Question

A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 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

@JFraser

Answer 2

@thomaster

Answer 3

what's the equation for freezing-point depression?

Answer 4

I think it's
ΔTF = KF · b · i

Answer 5

Am I right or is that the wrong equation? :/
I tend to mix things up sometimes

Answer 6

Would I by any chance use molality to solve this?

Answer 7

And after this I have one more similar question, I just really need help figuring this out @JFraser

Answer 8

you would use molality, that's what the lowecase \(m\) is for in your equation

Answer 9

I know that molality = the moles of solute/ the kilograms of the solvent

Answer 10

I juts don't know how to apply that to the question... And idk if that's what I would need to find the freezing point

Answer 11

*just

Answer 12

Can you show me what the first step would be? Like how to step up the problem and stuff?

Answer 13

you need to turn the \(grams\) of glucose into \(moles\) of glucose, to use in the formula for molality

Answer 14

I got that 15.5 grams of glucose is 0.08603660341253
which is really long, so would I shorten it to 0.086 moles of glucose?

Answer 15

And now I have to convert 245 grams of water to kilograms right?

Answer 16

So that's 0.245 kg
So now I just plug those things in to the molality formula...
0.086 moles/0.245kg is the formula now
What do I do next?

Answer 17

I solved and got that the molality is 0.351
So what do I do next? @JFraser

Answer 18

plug that molality into the formula and solve for \(\Delta T\)

Answer 19

the value of \(K_f\) is given, and you've just found the molality. The value of \(i\) is one, because glucose is a molecular compound and doesn't dissociate when it dissolves

Answer 20

So now I would just multiple the Kf by the molality times 1 and that's my answer?

Answer 21

yep