Chemistry
posted by K on .
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

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

Sorry wrong post

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 
So nacl is effective in both cases

I believe that is a valid assumption.

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

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.