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Grams of solute needed to lower vapor pressure of solvent?

  1. Feb 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Some KCl is dissolved in water at 25°C, where it completely dissociates. The vapor pressure of pure water at 25°C is 28.3 mmHg. Estimate the mass in grams of KCl needed per liter of pure water to reduce the vapor pressure of water at 25°C by 5%.

    2. Relevant equations


    3. The attempt at a solution
    I'm really not sure how to go about this problem. First I thought I could maybe use raoult's law, but then I don't think that would work to get the mass of the solute. Then I thought maybe if the vapor pressure is lowered by 5%, then the solvent is also lowered by 5%, which would mean that for 1 L of water I would need 0.05 L of KCl to lower the vapor pressure by 5%, but I don't think this is right either.
     
  2. jcsd
  3. Feb 17, 2015 #2
    Use the equation (p-p°)/p° = x where p is vapour pressure of solution after kcl is added,p° V.P of pure water, and x is mole fraction of kcl in water.
    The above equation's derivation comes from raoult' law which you can google.
    Now you can substitute the values.
     
  4. Feb 17, 2015 #3
    so 5% of 28.3mmHg would be 26.885mmHg, then 26.885mmHg= X(mole fraction) * 28.3mmHg------> X=0.95

    1L H2O * (1000mL/1L) *(1g/1mL) * 1mol/18.02g = 55.49 mol H2O......mol KCl= Z

    (55.49 mol H2O)/(55.49mol H2O + Z) = 0.95

    Z= (55.49mol)/(0.95) - 55.49mol = 2.92 mol KCl * 74.55g/mol = 218g KCl

    is this right?
     
  5. Feb 17, 2015 #4
    What's the molecular weight of kcl?
     
  6. Feb 18, 2015 #5
    74.55g/mol
     
  7. Feb 18, 2015 #6
    You are wrong.
    The answer coming is 207 g.
    I got x=0.05.
     
  8. Feb 18, 2015 #7
    Why are you not applying the relative lowering of vapour pressure formula,
    And directly the Raoult's law?
    Do you not know that formula in post #2?
     
  9. Feb 18, 2015 #8
    No I am not familiar with the formula from the second post. Why can't I directly use the Raoult's law? Doesn't it make sense the way I am doing it?
     
  10. Feb 18, 2015 #9
  11. Feb 18, 2015 #10
    I look up the vapor pressure lowering formula you mentioned, and the mole fraction in that formula belongs to the solute. In the Raoult law, the mole fraction belongs to the solvent. If I use the vapor pressure lowering formula i would get z/(z+55.49 mol H2O)=0.05, so I think it is correct to use the Raoult law
     
  12. Feb 18, 2015 #11
    Actually vapor pressure lowering formula is a manipulation of Raoult's law. So the main credits to Raoult's law. Now I tell you the whole story-
    If we put solute in a pure solvent
    P1=xsolvent*P1°
    here P1 is vapour pressure of solution and P1° the V.P of pure solvent.
    Now when we add solute there is lowering in V.P of pure solvent.
    ## \Delta ##P1=P1°- P1= P1°- P1°*xsolvent = P1°(1- xsolvent)
    Also xsolute = 1- xsolvent
    So,
    ## \Delta ##P1= xsolute*P1°
    Now
    ##\frac{ \Delta P1}{P1°}## = ##\frac{ P1°- P1}{P1°}## = xsolute= ##\frac{ n_ {solute}}{n_{solvent}+ n_{solute}}## Since nsolute<< nsolvent
    So
    ##\frac{ P1°- P1}{P1°}## = ##\frac{ n_ {solute}}{n_{solvent}}##
    or more simply
    ##\frac{ P1°- P1}{P1°}## = ##\frac{ M_ {solvent}*w_{solute}}{M{solute}*w_{solvent}}##
     
    Last edited: Feb 18, 2015
  13. Feb 18, 2015 #12

    Borek

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    Staff: Mentor

    It is about using Raoult's law AND the fact that sum of molar fractions of all substances present in the mixture equals 1.

    Raghav: while your approach is correct, you seem to be making it unnecessarily complicated by stating "use the formula". It hides the logic behind. See paragraph above.

    As far as I can tell you are both ignoring the fact KCl is dissociated.
     
    Last edited: Feb 19, 2015
  14. Feb 18, 2015 #13
    Yeah, early I was saying to use the formula, but after that I have also given proof of it by logic. Now what logic I am hiding?

    We are ignoring the dissociation of KCl because the vant hoff factor is not given in question, otherwise that's true that observed colligative property = i * calculated colligative property.
    i = 2 for KCl for complete dissociation but we don't know that it has completely dissociated.
    I think OP is asking a school level topic wise problem.
     
    Last edited: Feb 18, 2015
  15. Feb 18, 2015 #14
    Sorry, Sorry. I have made a grave mistake in above post.
    Yeah in question complete dissociation is written.
    So i=2 as
    KCl is in equilibrium with K+ + Cl-.
    Borek you are correct.
    I was taking the OP in complete trouble.
    That's why you are a mentor, as experience is also needed and I have few posts than you.
     
  16. Feb 18, 2015 #15
    I am confused now, so what is the proper way to solve the problem?
     
  17. Feb 19, 2015 #16
    That's correct.
    That X is Xsolvent
    Now X for solute is 1- Xsolvent.
    Then Xsolute = nsolute/nsolvent since nsolute<< nsolvent.
    From here you get mass of KCl.
    But since KCl is completely dissociated.
    i = 2 ( See vant hoff factor)

    Actual mass of KCl = i * calculated mass
     
  18. Feb 19, 2015 #17

    Borek

    User Avatar

    Staff: Mentor

    Nope. Think it over.
     
  19. Feb 19, 2015 #18
    Hmm--- So many mistakes. :H
    So Actual concentration of KCl = i * calculated concentration of KCl

    Now calculated concentration is moles of KCl/ Volume of solution in liters
    Assume the volume is negligibly changed by adding KCl
     
  20. Feb 19, 2015 #19

    Borek

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    Staff: Mentor

    I am not convinced you got it right this time. Not sure what you mean by "actual" and "calculated" - but if the one you got from the pressure information is "calculated", it is still wrong way around.

    That's not necessary, molar fractions depend only on the amount of mixed substances, volume doesn't play any role. The only assumption you need is the one about the water density (1 L = 1 kg).
     
  21. Feb 19, 2015 #20
    Not got it.
    I think when finding concentration in molarity
    Molarity = no of moles of solute / volume of solution in liters
    Since we don't know density of solute, we cannot tell what will be the volume of solution and that's why we consider volume of water only.
     
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