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I've received several questions about the same end-of-chapter
exercises. <br>
Here's partial answers to those questions to get your started on
them if<br>
you haven't tried them or are still having problems with them. This<br>
seemed to be most peoples problems with them, not knowing how <br>
to start them. <br>
<br>
13.57: <br>
This is about an alloy of 80% Cu and 20% Zn. The primary question <br>
was how to start this. I would say to do this how I suggested. You
<br>
have a solid solution. You're given % composition. Assume 100 g <br>
of soln (metal). That means you have 80 g Cu and 20 g Zn. This <br>
is what you would have done in 1210 to determine an empirical
formula. <br>
Then use the density of the soln (metal) to convert 100 g sol to
volume <br>
of soln. By the way, 1 m<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>3</sup>
= 10<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>3</sup>
L (similar to 1 cm<sup>3</sup> = 1 mL). <br>
<br>
13.60: <br>
The primary problem I've seen with this one is people not
knowing for<br>
gases a volume % is the same as a mole % at a constant P and T.
Thus a<br>
mixture with a vol % of 4 % CO2 has a mole % of 4 % CO2. Also, <br>
mole % = mole fraction * 100. <br>
<br>
I've also seen people having trouble remembering that for a gaseous
mixture <br>
the relationship between partial pressure of a gas and total
pressure. The <br>
partial pressure of gas A is given by the following, <br>
<br>
P_A = X_A * P_total <br>
<br>
You need this for VP problems to calculate the mole fraction of a
substance in <br>
the gas phase above a soln. <br>
<br>
13.98 <br>
This question is a conc. conversion question. You're given a 1.80 M
LiBr soln <br>
in the solvent acetonitrile (CH3CN) and the density of the soln.
You're then <br>
asked to convert this to molality, mole fraction of LiBr and mass %
(of CH3CN). <br>
<br>
I approached this type of problem in class by listing what the given
conc means <br>
first and then assuming what's in the denominator, <br>
<br>
1.80 mol LiBr <br>
1.80 M LiBr = ------------------- <br>
1 L soln <br>
<br>
Assume you have a beaker with 1 L soln (what's in the denominator).
That then <br>
also gives you what's in the numerator, 1.80 mol LiBr. <br>
<br>
Then for each conc. you're trying to get write down it's definition
and then<br>
determine how to get to that conc. unit based on what you have. <br>
<br>
Don't forget, to from molarity to any of the others (and vice versa)
you will need <br>
density of the soln to convert from L soln to mass of soln (or vice
versa). <br>
<br>
Hopefully this helps with any lingering questions. <br>
<br>
Dr. Zellmer
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