"mistake" in solutions manual for 13.52(b)

[robert zellmer]

The solutions manual has a "mistake" for 13.52(b) in the 13th & 14th
editions.  This is the same problem in the 12th ed. This is similar to
13.48(b) in the 11th ed.

The problem states: "Calculate the number of moles of solute present
in 50.0 mg of an aqueous solution that is 1.50 /*m*/ NaCl (molality)".

To make the problem "easier" the solutions manual assumed the mass
of the solvent and mass of solution are pretty much the same.  This
approximation would be okay for dilute aqueous solutions and the more
dilute it is the better the approximation.  It's not so much a mistake as it
wasn't necessary.  The proper way to do the problem would be the
following.

First, write down what you're being asked,

? mol NaCl = 50.0 mg soln    (i.e. How many moles of NaCl are
                                                       in 50.0 mg of soln?)

Now, what are you given that could be used?  The molality.  Write
down what it means,

                           1.50 mol NaCl
1.50 */m /*NaCl  = ---------------------
                              1 kg H_2 O

Lets assume we have enough solution so that there's 1 kg of H_2 O
in the solution.  This then means we have 1.50 mol of NaCl in the soln.

We need the mass of solution.  We have the mass of solvent, H_2 O,
so we need the mass of NaCl.  We can get that from the moles using
molar mass.

? g NaCl = 1.50 mol NaCl * (58.44 g NaCl)/(1 mol NaCl) = 87._*6*_6 g NaCl

So the total mass of solution is

     g soln = 1000 g H_2 O  +  87._*6*_6 g NaCl =  1087._*6*_63 g soln

Now we have a new conversion factor, the moles of NaCl in mass of solution,

        1.50 mol NaCl
      -----------------------
       1087._*6*_63 g soln

We can now complete the problem using dimensional analysis,

                                                     1 g soln            
1.50 mol NaCl
? mol NaCl = 50.0 mg soln *  -------------------- * -----------------------
                                                 1000 mg soln 
1087._*6*_63 g soln

                     = 6.8955 x 10^-5 mol NaCl

                     = 6.90 x 10^-5 mol NaCl

The solutions manual gives 7.50 x 10^-5 mol NaCl.  This is an 8.8% error.
That would normally be way too much error and there was no reason to
make this assumption.

Dr. Zellmer
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