Grams of solute needed to lower vapor pressure of solvent?

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Discussion Overview

The discussion revolves around calculating the mass of KCl required to lower the vapor pressure of water by 5% at 25°C. Participants explore the application of Raoult's law and the concept of vapor pressure lowering, considering the complete dissociation of KCl in solution.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using Raoult's law to relate the vapor pressure of the solution to the mole fraction of KCl.
  • Another participant calculates the mole fraction and derives the mass of KCl needed, arriving at 218g, but questions whether this is correct.
  • There is a discussion about the molecular weight of KCl, which is confirmed as 74.55g/mol.
  • Some participants express confusion over the application of the relative lowering of vapor pressure formula versus Raoult's law.
  • One participant points out that the dissociation of KCl must be considered, noting that the van 't Hoff factor is 2 for KCl.
  • There are disagreements about the definitions of "actual" and "calculated" concentrations in the context of the problem.
  • Participants debate whether the volume change from adding KCl can be neglected in the calculations.
  • One participant emphasizes that molar fractions depend only on the amounts of substances mixed, not on volume.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem. There are multiple competing views on the application of Raoult's law, the treatment of dissociation, and the definitions of concentration terms.

Contextual Notes

Some participants acknowledge the complexity of the problem due to the dissociation of KCl and the assumptions regarding the volume of the solution. There is also uncertainty regarding the correct interpretation of the formulas and their derivations.

  • #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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