A solution of `0.2 g` of a compound containing `Cu^(2+)` and `C_(2)O_(4)^(2-)` ions on titration with `0.02M KMnO_(4)` in presence of `H_(2)SO_(4)` consumes `22.6mL` oxidant. The resulting solution is neutralized by `Na_(2)CO_(3)`, acidified with dilute `CH_(3)COOH` and titrated with excess of `KI`. The liberated `I_(2)` required `11.3 mL "of" 0.05M Na_(2)S_(2)O_(3)` for complete reduction. Find out mole ratio of `Cu^(2+)` and `C_(2)O_(4)^(2+)` in compound.

Explanation

To solve the problem, we need to determine the mole ratio of \( Cu^{2+} \) and \( C_2O_4^{2-} \) in the compound. We will follow these steps:

### Step 1: Calculate the moles of \( KMnO_4 \) used in the reaction.
Given the molarity of \( KMnO_4 \) is \( 0.02 \, M \) and the volume used is \( 22.6 \, mL \):
\[ \text{Moles of } KMnO_4 = \text{Molarity} \times \text{Volume (L)} = 0.02 \, \text{mol/L} \times \frac{22.6 \, \text{mL}}{1000} = 0.000452 \, \text{mol} \]

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