Osmolality and Freezing Point

I'm trying to calculate the Osmolality and Freezing Point of the following problems with the given information: Water Kf = 1.86 C/m , F. pt. = 0.00 Celsius degrees

1) 2.00 m Na2SO4
2) 2.00 m Na3PO4

I know the formula for Osmolality is pi = (i) x (M) x (R) x (T) and for Tf = (i) x (Kf) x (m).

But from there on I'm confused. I know I can easily plugin for Tf ---> (2.00m) x (1.86 C/m), but I don't know what to use for (i) for Tf and Osmolality.

If I can have someone tell me what that variable would be, I can solve these problems.

Please and Thank you.

That formula you gave (pi = iMRT) is for osmotic pressure.

molality = (moles of solute)/(kg of solvent)

i = the number of ions that dissociate when added to water
* its also known as a proportionality constant

1) i = 3 ( Na+, Na+, SO4^-2)
2) i = 4 (Na+, Na+, Na+, PO4^-3)

These are colligative properties so as you add compounds that dissociate it will DECREASE or "depress" the freezing point.

1) Tf = (3) x (1.86 C/m) (2.00m) = 11.16 C
so since your normal freezing point is 0 C and this is freezing point DEPRESSION
Freezing point of water with 2.00 m Na2SO4 is:

Tf = -11.16 C

2) Tf = (4) x (1.86 C/m) (2.00m) = 14.88 C
and again since your normal freezing point is 0 C and this is freezing point DEPRESSION
Freezing point of water with 2.00 m Na3SO4 is:

Tf = -14.88 C

edit: fixed a typo i made