A solution is prepared by dissolving m g of CuSO4·5H2O (M=249.7 g mol^-1) in water and making the volume 500 mL. To 25.0 mL of it, excess KI is added…

Copper(II) oxidises iodide: $2\mathrm{Cu^{2+}}+4\mathrm{I^-}\to2\mathrm{CuI}+\mathrm{I_2}$, and the iodine is titrated: $\mathrm{I_2}+2\mathrm{S_2O_3^{2-}}\to2\mathrm{I^-}+\mathrm{S_4O_6^{2-}}$. One $\mathrm{I_2}$ comes from two $\mathrm{Cu^{2+}}$ and consumes two thiosulfate, so $1\ \mathrm{Cu^{2+}}\equiv1\ \mathrm{S_2O_3^{2-}}$. Moles of thiosulfate $=20.0\times0.100=2.00\ \mathrm{mmol}$, so each $25.0\ \mathrm{mL}$ aliquot contains $2.00\ \mathrm{mmol}$ of $\mathrm{Cu^{2+}}$. The full $500\ \mathrm{mL}$ holds $2.00\times20=40.0\ \mathrm{mmol}$, giving $m=0.0400\times249.7=9.99\ \mathrm{g}$. Taking a $2:1$ ratio of $\mathrm{Cu^{2+}}$ to thiosulfate halves the answer ($4.99$); doubling gives $19.98$; forgetting the twenty-fold dilution gives $0.50$. The end point is detected when the blue starch-iodine colour disappears, and precipitation of CuI drives the otherwise unfavourable $\mathrm{Cu^{2+}/I^-}$ reaction to completion.