What volume of 0.1 M $KMn{O}_4$ is needed to oxidize 100 mg of $Fe{C}_2{O}_4$ in acid solution?

The correct option is A 4.1 mL
The given reaction is,

The oxidation of oxalate ions to carbon dioxide involves 2 electrons.

$C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}$

The oxidation of ferrous ions to ferric ions involves 1 electron.
$F{e}^{2+}\to F{e}^{3+}+e^{-}$

Total number of electrons involved per molecule oxidation of $Fe{C}_2{O}_4$ to $F{e}^{3+}$ and $C{O}_2$ is 3.

So, n- factor of $Fe{C}_2{O}_4=3$

In acidic solution, 1 mole of $KMn{O}_4$ involves 5 moles of electrons.

So, n- factor of $KMn{O}_4=5$

Thus, 3 moles of $KMn{O}_4$ will oxidize 5 moles of ferrous oxalate.

The molar mass of ferrous oxalate is 143.91 g/mol.

Using ,
Equivalents of $KMn{O}_4=$Equivalents of $Fe{C}_2{O}_4$

$\Rightarrow$ Moles $\times$ n factor of $KMn{O}_4$= Moles $\times$ n factor of $Fe{C}_2{O}_4$

$\Rightarrow$ $0.1\times \text{volume}\times 5=\frac{\text{mass}\times \left(\text{n factor}\right)}{\text{molar mass}}$


putting the values,
$\Rightarrow$ $0.1\times \text{volume}\times 5=\frac{100\times 3}{1000\times \left(143.91\right)}$

$\Rightarrow$ $\text{volume}=\frac{100\times 3}{1000\times \left(143.91\right)\times \left(0.1\right)\times \left(5\right)}$

$\Rightarrow$ volume= $4.1\times {10}^{-3}L=4.1mL$