Answer. given concentration of ferric nitrate is
8.1x10-2 M and volume is 8.5 ml.  Â
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first finding the mass of ferric ions in the given
concentration
.mass of Fe+3 ions m=(molarity*volume*molar
mass)/1000
                               Â
m=(8.1x10-2*8.5*55.85) /1000
                         Â
m=3.85x10-2 g/mol.
Now finding the concentration of ferric ions in the
mixture.total volume of the mixture=8.5+11+19.9=39.4 ml.
molarity =(3.85x10-2
x1000)/(55.85x39.4)=38.5/2200.49=0.02M
concentration of Fe+3=0.02 M
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