KMnO_4 reacts with oxalic acid according to t | vedclass

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(C) From the balanced chemical equation,$2 \ 	ext{moles}$ of $KMnO_4$ react with $5 \ 	ext{moles}$ of $H_2C_2O_4$.Using the concept of milliequivale

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$50 \ mL$ of $0.1 \ M \ H_2C_2O_4$

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(C) From the balanced chemical equation,$2 \ 	ext{moles}$ of $KMnO_4$ react with $5 \ 	ext{moles}$ of $H_2C_2O_4$.Using the concept of milliequivalents (Meq):$	ext{Meq of } KMnO_4 = 	ext{Molarity} 	imes n	ext{-factor} 	imes 	ext{Volume (mL)}$For $KMnO_4$,the $n$-factor is $5$ $(Mn^{+7} 	o Mn^{+2})$.$	ext{Meq of } KMnO_4 = 0.1 	imes 5 	imes 20 = 10 \ 	ext{Meq}$.For $H_2C_2O_4$,the $n$-factor is $2$ $(C_2^{+3} 	o 2C^{+4} + 2e^-)$.We need $10 \ 	ext{Meq}$ of $H_2C_2O_4$.$	ext{Meq} = 	ext{Molarity} 	imes 2 	imes 	ext{Volume (mL)} = 10$.Checking option $(C)$: $50 \ mL$ of $0.1 \ M \ H_2C_2O_4$:$	ext{Meq} = 0.1 	imes 2 	imes 50 = 10 \ 	ext{Meq}$.Thus,$20 \ mL$ of $0.1 \ M \ KMnO_4$ is equivalent to $50 \ mL$ of $0.1 \ M \ H_2C_2O_4$.

$KMnO_4$ reacts with oxalic acid according to the equation:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Here $20 \ mL$ of $0.1 \ M \ KMnO_4$ is equivalent to:

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$KMnO_4$ reacts with oxalic acid according to the equation:$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$Here $20 \ mL$ of $0.1 \ M \ KMnO_4$ is equivalent to: