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In the reaction of $\mathrm{KMnO}_{4}$ with an oxalate in acidic medium, $\mathrm{MnO}_{4}^{-}$ is reduced to $\mathrm{Mn}^{2+}$ and $\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ is oxidised to $\mathrm{CO}_{2}$. Hence, $50 \mathrm{~mL}$ of 0.02 $\mathrm{M} \mathrm{KMnO}_{4}$ is equivalent to
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The correct answer is:
$50 \mathrm{~mL}$ of $0.05 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$
eg. of \(\mathrm{KMnO}_{4}\)
\(\begin{aligned}
&=5 \times 0.02 \times 50 \\
&=250 \times 0.02 \\
&=509 .
\end{aligned}\)
eq. of \(\mathrm{KAnO}_{4} \equiv \mathrm{eq}_{1}\) of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }_{2}-\)
\(\begin{aligned} 5=\operatorname{option} 2 &=50 \times 0.05 \times 2 \\ &=589 . \end{aligned}\)
\(\begin{aligned}
&=5 \times 0.02 \times 50 \\
&=250 \times 0.02 \\
&=509 .
\end{aligned}\)
eq. of \(\mathrm{KAnO}_{4} \equiv \mathrm{eq}_{1}\) of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }_{2}-\)
\(\begin{aligned} 5=\operatorname{option} 2 &=50 \times 0.05 \times 2 \\ &=589 . \end{aligned}\)
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