For starters I found the molar mass Mg(NO3)2 to be 148.149. Did you use a hydrated form of Mg(NO3)2 or did you just add the numbers wrong.
Second, I find with these equations that it is too easy to make a math error with all the slashes for divisions, especially in the denominator. If I use your numbers (with the 210.33 for the molar mass) I found 4.32. Using my value of 148.15 I found 3.047. I think you would be well advised to do the problem piece-meal. Like so.
Delta T = 1.9 as you have.
mols Mg(NO3)2 = 9.2/148.15 = 0.06210
m = mols/kg = 0.06210/0.05129 = 1.2108
i = delta T/(Kb*m) = 1.9/(0.515*1.2108) = 3.047. You may not be allowed that many significant figures but you can check your work with my readings. Check my work. Check my thinking.
[Putting it all together, I find the final equation to be like this but check me out.
i = (delta T*kg solvent*GMW)/(Kb*grams solute) = 3.047.
For a lab, I have to find the experimental values of i (Van T'Hoff factor).
so far this is what i have, but it's incorrect. could you please tell me what i'm doing wrong??? thanks.
given info:
the lab varied the concentration of Mg(NO3)2
the theoretical value of i is 3
the kg of water is .05129 kg
the delta T is 1.9 (101.9-100)
the g of Mg(NO3)2 is 9.20 g
i = kb*g*deltaTb/GMW*kg of solvent
i = (.515 degreesCkg/mol*9.20g*1.9 degrees C)/(210.33g)(.05129)
i=10?!?!?! ...i stopped bc ten is wayyy off.
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thank you so much! the molar mass was a major problem...i didn't realize i was so off. thank you for pointing that out.
1 answer