The volume of \(0.02 \mathrm{M}\) acidified permanganate solution required for complete reaction of \(60…

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
  1. 60
  2. 20
  3. 40
  4. 6

Solution

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 \(\mathrm{I}\) \(\begin{aligned} 0.02 \times V \times 5 & =60 \times 0.01 \times 2 \\ V & =6 \mathrm{~mL} \end{aligned}\)

Asked in: AP EAMCET 2019 (20 Apr Shift 1)

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