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Posted by on Tuesday, February 26, 2013 at 8:31pm.

I have o.236 g for urea and 0.235g for NaCl.
And the question is
On a gram basis, which is more effective in depressing the freezing point, urea or NaCl? On a mole basis, which Is more effective?
I am confused on what depressing the freezing point me and idk how to explain the answer

  • Chemistry - , Tuesday, February 26, 2013 at 10:46pm

    The Freezing point depression equation is ΔT = i*Kf*m

    Where m=molality=moles of solute/kg of solvent (H20)

    i= van 't Hoff factor

    Kf=constant

    So, lets get rid of some of this stuff and equate ∆T= moles of solute/kg of solvent.

    the one with the greater number of moles, assuming the kg of H20 stays constant, will cause an increase in ∆T.

  • Chemistry - , Wednesday, February 27, 2013 at 12:05am

    I found molality of urea but I am not sure how to do after

  • Chemistry - , Wednesday, February 27, 2013 at 12:07am

    Sorry wrong post

  • Chemistry - , Wednesday, February 27, 2013 at 12:16am

    For NaCl, Van 't Hoff factor=2. Find the molality like you did for Urea. But remember, the question is asking how does ∆T change with different mass or mole ratios.

    grams of substance*( 1 mole/molecular weight)= moles of substance

  • Chemistry - , Wednesday, February 27, 2013 at 12:40am

    So nacl is effective in both cases

  • Chemistry - , Wednesday, February 27, 2013 at 4:43am

    I believe that is a valid assumption.

  • Chemistry - , Wednesday, February 27, 2013 at 10:29am

    My teacher say I don't have to show any calculation so how do I explain it using the graph I have for both??

  • Chemistry - , Wednesday, February 27, 2013 at 4:32pm

    That is what i was conveying in my second post; you need to show how the freezing point changes. Looking at the equation, you will see that it depends on moles not mass. The compound with the higher number of moles, assuming H2O is constant for both, will the greater effect on the freezing point.

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