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$\mathrm{KMnO}_4$ reacts with ferrous sulphate according to the equation $\mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}$ Here $10 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{KMnO}_4$ is equivalent to
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$50 \mathrm{ml}$ of $0.1 \mathrm{~M~} \mathrm{FeSO}_4$
In this reaction
$\mathrm{MnO}_4^{-}+5 \mathrm{Fe}_5^{2+}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}$
5 time quantity of $\mathrm{Fe}^{2+}$ consumed.
So 5 time of $\mathrm{FeSO}_4$ will be equivalent to $50 \mathrm{ml}$
$\mathrm{MnO}_4^{-}+5 \mathrm{Fe}_5^{2+}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}$
5 time quantity of $\mathrm{Fe}^{2+}$ consumed.
So 5 time of $\mathrm{FeSO}_4$ will be equivalent to $50 \mathrm{ml}$
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