View Full Version : Molar Mass From Freezing Point Depression
TrueStar
Oct28-09, 12:05 AM
1. The problem statement, all variables and given/known data
Find the molar mass of a nonpolar molecular compound if 5.52 grams dissolved in 36.0 grams of benzene begins to freeze at -1.87 C? The freezing point of pure benzene is 5.50 C. The freezing point depression constant is -5.12.
2. Relevant equations
(Change of temperature)=(FP constant)(molality)
MM=moles/gram
m=moles of solute/kg of solvent
3. The attempt at a solution
The change in temperature is 5.50-(-1.87) which is 7.37.
7.37=5.12x --- x will equal .07. This equates to .07 moles of solute per solvent.
I'm not sure what to do next to get the molar mass of the compound. Would I divide 5.52 by .07 moles?
Thank you!
I haven't checked your numbers, but
Molar mass = grams/moles. Divide grams by moles.
the pro
Oct28-09, 12:39 AM
n=m/M then n/1=m/M then m=n x m
TrueStar
Oct28-09, 12:47 AM
I know molar mass is grams/mole. I just am not sure if it's OK to divide the 5.16 grams by the .07 moles because it's actually .07 moles per Kg. (molality).
You doubts are not unfounded. You don't have 0.07 moles of substance - that you would have in 1000 g of bezene. But there is only 36 g of benzene, so obviously number of moles is much smaller.
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TrueStar
Oct28-09, 05:46 PM
OK, here's what I've done on the second try.
The problem setup with the freezing point depression equation is 7.37=5.12x. x will equal to 1.44 which means 1.44 moles of solute per kg of solvent. I don't have that much solvent, just 36g. After changing kg to grams and then multiplying that number of moles by 36, I have .05184 moles.
I know there is 5.52 g of solvent in the solution. That's 5.52 g per .05184 moles. That translations to 106.48 g per mole...which is the molar mass and the answer.
Am I doing this correctly?
Looks OK to me.
Beware - you have solved the same equation twice (7.37=5.12x) getting two different results - 0.07 and 1.44. that's not good.
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www.titrations.info (http://www.titrations.info) - all about titration (http://www.titrations.info) methods
TrueStar
Oct28-09, 06:30 PM
Yes I noticed that in my first post. 1.44 should be the right answer. I divided the wrong way last night. I was tired.
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