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In an oxidation-reduction reaction, $\mathrm{MnO}_4^{-}$ion is converted to $\mathrm{Mn}^{2+}$. What is the number of equivalents of $\mathrm{KMnO}_4$ (mol. wt. $=158$) present in 250 mL of $0.04 \mathrm{M~} \mathrm{KMnO}_4$ solution ?
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Verified Answer
The correct answer is:
0.05
$\begin{gathered}
\text { Change in oxidation number }=(+7)-(+2)=5 \\
\therefore \text { Normality of solution }=5 \times 0.04=0.20 \mathrm{~N} \\
\text { Volume }=250 \mathrm{~mL}
\end{gathered}$
$\therefore$ Number of equivalents of $\mathrm{KMnO}_4$
$=0.20 \times \frac{250}{1000}=0.05$
\text { Change in oxidation number }=(+7)-(+2)=5 \\
\therefore \text { Normality of solution }=5 \times 0.04=0.20 \mathrm{~N} \\
\text { Volume }=250 \mathrm{~mL}
\end{gathered}$
$\therefore$ Number of equivalents of $\mathrm{KMnO}_4$
$=0.20 \times \frac{250}{1000}=0.05$
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