Question

The volume of $0.02M$ acidified permanganate solution required for complete reaction of $60mL$ of $0.01M{I}^{-}$ ion solution to form $I_2$ in $mL$ is

• A. 60
• B. 20
• C. 40
• D. 6

Answer 1

Answer: (D)

Reaction of $KMn{O}_4$ with $I^{-}$ converting $I_2$ is

$2KMn{O}_4+10I^{-}⟶5I_2+2M{n}^{+2}$

$n$ -factor of $KMn{O}_4=5$ (in acidic medium)

$I^{-}$ convert into $I_2$ then, $n$ -factor $=2$

Number of equivalent of $KMn{O}_4$

= Number of equivalent of $I^{-}$

$0.02\times V\times 5=60\times 0.01\times 2$

$V=6mL$