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Packing Fraction and Efficiency

Volume of 0...

Question

Volume of $0.02$ M ${M n O_{4}}^{-}$ solution required to oxidise $40.0$ mL of $0.1$ M $F e^{2 +}$ solution is :

200 mL

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100 mL

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40 mL

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20 mL

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Solution

The correct option is C 40 mL
${M n O_{4}}^{-} + 5 F e^{2 +} \to 5 F e^{3 +} + M n^{2 +}$
$M_{1} V_{1} ({M n O_{4}}^{-}) = 5 M_{1} V_{1} ({M n O_{4}}^{-})$ milliequivalents
$M_{2} V_{2} (F e^{2 +}) = M_{2} V_{2} (F e^{2 +})$ milliequivalents
$N_{1} V_{1} ({M n O_{4}}^{-}) = N_{2} V_{2} (F e^{2 +})$
$5 \times 0.02 \times V_{1} = 40 \times 0.1$
$V_{1}$ $= 40$ mL.

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