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$\mathrm{KMnO}_4$ reacts with ferrous sulphate according to the following equation,
Here, 10 mL of $0.1 \mathrm{MKMnO}_4$ is equivalent to
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Here, 10 mL of $0.1 \mathrm{MKMnO}_4$ is equivalent to
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Verified Answer
The correct answer is:
50 mL of $0.1 \mathrm{M} \mathrm{FeSO}_4$
For the reaction,
$\mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow$
$\mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}$
5 times quantity of $\mathrm{Fe}^{2+}$ consumed than $\mathrm{KMnO}_4$.
Thus, 10 mL of $0.1 \mathrm{M} \mathrm{KMnO}_4$ is equivalent to 50 mL of $0.1 \mathrm{M} \mathrm{FeSO}_4$.
$\mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow$
$\mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}$
5 times quantity of $\mathrm{Fe}^{2+}$ consumed than $\mathrm{KMnO}_4$.
Thus, 10 mL of $0.1 \mathrm{M} \mathrm{KMnO}_4$ is equivalent to 50 mL of $0.1 \mathrm{M} \mathrm{FeSO}_4$.
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