What volume of 0.2 N `KMnO_4` is required to oxidise 10 " mg of " ferrous oxalate in acidic medium? (Molecular Weight of `FeC_2O_4` is 144 g)

Explanation

To solve the problem of determining the volume of 0.2 N KMnO₄ required to oxidize 10 mg of ferrous oxalate (FeC₂O₄) in an acidic medium, we will follow these steps:

### Step 1: Calculate the number of moles of ferrous oxalate (FeC₂O₄)
Given that the molecular weight of FeC₂O₄ is 144 g/mol, we can calculate the number of moles in 10 mg of ferrous oxalate.

\[ \text{Mass of FeC}_2\text{O}_4 = 10 \text{ mg} = 10 \times 10^{-3} \text{ g} \]
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