How much 0.02 m kmno4 solution should be needed if the solutions tested have a composition of 3% h2o2 by mass?

When the concentration is expressed in mass percentage, that means there is 3 g of solvent H₂O₂ in 100 grams of the solution. Then, that means the remaining amount of solute is 97 g. We use the value of molarity (moles/liters) to determine the amount of solution in liters, denoted as V. The solution is as follows:

0.02 mol KMnO4/L solution * 158.034 g KMnO4/mol * V = 97 g KMnO4

Solving for V,
V = 30.69 L

Idon't see the pic?