Some Basic Concepts of Chemistry 2 Question 48
48. $2.68 \times 10^{-3}$ moles of a solution containing an ion $A^{n+}$ require $1.61 \times 10^{-3}$ moles of $\mathrm{MnO}{4}^{-}$for the oxidation of $A^{n+}$ to $A \mathrm{O}{3}^{-}$ in acidic medium. What is the value of $n$ ?
$(1984,2 \mathrm{M})$
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Solution:
- For the oxidation of $A^{n+}$ as :
$$ A^{n+} \longrightarrow A \mathrm{O}_{3}^{-} \quad n \text {-factor }=5-n $$
$\Rightarrow$ Gram equivalent of $A^{n+}=2.68 \times 10^{-3}(5-n)$
Now equating the above gram equivalent with gram equivalent of $\mathrm{KMnO}_{4}$ :
$$ \begin{aligned}
