for a while!

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 1

Freezing point is a colligative property. Did you look up the formula for freezing point depression?

Answer 2

Oh dear friend... just finished a test on solutions and freezing points, and I've gotta tell ya, that it sure isn't an easy subject...

Answer 3

But I am sure this website will help! http://chemistry.about.com/od/workedchemistryproblems/a/Freezing-Point-Depression-Example-Problem.htm

Answer 4

Good luck, fellow chemist

Answer 5

the formula for freezing point depression is deltaTf = (Freezing point of pure solvent) - (Freezing point of solution) right?

Answer 6

Yes

Answer 7

Sorry, i did not see your response before

Answer 8

\[\Delta~t = i*k_{b}*m \]

where i is the number of particles produced as your solute is dissolved in the solution
kb is our constant

and m is the molality moles of solute per kilogram of solution.

Answer 9

Step #1 convert grams of glucose to moles
Step #2 then convert grams of water to kilograms.
Step #3 take (answer from step #1)/(answer from step #2) = molality

Then

multiply that by the constant to find the freezing point depression.

Answer 10

@farranheit