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I Parts per million of vapor above a liquor in a sealed jar

  1. Oct 26, 2016 #1
    How would you calculate the concentration of a vapor above a liquid in a sealed jar in parts per million? Supposing that the liquid is a 10% solution of X with known vapor pressure.

    We've gotten to this point:
    1ppm = 1mgX/ g solution
    1ppm = 1mgX/1mL solution (assuming it's water and water has 1g/mL)
    1ppm = 1000mg / liter = 1 gm/liter

    A tougher issue is how can you calculate the concentration of a vapor above a solid in a sealed jar.

    Any help is appreciated!
  2. jcsd
  3. Oct 26, 2016 #2


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    ppm are parts per million, so not gram/kilogram, but milligram per kilogram.
    If you know the vapour pressure, you can use pV = nRT to calculate the concentration n/V.
  4. Oct 27, 2016 #3
    Can you please provide an example of a specific solution that you might be interested in.
  5. Oct 27, 2016 #4

    Given 10mL of 10% Citral solution (dissolved in diethyl pthalate) ( vapor pressure 0.2 mm of Hg (@ 20°C)) in a closed 20mL vial, what is the concentration of the air above the liquid in parts per million.
  6. Oct 27, 2016 #5
    After getting to moles per liter how do you convert to ppm?
  7. Oct 27, 2016 #6
    Do you want the mass concentration or the molar concentration? Does the solution of Citral in diethyl pthalate represent something close to an "ideal solution" (so that Raoult's law can be applied), or, in your judgment, is it non-ideal? Is that 10 % solution by mass or is it molar? Was the air in the vial initially at 1 atm?
  8. Oct 27, 2016 #7
    We would like the mass concentration, and then to eventually convert this to ppm.

    The solution can be considered ideal.

    10% is by mass (1mL pure Citral in 10 total mL of Diethyl phthalate)

    The air in the vial was initially at 1atm
  9. Oct 27, 2016 #8
    So, if you had 10 grams of solution, how many grams would you have each of Citral and Diethyl phthalate?
    How many moles would you have each of Citral and Diethyl phthalate?
    What would the total number of moles be for 10 gm of solution?
    What would be the mole fraction of Citral?

    Is this a homework problem?
  10. Oct 27, 2016 #9
    This is not a homework problem. General curiosity and tangentially related to some work we are doing with citral.

    Working through your questions now.
  11. Oct 27, 2016 #10
    Using 893kg/m3 as the density for citral, and 1.12g/cm3 as the density for diethyl phthalate, I have:

    5.866*10^-3 moles citral and 4.536*10^-2 moles diethyl phthalate in 10mL of 10% solution

    The mole fraction for citral is 0.1145

    What do I need the moles of citral in 10grams of solution for?
  12. Oct 27, 2016 #11
    We are using 10 gm as a "basis" for the calculation. You don't need to know the densities. If you have 10 gm of solution, then you have 1 gm of Citril and 9 gm of diethyl phthalate. The molecular weight of diethyl phthalate is 229, and the molecular weight of Citral is 152. So the number of moles are
    Citral: 0.00659 moles
    Diethyl phthakate: 0.0393
    Total moles = 0.0459
    mole fraction citral = 0.144
    From Raolt's law, if the mole fraction Citral is 0.144 and the equilibrium vapor pressure of pure Citral is 0.2 mm Hg, what is the partial pressure of Citral in the head space? What is the total pressure of the gas in the head space?
  13. Oct 27, 2016 #12
    A 10% solution usually refers to the mass percent, not the volume percent.
  14. Oct 27, 2016 #13


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    Somewhat contradictory ! (So Chet goes that way)

    So your citral has a vapour pressure of 0.22 mm Hg at 20 C. And the diethylphtalate 0.002 at 25 C. Ignore the latter and work out how many moles there are in the gas volume (20 ml minus the 11 ml liquid volume). Idem air. And I don't assume you want
    but instead the ppm citral, so 106 times mass of citral / mass of air
    Last edited: Oct 27, 2016
  15. Oct 27, 2016 #14


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    And (lazy me) you don't have to work it out completely: with m = nM and pV = nRT (both for air and for citral) V, R and T divide out and things become simple.

    (note I ignored the much lower partial pressure from the solvent)
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