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Stoichiometric Calculations

100 mL of H...

Question

$100$ mL of $H_{2} O_{2}$ is oxidised by $100$ mL of $0.01 M$ $K M n O_{4}$ in acidic medium ( $M n O_{4}^{⊝}$ reduced to $M n^{2 +}$ ). $100$ mL of the same $H_{2} O_{2}$ is oxidised by $V$ mL of $0.01 M$ $K M n O_{4}$ in basic medium ( $M n O_{4}^{⊝}$ reduced to $M n O_{2}$ ). Hence, $V$ is :

A

500

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B

100

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C

\frac{100}{3}

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D

\frac{500}{3}

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Solution

The correct option is D $\frac{500}{3}$
As given,
$100 \times 5 N (M n O_{4}^{⊝}) \equiv V \times 3 N (M n O_{4}^{⊝})$
$V = \frac{500}{3} m L$

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