what is the ideal melting and freezing pt of .15 mole of Mg SO4 dissolved in 800 g H2O?

Question

Formulas to work with?
delta Tf = (Kf(mole of solute)) / (kg solvent)
delta Tb = Kb (same as above)
Freezing pt. of solution = 0 celcius - delta Tf
boiling pt. = 100 celcius + delta Tb

Kf of H2O = 1.86
Kb of H2O= 0.52
do you use Kb and Kf of H2O?

Answer 1

the formulas you have are all that you need. The critical point to know is that MgSO4 breaks up into 2 ions. So you really have 0.30 moles of particles in your solution.

For the freezong point use delta Tf = Kf (moles of ions in the solute) / kg of solvent.

Kf = 1.86
moles = .30
Solvent = .800 Kg.

Solve for delta T and subtract from 0 C


The boiling point is the same set up, except use the Kb of .52

The rest is the same

Answer 2

It s the same set up as in yur MgSO4 problem, except you only have .15 moles of solute because there are no ions here.