Molar Mass From Freezing Point Depression

Click For Summary

Discussion Overview

The discussion revolves around calculating the molar mass of a nonpolar molecular compound using the freezing point depression method. Participants are working through a homework problem involving the freezing point of benzene and the mass of solute added.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates the change in temperature due to freezing point depression and attempts to find the molar mass by determining the number of moles of solute.
  • Another participant confirms that molar mass is calculated as grams divided by moles, but raises a concern about the correct interpretation of molality in the context of the problem.
  • A participant expresses uncertainty about whether it is appropriate to divide the mass of solute by the calculated moles, given that the moles are based on a different mass of solvent.
  • One participant corrects the previous calculations, stating that the number of moles should be adjusted based on the actual mass of solvent used (36 g instead of 1000 g).
  • A later reply indicates that the participant has recalculated and arrived at a different number of moles, leading to a proposed molar mass, but questions whether the calculations are correct.
  • Another participant points out the inconsistency in the number of moles derived from the same equation, highlighting a potential error in the calculations.
  • One participant acknowledges the mistake in their earlier calculation and asserts that the corrected value should be used.

Areas of Agreement / Disagreement

Participants express differing views on the calculations, with some agreeing on the method but disagreeing on the results. There is no consensus on the final molar mass due to the discrepancies in the calculations presented.

Contextual Notes

Participants are navigating issues related to the conversion of units, the interpretation of molality, and the application of the freezing point depression equation. There are unresolved mathematical steps and assumptions regarding the solvent mass and its impact on the number of moles calculated.

TrueStar
Messages
94
Reaction score
0

Homework Statement



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.

Homework Equations



(Change of temperature)=(FP constant)(molality)
MM=moles/gram
m=moles of solute/kg of solvent


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!
 
Physics news on Phys.org
I haven't checked your numbers, but

Molar mass = grams/moles. Divide grams by moles.
 
n=m/M then n/1=m/M then m=n x m
 
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.

--
methods
 
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.

--
methods
 
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.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
6K
Replies
1
Views
2K
Replies
2
Views
17K
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K