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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">The solutions manual has a "mistake" for 13.52(b) in the 13th & 14th<br>
editions. <br>
<br>
The problem states: "Calculate the number of moles of solute present<br>
in 50.0 mg of an aqueous solution that is 1.50 <b><i>m</i></b> NaCl (molality)".<br>
This is not a concentration conversion problem.<br>
<br>
To make the problem "easier" the solutions manual assumed the mass<br>
of the solvent and mass of solution are pretty much the same. This<br>
approximation would be okay for dilute aqueous solutions and the more<br>
dilute it is the better the approximation. It's not so much a mistake as it<br>
wasn't necessary. The proper way to do the problem would be the<br>
following.<br>
<br>
First, write down what you're being asked,<br>
<br>
? mol NaCl = 50.0 mg soln (i.e. How many moles of NaCl are<br>
in 50.0 mg of soln?)<br>
<br>
Now, what are you given that could be used? The molality. Write<br>
down what it means,<br>
<br>
1.50 mol NaCl<br>
1.50 <b><i>m </i></b>NaCl = ---------------------<br>
1 kg H<sub>2</sub>O<br>
<br>
Lets assume we have enough solution so that there's 1 kg of H<sub>2</sub>O<br>
in the solution. This then means we have 1.50 mol of NaCl in the soln.<br>
<br>
We need the mass of solution. We have the mass of solvent, H<sub>2</sub>O,<br>
so we need the mass of NaCl. We can get that from the moles using<br>
molar mass.<br>
<br>
? g NaCl = 1.50 mol NaCl * (58.44 g NaCl)/(1 mol NaCl) = 87.<b><u>6</u></b>6 g NaCl<br>
<br>
So the total mass of solution is <br>
<br>
g soln = 1000 g H<sub>2</sub>O + 87.<b><u>6</u></b>6 g NaCl = 1087.<b><u>6</u></b>63 g soln<br>
<br>
Now we have a new conversion factor, the moles of NaCl in mass of solution,<br>
<br>
1.50 mol NaCl<br>
-----------------------<br>
1087.<b><u>6</u></b>63 g soln<br>
<br>
We can now complete the problem using dimensional analysis,<br>
<br>
1 g soln 1.50 mol NaCl<br>
? mol NaCl = 50.0 mg soln * -------------------- * -----------------------<br>
1000 mg soln 1087.<b><u>6</u></b>63 g soln<br>
<br>
= 6.8955 x 10<sup>-5</sup> mol NaCl<br>
<br>
= 6.90 x 10<sup>-5</sup> mol NaCl<br>
<br>
The solutions manual gives 7.50 x 10<sup>-5</sup> mol NaCl. This is an 8.8% error.<br>
That would normally be way too much error and there was no reason to<br>
make this assumption.<br>
<br>
Dr. Zellmer <o:p></o:p></span></p>
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