How many $m L$ of $0.125 M$ ${C r}^{3 +}$ must be reacted with $12.00 m L$ of $0.200 M$ $M n O_{4}^{-}$ if the redox products are $C r O_{7}^{2 -}$ and ${M n}^{2 +}$ ?

8 m L

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16 m L

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24 m L

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32 m L

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Solution

The correct option is C $32 m L$

${C r}^{3 +} + M n O_{4}^{-} ⟶ {M n}^{2 +} + \frac{1}{2} {C r}_{2} O_{7}^{2 -}$
Equivalent of ${C r}^{3 +} = 3 \times$ moles of ${C r}^{3 +}$

Equivalents of $M n O_{4}^{-} = 5 \times$ moles of $M n O_{4}^{-}$

Amount of ${C r}^{3 +} = 0.125 \times V$ millimole
$= 0.125 \times V \times m e q$

Amount of $M n O_{4}^{-} = 0.200 \times 12.00 \times 5$ milli equivalent

$0.125 \times V \times 3 = 0.200 \times 12.00 \times 5$

$V = 32 m L$

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