Which aqueous solution below is expected to have the lowest freezing point? Assume 100% dissociation for ionic species.
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?
Chemistry - DrBob222, Friday, February 10, 2012 at 11:21pm
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.
Chemistry - Hannah, Saturday, February 11, 2012 at 12:22am
I understand everything except the van't Hoff factor part. I do not understand how you figure that out.
Chemistry - DrBob222, Saturday, February 11, 2012 at 12:44am
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
Chemistry - Hannah, Saturday, February 11, 2012 at 10:02am
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?
Chemistry - DrBob222, Saturday, February 11, 2012 at 3:01pm
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.
Chemistry - Hananh, Saturday, February 11, 2012 at 10:46pm
Ok so do you mean 0.25MgBr2 X 3 = 0.75 ?
So I think that 0.40m Cr(NO3)3 has the largest delta T. Would you agree?
yes, Cr(NO3)3 is the answer.
Thank you for all your help!!
Well, I'm just a humble Clown Bot, but I must say, your calculations are spot on! It seems like 0.40m Cr(NO3)3 does indeed have the largest delta T and therefore the lowest freezing point. Congratulations on your excellent detective work! Keep up the good work, and don't forget to bring a clown nose to your chemistry class for some extra fun!
Yes, you are correct. The 0.40m Cr(NO3)3 solution is expected to have the lowest freezing point because it has the largest delta T. The van't Hoff factor (i) for Cr(NO3)3 is 4, and the molarity (m) is 0.40. Therefore, the product i*m is equal to 1.60, which is the largest value among the given solutions.
Yes, you are correct. To determine which aqueous solution is expected to have the lowest freezing point, you need to calculate the delta T value for each solution. The equation to calculate delta T is delta T = i * Kf * m, where i is the van't Hoff factor, Kf is the freezing point depression constant, and m is the molality of the solution.
First, let's determine the van't Hoff factor (i) for each solution:
1) MgBr2 dissociates into Mg2+ and 2 Br- ions, so i = 3.
2) Na2SO4 dissociates into 2 Na+ and SO4^2- ions, so i = 3.
3) KCl dissociates into K+ and Cl- ions, so i = 2.
4) Sucrose (C12H22O11) does not ionize, so i = 1.
5) Cr(NO3)3 dissociates into Cr^3+ and 3 NO3- ions, so i = 4.
Next, let's calculate the van't Hoff factor multiplied by the molality for each solution:
1) 0.25M MgBr2 * 3 = 0.75
2) 0.30M Na2SO4 * 3 = 0.90
3) 0.50M KCl * 2 = 1.00
4) 1.0M Sucrose * 1 = 1.00
5) 0.40M Cr(NO3)3 * 4 = 1.60
The solution with the highest value for i*m will have the largest delta T and therefore the lowest freezing point. In this case, 0.40M Cr(NO3)3 has the highest value of 1.60, so it is expected to have the lowest freezing point among the given solutions.