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ΔTF = KF × b × i freezing point saltwater

  1. Feb 24, 2012 #1
    1. The problem statement, all variables and given/known data

    I think this is the right formula: ΔTF = KF × b × i
    Freezing-point depression.
    I have the amount of salt added to the water but how am I going to do to calculate the difference in temperature? Or with other words the freezing point of water saturated with salt (NaCl)? Or am I totally gone here? >_<
    And KF is 1.853 K·kg/mol.

    But b and i, what should they include? Be nice, I'm an amateur :p
     
  2. jcsd
  3. Feb 24, 2012 #2

    Borek

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    Staff: Mentor

  4. Feb 24, 2012 #3
    I don't really know the values to use. What should I use instead if concentrated ones doesn't work with that formula?
     
  5. Feb 24, 2012 #4

    Borek

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    They should be calculated. Although in this paritcular case...

    ...freezing point tables (determined experimentally) sound like a better solution.
     
  6. Feb 24, 2012 #5
    Okay. Now let's say I want to know when salt water freeze when I've 1.5 g NaCl in 0.25 L water.
    ΔTF=1.853 * 0.025667 (mol NaCl) * 2 (ions Na+ & Cl-)
    = gives me 0.0951219. Which is totally wrong I think. It's not that little in temperature difference if I had salt into water.. Please explain....
    -___- ' I would be very glad :)
     
  7. Feb 24, 2012 #6

    Borek

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  8. Feb 25, 2012 #7
    Now I think I got it:
    E.g. Kf=1.867 * 0.25/ 0.01388 * 2 * 0.1368= ~9.13 °C => Freezing point is reduced to -9.13 °C
    I think it should be right? If b is (2g/58.44)/0.25= 0.1368
    Which is 0.25 L water & 2 grams of salt. xD
     
    Last edited: Feb 25, 2012
  9. Feb 25, 2012 #8

    Borek

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    I have no idea what is what of what in what you wrote - but it is wrong.

    This is a simple plug and chug. The only thing you have to do is to correctly calculate concentration. You don't have to rearrange the equation to calculate Kf, as Kf is given, besides, Kf is not what you are looking for - ΔTF is.
     
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