To prepare a very dilute solution, it is more accurate to make up a more concentrated standard solution, and carry out a series of successive dilutions, than to weigh out a very small mass of the solute.

A solution was made by dissolving $0.587 \mathrm{~g}$ of $\mathrm{KMnO}_{4}$ in dilute sulfuric acid and making the volume of solution up to $1 \mathrm{dm}^{3}$ in a volumetric flask. $10.0 \mathrm{~cm}^{3}$ of this solution were transferred to a second $1 \mathrm{dm}^{3}$ volumetric flask and diluted to the mark with water. The dilution process was then repeated once, that is, $10.0 \mathrm{~cm}^{3}$ of this solution were transferred to a $1 \mathrm{dm}^{3}$ volumetric flask and diluted to the mark with water. (Section 1.5)

(a) What mass (in $\mathrm{mg}$ ) of $\mathrm{KMnO}_{4}$ would you have had to weigh out to make $500 \mathrm{~cm}^{3}$ of a solution with the same concentration as the final dilute solution?

(b) What is the concentration of the final dilute $\mathrm{KMnO}_{4}$ solution in mol $\mathrm{dm}^{-3}$ ?