Posted by Hannah on Friday, February 10, 2012 at 10:58pm.
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.
I understand everything except the van't Hoff factor part. I do not understand how you figure that out.
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
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?
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.
Ok so do you mean 0.25MgBr2 X 3 = 0.75 ?
i equals 3 for MgBr2 correct?
So I think that 0.40m Cr(NO3)3 has the largest delta T. Would you agree?
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