Describe how you would prepare exactly 250 mL of .180 M K2S2O8 solution for use. Assume that you have a supply of solid K2S2O8, deionized water, a balance, and a volumetric flask. Also, how many grams of K2S2O8 would you need?
The Attempt at a Solution
.180 M K2S2O8 / 270.32 g/mol = 6.66 x 10^-4 g K2S2O8
Pour 250 mL of DI water into a volumetric flask. Balance out 6.66 x 10^-4 g of K2S2O8. Add the two together.