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Author Topic: Freezing point depression and molecular weight  (Read 6809 times)

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TheWall04

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Freezing point depression and molecular weight
« on: November 12, 2007, 07:29:16 AM »

Okay so I have a question regarding freezing point depression and molecular weight. I can get to a certain point...but i dont know where to go after that :( so if someone could help me out that would be greatly appreciated. The question is;

A compound containing only Boron, Nitrogen, and Hydrogen was found to be 40.3% B, 52.2% N, and 7.5% H by mass. If 3.301g of this compound is dissolved in 50.00g of benzene, the solution produced freezes at 1.30 degrees Celcius. If Kf for benzene is 5.12 degrees celcius m^-1 and the freezing point of pure benzene is 5.48 degrees celcius, what is the molecular weight of this compound?

The change in freezing point is given by the expression;
DeltaTf = Kf x m

What I have so far is..

Kf = 5.12 degrees celcius m^-1
DeltaTf = 5.48 - 1.30 = 4.18 degrees celcius
I think plugged these values into the eqn above
(4.18 Degrees celcius) = (5.12 degrees celcius m^-1) (m)
m = 4.18/5.12
m = 0.816

But where do i go from there?!? Can someone please help me :(
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ARGOS++

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Re: Freezing point depression and molecular weight
« Reply #1 on: November 12, 2007, 09:50:35 AM »


Dear TheWall04;

Take a look for the definition of your “m” in your equation:
ΔTf = Kf * m

But take care: There is one spelled with a ‘R’ and one spelled with a ‘L’. Take the right one and rearrange “its definition”. That will solve your problem.


Good Luck!
                   ARGOS++
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Formulas, in its best sense,  ARE ONLY Recipes for “A Picture”,  —
     If you DON’T catch “The Picture”, you are lost, - for ever!      (A++)

There is ONLY one correct Formula for the “Hydrogen”:  —
     The Atom/Molecule, ITSSELF!                          (Dr. R. Mory  1968)

"Make it AS SIMPLE AS possible,  — BUT NEVER SIMPLER!"  (A. Einstein)

TheWall04

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Re: Freezing point depression and molecular weight
« Reply #2 on: November 12, 2007, 12:24:59 PM »

So I really have no idea what you just said there.
The definition of "m" is the molarity of the solute (i.e. the moles of solute in one kilogram of solvent). Is that not what i found? And what is this "L" and "R" buisness?

And i never mentioned before that DeltaTf is the amount by which the freezing point is lowered or "depressed". And Kf is the freezing point depression constant, or "cryoscopic constant"

Could anyone help me out please?
thanks :)
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ARGOS++

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Re: Freezing point depression and molecular weight
« Reply #3 on: November 12, 2007, 03:53:43 PM »

Dear TheWall04;

Sorry! – You selected the wrong 50%, because you must use the molaLity and not the molaRity.

And the MolaLity is defined as the “numbers” of moles per 1000 grams.
So from the definition of molaLity it follows for your “m” in your case:
 
m” = (m? / M?) /  mB * 1000g

With:  m?   = mass of the unknown;    M?  = Molecular weight of the unknown.
         mB   = mass of Benzene

Now you have only to rearrange this equation to M? = … and you can calculate your preferred Molecular weight. With the use of the ratios (from the element analysis) you can calculate finaly the summformula and tell the name of the unknowned compound.

Do you think it is still too difficalt for you?
(Is now the difference between ‘R’ and ‘L’ Business also clear to you?)

Good Luck!
                   ARGOS++
« Last Edit: November 12, 2007, 04:15:02 PM by ARGOS++ »
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Formulas, in its best sense,  ARE ONLY Recipes for “A Picture”,  —
     If you DON’T catch “The Picture”, you are lost, - for ever!      (A++)

There is ONLY one correct Formula for the “Hydrogen”:  —
     The Atom/Molecule, ITSSELF!                          (Dr. R. Mory  1968)

"Make it AS SIMPLE AS possible,  — BUT NEVER SIMPLER!"  (A. Einstein)

TheWall04

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Re: Freezing point depression and molecular weight
« Reply #4 on: November 12, 2007, 05:22:02 PM »

So how do you go about solving it? I'm finding it hard to follow what you are saying. Could you possibly start off the problem for me so i can see where youre going with it. Im more of a visual learner. Have i done anything right so far?
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LQ43

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Re: Freezing point depression and molecular weight
« Reply #5 on: November 12, 2007, 05:49:41 PM »

ARGOS++ is giving you good clues but maybe it is easier to see like this

m = molality which you calculated as 0.816m

which means 0.816 moles solute / kg of benzene but you have 50.00 g of benzene (convert to kg)

how many moles of solute?

now you have moles of solute, your problem give you gms of solute

solve for MW

btw they also give % composition, the next part must be what is the molecular formula

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TheWall04

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Re: Freezing point depression and molecular weight
« Reply #6 on: November 12, 2007, 07:02:37 PM »

Ohh okay got you! thank you so much!
But how do i solve for MW?
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LQ43

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Re: Freezing point depression and molecular weight
« Reply #7 on: November 12, 2007, 07:06:19 PM »

MW = g/mol

If 3.301g of this compound
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TheWall04

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Re: Freezing point depression and molecular weight
« Reply #8 on: November 13, 2007, 05:43:44 AM »

Okay so i think i have it now.. Could someone tell me if i did it correctly
so we have m = 0.816 moles / kg of benzene
but we have 50.00g of benzene
So 50.00g  x  1kg/1000g = 0.05kg
now we figure out that
0.816mol x 0.05g = 0.0408 moles of solute

MW = g/mol
The given mass is 3.301g so

MW = 3.301g / 0.0408mol  = 80.91g/ mol

 Is this correct??                 
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ARGOS++

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Re: Freezing point depression and molecular weight
« Reply #9 on: November 13, 2007, 06:05:38 AM »


Dear LQ43;

Thank you very much, but from TheWall04’s first correct result I didn’t realase that I face a real beginner (to TheWall04: There is absolutely nothing bad about that.) and so I gave only the next clues and not the step by step procedure.


Dear TheWall04;

Excellent!
Now can you calculate how much of this molecular weight (MW) the Boron will consume by using the given percentage? And when you finally divide the got part of the “consumed” moleculare weight by the Boron by the MW of Boron you will got how many “atoms”  of Boron your molecular formula will contain. (Maybe you have a very small little bit to round)

And then do the same for the Nitrogen and for the Hydrogen too?
This results in the whole molecular formula of your unknown.

If you are missing any imaginaten how the compound is called you have a little bit to search on the left top side of this page under the rubric: “Resources

I hope it will be of help to you.

Good Luck!
                   ARGOS++


P.S.: To TheWall04 :  Thank you for your present. I gave you also one, - only with opposite sign. Thank you anyway.
.
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Formulas, in its best sense,  ARE ONLY Recipes for “A Picture”,  —
     If you DON’T catch “The Picture”, you are lost, - for ever!      (A++)

There is ONLY one correct Formula for the “Hydrogen”:  —
     The Atom/Molecule, ITSSELF!                          (Dr. R. Mory  1968)

"Make it AS SIMPLE AS possible,  — BUT NEVER SIMPLER!"  (A. Einstein)

TheWall04

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Re: Freezing point depression and molecular weight
« Reply #10 on: November 13, 2007, 06:36:27 AM »

Thank you Argos++ and LQ43 for your assistance I appreciate it.  ;D
I now know what I'm doing.
Take care!
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