`3.90xx10^(-3)` moles of a solution containing an ion `A^(n+)` require `2.34xx10^(-3)` moles of `MnO_4^(-)` for the oxidation of `A^(n+) " to " AO_3^(-)` in acidic medium . What is the value of n ?

Explanation

To solve the problem, we need to determine the value of \( n \) in the oxidation reaction of \( A^{n+} \) to \( AO_3^{-} \) using \( MnO_4^{-} \) as the oxidizing agent. Here’s the step-by-step solution:

### Step 1: Identify the change in oxidation states
The ion \( A^{n+} \) is being oxidized to \( AO_3^{-} \). We need to determine the oxidation state of \( A \) in \( AO_3^{-} \).

1. The charge on \( AO_3^{-} \) is -1.
2. The oxidation state of oxygen (O) is -2. Since there are 3 oxygen atoms, the total contribution from oxygen is \( 3 \times (-2) = -6 \).
3. Let the oxidation state of \( A \) be \( x \). Therefore, we can set up the equation:
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