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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!

Bohrok
Oct28-09, 12:16 AM
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).

Borek
Oct28-09, 04:32 AM
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?

Borek
Oct28-09, 06:18 PM
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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chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), concentration calculator (http://www.chembuddy.com/?left=CASC&right=concentration_and_solution_calculator)
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.