Which aqueous solution below is expected to have the lowest freezing point? Assume 100% dissociation for ionic species.

Question

1)0.25m MgBr2
2)0.30m Na2SO4
3) 0.50m KCl
4) 1.0m sucrose
5) 0.40m Cr(NO3)3

I think that the answer is 0.30NaSO4 but I am not positive. Is this correct?

Answer 1

delta T = i*Kf*m
You want the largest delta T (which will give the lowest freezing point).
Kf is constant so we can forget that. The two that matter are i and m
So multily i*m for the 5 and the largest number wins. Remember i is the van't Hoff factor which is the number of particles produced when the materials are placed in solution. sucrose is l. KCl is 2, etc.

Answer 2

I understand everything except the van't Hoff factor part. I do not understand how you figure that out.

Answer 3

That how the material ionizes.
KCl ==> K^+ + Cl^- and i = 2
MgCl2 ==> Mg&2+ + 2Cl^- and i = 3
Na2SO4 == 2Na^+ + SO4^2- and i = 3
Cr(SO4)3 ==> Cr^3+ + 3SO4^2- and i = 4
Sugar doesn't ionize so it is just 1 particle
Sugar ==> sugar(aq) and i = 1

Answer 4

Ok so then I just have to find the mass of each solution

so KCl = 74.55 X 2
MgBr2 = 184.11 X 3
Na2SO4 = 94.05 X 3
Cr(SO4)3 = 340.21 X 4

So Cr(SO4)3 would have the largest delta T, so the lowest freezing point.
Is this correct?

Answer 5

Why do you need the molar mass?
delta T = i*Kf*m
You know i from my previous response and you know m from the problem. i*m that gives the largest number will be the one that gives the largest delta T.

Answer 6

Ok so do you mean 0.25MgBr2 X 3 = 0.75 ?

Answer 7

i equals 3 for MgBr2 correct?

Answer 8

So I think that 0.40m Cr(NO3)3 has the largest delta T. Would you agree?