The volume of 0.0200 M KMnO4 needed to titrate 1.0 mL of 3% H2O2?

Answer 1

3% H2O2 means 3 grams of H2O2/100 mL of solution. So 1 mL of that would contain 0.03 grams H2O2. Convert that to moles using the molar mass of H2O2. Then you'll need to have the balanced equation for the reaction. With the coefficients in the balanced equation you can convert moles of H2O2 to moles of KMnO4. Then you can divide by the molarity of that solution to give you the volume (in L) of the KMnO4 solution that you need.