Grams of solute needed to lower vapor pressure of solvent?

[Raghav Gupta]

You are welcome. I hope you would do the same for somebody but in a short way.

 

[Borek]

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.