# Grams of solute needed to lower vapor pressure of solvent?

• Elvis 123456789
In summary: Ok, I think I got it now. So the "calculated concentration" would be the concentration you get from using the mole fraction in the Raoult's law equation and the "actual concentration" would be the concentration you get from using the mole fraction in the vapor pressure lowering formula?Yes, that's correct. And from the "actual concentration" we can calculate the actual mass of KCl needed, taking into account the dissociation and the vant hoff factor.
You are welcome. I hope you would do the same for somebody but in a short way.

Elvis 123456789 said:
Why is it necessary to get the mole fraction of the solute? Isn't it enough to have the mole fraction of the solvent "X=0.95"

(55.51mol H2O)/(2z + 55.51mol H2O)=0.95--------> z= 1.46mol KCl * 74.55g/mol -----> z= 109g KCl

Is this way valid? Or is it necessary to use the other formula?

Both approaches are perfectly valid and equivalent.

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