Grams of solute needed to lower vapor pressure of solvent?

[Raghav Gupta]

I also found this key for an exam and the problem is solved the way i had originally done it? ...http://rogersal.people.cofc.edu/Fire Alarm Test.pdf

Okay that's also correct by using only mole fraction of solvent.
But the link is ignoring the fact of complete dissociation of KCl.

 

[Raghav Gupta]

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?

The answer you are getting matches with Borek and here you are taking in account complete dissociation of KCl.
But why the link is not showing that. It is giving the answer 221g?

 

[Elvis 123456789]

Yah I don't know what's up with that. The key was apparently made by someone with a Phd in chemistry or a closely related field.

 

[Raghav Gupta]

Anyways I realize that you are correct.

 

[Elvis 123456789]

Well this has been a pretty lengthy thread. Thank you for staying with me and helping me out, its been much appreciated.

 

[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.