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0.144 g of pure ${FeC}_{2} O_{4}$ was dissolved in dilute $H_{2} {SO}_{4}$ and the solution was diluted to 100 ml. What volume in ml of M ${KMnO}_{4}$ will be needed to oxidise ${FeC}_{2} O_{4}$ solution?

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answer is 6.

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Detailed Solution

${meq. of FeC}_{2} O_{4} = {meq of KMnO}_{4} \frac{W}{E} \times 1000 = N.V$

Or, $\frac{0.144}{144 / 3} \times 1000 = 0.1 \times 5 \times V$
Or V=6 ml

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