$20 \mathrm{~mL}$ of $0.02\, \mathrm{M}$ hypo solution is used for the titration of $10 \mathrm{~mL}$ of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of $\mathrm{Cu}^{2+}$ is found to be ____________ $\times 10^{-2} \,\mathrm{M}$. [nearest integer]

Given : $$2 \,\mathrm{Cu}^{2+}+4 \,\mathrm{I}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2}$$

$$ \mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-} $$

$2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{l}_{2}+\mathrm{I}_{2}$ <br/><br/> $\mathrm{I}_{2}+\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}$ <br/><br/> Milliequivalents of hypo solution $=0.02 \times 20=0.4$ <br/><br/>Milliequivalents of $\mathrm{Cu}^{2+}$ in $10 \mathrm{~mL}$ solution = <br/><br/> Milliequivalents of $\mathrm{I}_{2}=$ Milliequivalents of hypo = 0.4 <br/><br/> Millimoles of $\mathrm{Cu}^{2+}$ ions in $10 \mathrm{~mL}=0.4$ <br/><br/> Molarity of $\mathrm{Cu}^{2+}$ ions $=\frac{0.4}{10}=0.04 \,\mathrm{M}$ <br/><br/> $=4 \times 10^{-2} \,\mathrm{M}$