1. The problem statement, all variables and given/known data "A scientist prepares 100 mL of an iron (III) nitrate solution by weighing out 4.8372 grams of the solid iron (III) nitrate and dissolving it in deionized water using a 100 mL volumetric flask. She then pipettes 5 mL of that solution into a 20 mL volumetric flask and fills it to the mark, then dilutes the resulting solution by a factor of 100. What are the molarities of Fe3+ (aq) and NO3- (aq) in the final solution? Show your work." 2. Relevant equations Molarity = moles / liter M1V1 = M2V2 3. The attempt at a solution I believe there are two ways to solve this (correct me if I'm wrong) 1. Raw calculation: Well it starts off with 100 mL of deionized water added to 4.8372g of Fe(NO3)3. That solution will be 0.048372g Fe(NO3)3 / mL of H2O. She takes 5 mL of that solution (0.048372g/mL x 5 mL = 0.012093 g of Fe(NO3)3), and then dilutes it by a factor of 100. A dilution factor of 100 is the same as 99 mL of H2O and 1 mL of solution correct? Not sure how I would apply the dilution factor .. would it just be 0.012093g of Fe(NO3)3 / 99 mL H2O? 2. I imagine M1V1 = M2V2 could provide a shortcut..