Grams of solute needed to lower vapor pressure of solvent?

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SUMMARY

The discussion centers on calculating the mass of KCl required to lower the vapor pressure of water by 5% at 25°C, utilizing Raoult's Law and the concept of mole fractions. The vapor pressure of pure water is 28.3 mmHg, and the calculated mass of KCl needed is approximately 109 grams per liter of water when considering complete dissociation of KCl into K+ and Cl- ions. Participants clarify the importance of using the correct mole fraction for solute and solvent, leading to a consensus on the correct approach to the problem.

PREREQUISITES
  • Understanding of Raoult's Law and its application in vapor pressure calculations.
  • Knowledge of mole fractions and their significance in solution chemistry.
  • Familiarity with colligative properties and the van 't Hoff factor.
  • Basic skills in stoichiometry and molar mass calculations.
NEXT STEPS
  • Study the derivation and application of Raoult's Law in detail.
  • Learn about colligative properties and how they relate to ionic dissociation.
  • Explore the concept of mole fractions in various solution scenarios.
  • Practice problems involving vapor pressure lowering and the van 't Hoff factor.
USEFUL FOR

Chemistry students, educators, and anyone involved in solution chemistry or physical chemistry, particularly those focusing on vapor pressure and colligative properties.

  • #31
Elvis 123456789 said:
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.
 
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  • #32
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?
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?
 
  • #33
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.
 
  • #34
Anyways I realize that you are correct.
 
  • #35
Well this has been a pretty lengthy thread. Thank you for staying with me and helping me out, its been much appreciated.
 
  • #36
You are welcome. I hope you would do the same for somebody but in a short way.:smile:
 
  • #37
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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