40.0 g of magnesium chloride (MgCl2) in 250.0 g of water at 22 °C.Calculate a) the vapor pressure of this solution, b) the freezing point, c) the boiling point.

Question

I know it is something like this, but I continue to get the incorrect answer.

A) 40.og Mgcl2*(1mol mgcl2/ 95.2 mgcl2)*(3 mol mgcl2/ 1mol MgCl2)= 1.261
250.o H20*(1mol h20/ 18.02h20)= 13.87
(13.87)/(13.87+1.261)=.9166--> .9166(23.80)= 21.81. That's not the correct answer, though.

C) I get 100.8 for Boiling point
B) and negative one for freezing.

Can someone please tell me how to do this correctly?

Answer 1

freezing point.
mols MgCl2 = grams/molar mass = 40.095.2 = about 0.42 (that's estimated--you check ALL these numbers).
m = mols/kg solvet = 0.420/0.250 = about 1.68m
delta T = i*Kf*m
i for MgCl2 = 3
dT = 3*1.86*1.68 = about 9.4
freezing point = 0-9.4 = -9.4.

boiling point:
dT = i*Kb*m
i is same
Kb = 0.51
m is same. Solve for dT and add to 100. I think your error here is you didn't use the i factor of 3.

Answer 2

For the vapor pressure part of the problem it looks ok to me EXCEPT I think you used vapor pressure H2O at 25 C and not 22C. According to my sources, v.p. H2O @ 22C = 19.8 mm Hg.

Answer 3

Thank You Dr.Bob. I have one more question, how exactly do you is "i" 3? is it a given?

Answer 4

Yes and no.
i is the van't Hoff factor and it is the number dissolved particles.
So for NaCl you substitute i = 2.
For MgCl2 it is 3.
For MgSO4 it is 2
For SnCl4 it is 5 etc.
Technically, this number we substitute is a limiting number; i.e., it won't be greater than 3,2,5,etc but it may be less. We usually ignore all of those things that make it less and use the limiting number. When you get into activities you will see how that affects it.

Answer 5

I see it mainly deals with Ionic compounds correct? MgCL2 is two because when it is separated it makes 3 moles of MgCl2 for every one mole of MgCl2?

Answer 6

Yes, except for the typo you made. MgCl2 is THREE because when it is separated it makes 3 mols .....
Covalent compounds have i = 1.