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Vapour pressure problem (simple)

  1. Jun 9, 2009 #1
    The vapour pressure of pure chloroform at 70.0oC is 135.7 kPa. What mass of iodine (I2) should be dissolved in 1 L of chloroform, CHCl3 (density = 1.49 g/cm3) to lower the vapour pressure by 13.3 kPa?


    hey guys i really cant do this question, i think in involves raults law, P=XPA
    anyhelp would be good. thanks



    p.s. answer is 1.32 x 101 g
     
  2. jcsd
  3. Jun 10, 2009 #2

    Borek

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    Staff: Mentor

    No idea if you are to take into account iodine pressure or not.

    Assuming it has to be ignored - at what x the pressure of chloroform vapour will be lowered by 13.3 kPa?
     
  4. Jun 10, 2009 #3
    thnks for the reply, umm 122.4kpa
     
  5. Jun 17, 2009 #4
    Hey,

    Here is the solution I got, its different to your answer though...

    Psolution=xsolvent.Psolvent

    so Psolution=135.7-13.3 kpa (the new pressure of the solution)

    now in 1L =1000cm3 there is 1490 grams = 12.482 moles

    so re-arrange the equation and you get=> x = (12.482 - 11.26)/(3.55x10-3)

    =344.23 grams

    if 133.32 is the correct answer as you suggest I dont know where went wrong.
     
  6. Jun 17, 2009 #5
    hey mate i checked with the teacher and ur right(stupid internet tests). just wondering if you could show me where you got a few of the those numbers from, ive tried but get lost. thanks
     
  7. Jun 20, 2009 #6
    Ok,

    so Xsolvent=molsolvent/(molsolute+molsolvent), we are looking to find molsolute (I2) so we can multiply by the RFM (253.8 g.mol-1) to get the mass needed.

    1L of chloroform has mass 1490g = 12.482mol

    Xsolvent=(135.7kPa-13.3kPa)/135.7kPa=.902

    Xsolvent=12.482/(12.482+x)=.902

    re-arranged x=1.356 mol

    therefor mass I2 = 344g
     
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