boiling point Temp = molality x Kb... not molarilty... do you guys have your math right?

I am VERY confused...

Dude, I know molality is what you want. However, for osmotic pressure it is Molarity. I did it pretty quickly. What the heck is IP? I thought my work was clear.

4e-6 Moles of MgCl2 in 40 ml. We must convert to Liters so 40 ml= 40e-3 which is equal to 4e-2 L. You need to calculate the molar solubility. The ksp is 4.1e-20=4x^3. Why, since you have two moles of CL for every mole of Mg, the equation calls that ksp = x(2x)^2

Now, (4.1e-20)/4=x^3= 1.025e-20=x^3

Now, we need to make the power a factor of 3 so we move the base over by one so it becomes 10.25e-21=x^3

x=2.25e-7 moles/Liter.

If the concentration of MgCl2 is greater than this then we will get a precipitate. Well, 4e-6 Moles/4e-2 Liters= 1e-4 M/L This is significantly greater than the molar solubility so most won't dissolve. The amount dissolved is the molar solubility.

To find the amount of moles in solution, we take 2.25e-7 * 4e-2= 9e-9moles. I messed this up before. Why? The molar solubility is for one liter so we have to calculate how much is in the reduced volume. Now we take 9e-9 moles/4e-2 Kg of solvent.

Why, well g/ml are equivalent so 4e-2 Liters=4e-2 Kg 9e-9/4e-2= 2.25e-7 molal for MgCl2.

The boiling elevation is .5(2.25e-7)(3)=3.375e-7 +100. The amount of insignificant.

For freezing point depression it's 0-1.9(2.25e-7)(3)

osmotic pressure is .5(2.25e-7)(3)(.0821)(298k).

My major issue is that there is a minute difference between molarity and molality. We aren't given density so for all intents and purposes, the are the same.

I have no idea about vapor pressure as I only know Raoult's law for two liquids.

Nacl has a one to one ratio so ksp=x*x= 3.6e-11.

We have to move it a number divisible by 2, so 36e-12=x^2

x=6e-6 this is the molar solubility

The Molarity is 4e-6/5e-2= .8e-4=8e-5 M/L

The concentration is greater so it will precipitate. The amount dissolved is the molar solubility. The molality is equal to the the molar solublity as demonstrated above with the calculation of the number of moles in 40 ml of MgCl2.

The reason I use Molarity is that determines how much is dissolved. Since the molar solubility is 6e-6. If the concentration was 6e-8 then that would be the amount dissolved since it's less than the maximum. To solve for the varying points you need the MOLALITY of DISSOLVED whereas before you use the MOLARITY ASSUMING 100% DISSOCIATION.

Now, for boiling= 100 +.5(6e-6)(2)

Freezing =0-1.9(6e-6)(2)

Osmotic= 6e-6*2*(.0821)(298K). This is assuming room temperature.

The difference in molality is kg of solvent vs Liters of solution. You can distinguish if you don't have the density. In many of these problems the molarity and molality are the same unless they give you specific density to calculate molarity.

An ionic solution has little impact. Raoult's law is in effect. The normal vapor pressure of pure water is 20 torr. However, since the mole fraction is over 99% water, the vapor pressure is pretty much the same. The boiling point of water goes up by 1 degree Celcius for every Mole of NaCl!! You have an insignificant amount. HTH