A 0

For this question, you have two dilutions (from what I understand). A 10:1 dilution, and a 5:1 dilution. We have the final concentration of 1.25 x10^-5 M.

We need to back dilute (undilute).

using C1V1 = C2V2 we should be able to accurately do this.

C1 = concentration 1
V1 = volume 1

etc.

but we do this 2x

the first time....

C1 = 1.25 x 10^-5 M
V1 = 250 mL

C2 = X
V2 = (how much was "quantitatively trnasferred to 250-mL volumetric flasks)

Then we do it again

C1 = X
V1 = 500

C2 = Y
V2 = 50 mL

Note: we assume that the oxidation reaction is complete, and only factor in the added reagents as a volume change.

Now we take our C2 and multiply it by 0.250 L (since that was the original volume of the concentrated stuff) and find themoles of Mn in our sample. This number of moles is then multiplied by the molar mass of manganese (check your periodic table) and then you will have the mass of manganese in you 0.500 g sample... do a percent by mass and you have the answer...