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The volume of \(0.02 \mathrm{M}\) acidified permanganate solution required for complete reaction of \(60 \mathrm{~mL}\), of \(0.01 \mathrm{M} \mathrm{I}^{-}\)ion solution to form \(\mathrm{I}_2\) in \(\mathrm{mL}\) is
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The correct answer is:
6
Reaction of \(\mathrm{KMnO}_4\) with \(\mathrm{I}^{-}\)converting \(\mathrm{I}_2\) is
\(2 \mathrm{KMnO}_4+10 \mathrm{I}^{-} \longrightarrow 5 \mathrm{I}_2+2 \mathrm{Mn}^{+2}\)
\(n\)-factor of \(\mathrm{KMnO}_4=5\) (in acidic medium)
\(\mathrm{I}^{-}\)convert into \(\mathrm{I}_2\) then, \(n\)-factor \(=2\)
Number of equivalent of \(\mathrm{KMnO}_4\)
\(=\) Number of equivalent of \(\mathrm{I}\)
\(\begin{aligned}
0.02 \times V \times 5 & =60 \times 0.01 \times 2 \\
V & =6 \mathrm{~mL}
\end{aligned}\)
\(2 \mathrm{KMnO}_4+10 \mathrm{I}^{-} \longrightarrow 5 \mathrm{I}_2+2 \mathrm{Mn}^{+2}\)
\(n\)-factor of \(\mathrm{KMnO}_4=5\) (in acidic medium)
\(\mathrm{I}^{-}\)convert into \(\mathrm{I}_2\) then, \(n\)-factor \(=2\)
Number of equivalent of \(\mathrm{KMnO}_4\)
\(=\) Number of equivalent of \(\mathrm{I}\)
\(\begin{aligned}
0.02 \times V \times 5 & =60 \times 0.01 \times 2 \\
V & =6 \mathrm{~mL}
\end{aligned}\)
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