[CHM 112] Question

[Andrew J. Pounds]

I knew that this was an "i" type problem because it gave me all of the 
other freezing point "stuff" but also asked for what fraction was 
associated as dimers.

Now remember, the "i" is telling us basically how many components of the 
solute are impacting the colligative property.  If the bromoform did not 
associate into dimers then the value of "i" would be one.   If the 
bromoform somehow broke into multiple pieces then the value of "i" would 
be greater than one.  The problem tells us that SOME of the bromoform 
will form dimers.  In other words, in solution we have some bromoform 
that does this...

\documentclass{article} \usepackage[utf8x]{inputenc} \pagestyle{empty} 
\usepackage{mhchem} \begin{document} \ce{CHBr3 + CHBr3 -> (CHBr3)2} 
\end{document}

So if the number of DIMERS formed is $x$ then the number of molecules 
that are left undimerized is $(1-2x)$. It is the sum of the undimerized 
and dimerized molecules that is equal to the value of "i" in the case.

$i = 0.6134 = (1-2x)+x$

so $x=0.3866$ (which is the fraction that are dimers).  Twice of this 
number would then be the number of molecules that formed dimers.  That 
is 0.7732, or 77.32%.




On 02/23/14 09:41, wrote:
>
> Hello Dr. Pounds,
>
> Could you explain how you did number 8 in additional problems for 
> chapter 13? Also, how did you know to solve for 'i' in this problem?
>
> Thanks,
>
>


-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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