100 ml of 0.1 N hypo decolourised iodine by the addition of x g of crystalline copper sulphate to excess of KI. The value of ‘x’ is (molecular wt. of ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}\mathrm{is}250)$

 $\begin{array}{l}\underset{1\mathrm{mole}}{\stackrel{+2}{\mathrm{Cu}}{\mathrm{SO}}_4}+2\mathrm{KI}\to {\mathrm{K}}_2{\mathrm{SO}}_4+\stackrel{+2}{\mathrm{Cu}}{\mathrm{I}}_2;\stackrel{+2}{2{\mathrm{CuI}}_2}\to \stackrel{+1}{{\mathrm{CuI}}_2}+{\mathrm{I}}_2\\ {\mathrm{I}}_2+\underset{1\mathrm{mole}}{{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3}\to 2\mathrm{NaI}+{\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6\\ \mathrm{Eq}.\mathrm{wt}.\mathrm{of}{\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}\text{\hspace{0.17em}=}\frac{\mathrm{Mol}.\mathrm{wt}}{\mathrm{change} \mathrm{in} \mathrm{oxidation} \mathrm{state} \mathrm{of} \text{'}\mathrm{Cu}\text{'}}=\frac{250}{1}=250\\ 100\mathrm{ml}\mathrm{of}0.1\mathrm{N}\mathrm{hypo}\equiv 100\mathrm{ml}\mathrm{of}0.1\mathrm{N}{\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}\\ \mathrm{mass} \mathrm{of} \hspace{0.17em}{\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}=\frac{\mathrm{N}\times \mathrm{E}\times \mathrm{V}}{1000}=\frac{250\times 0.1\times 100}{1000}=2.5\mathrm{g}\end{array}$