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Molality for compound with 2 particles

  1. Jun 28, 2013 #1
    Can someone explain why we take the number of particles into account after we have determined our molality? I feel like I'm just memorizing this and not understanding why you need to determine # of particles for molality in the problem below.

    The observed boiling point was elevated by 1.02 degrees C. What was the molality of the NaCl in the sample? (The molal boiling point elevation constant of water is 0.51 degrees C/m. The answer is 1m

    I understand you divide 1.02 by 0.51 to get 2m, based on units, but then the answer says that we have 2 particles for NaCl making the molality 1m. Why do we need to break it up like this. Can someone please explain conceptually what is going on. I feel like I'm just memorizing this.
     
  2. jcsd
  3. Jun 29, 2013 #2

    Borek

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    Staff: Mentor

    In different situations different things matter. In the case of boiling point elevation (or freezing point lowering) it is total concentration of all present molecules and ions that counts. If you dissolve some molecular compound - concentration of all molecules just equals its concentration. But in the case of ionic compounds number of molecules/ions will be usually higher when you take dissociation into account.

    Google for van 't Hoof factor.
     
  4. Jun 29, 2013 #3
    How do you know whether you multiply or divide by vant hoff factor. I think this is what is confusing me. I was able to calculate molality based on the units of the values they gave us which gave 2 moles/kg. I've seen some places that say that vant hoff factor isn't included in molality and some like the answer to this question from my book that say that it is included. So I'm not sure which is right. If it is included, how do we know whether we multiply or divide the molality by vant hoff factor. Also from boiling point elevation it seems vant hoff factor isn't included in molality it comes after, (change)T = kim, but the answer to my question suggests that vant hoff is included in molality. Is this is mistake in my book?
     
  5. Jun 29, 2013 #4
    ΔT = K · b(non-dissociated) · i

    you are given ΔT, K and the identity of the solute (NaCl) so you know that it dissociates into two particles, Na+ and Cl- so i=2. So now you solve for bsolute.

    b(non-dissociated)=1.02/(0.51 x 2)=1 (approximately)

    vant Hoff factor (i) is a ratio between the concentration of the dissociated and non-dissociated compound.

    b(dissociated) = b(non-dissociated) · i, so for NaCl

    b[NaCl]=(b[Na+] + b[Cl-]) · i
     
  6. Jun 29, 2013 #5
    Oh it makes perfect sense, I wasn't thinking about the equation. Thank you both so much!
     
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