KMnO_4 reacts with ferrous sulphate according | vedclass

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(D) From the balanced chemical equation:$MnO_4^- + 5Fe^{2+} + 8H^{+} 	o Mn^{2+} + 5Fe^{3+} + 4H_2O$$1 \ 	ext{mole of } KMnO_4 	ext{ reacts with } 5

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$50 \ mL$ of $0.1 \ M$ $FeSO_4$

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(D) From the balanced chemical equation:$MnO_4^- + 5Fe^{2+} + 8H^{+} 	o Mn^{2+} + 5Fe^{3+} + 4H_2O$$1 \ 	ext{mole of } KMnO_4 	ext{ reacts with } 5 \ 	ext{moles of } FeSO_4$.Given:$	ext{Moles of } KMnO_4 = 	ext{Molarity} 	imes 	ext{Volume (in L)} = 0.1 \ M 	imes 0.010 \ L = 0.001 \ 	ext{moles}$.Since $1 \ 	ext{mole of } KMnO_4 	ext{ reacts with } 5 \ 	ext{moles of } FeSO_4$,$	ext{Moles of } FeSO_4 	ext{ required} = 5 	imes 0.001 \ 	ext{moles} = 0.005 \ 	ext{moles}$.Now,$	ext{Volume of } FeSO_4 = \frac{	ext{Moles}}{	ext{Molarity}} = \frac{0.005 \ 	ext{moles}}{0.1 \ M} = 0.050 \ L = 50 \ mL$.Therefore,$10 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to $50 \ mL$ of $0.1 \ M$ $FeSO_4$.

$KMnO_4$ reacts with ferrous sulphate according to the equation:
$MnO_4^- + 5Fe^{2+} + 8H^{+} \to Mn^{2+} + 5Fe^{3+} + 4H_2O$
Here,$10 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to:

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$KMnO_4$ reacts with ferrous sulphate according to the equation:$MnO_4^- + 5Fe^{2+} + 8H^{+} \to Mn^{2+} + 5Fe^{3+} + 4H_2O$Here,$10 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to: