Question

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

• A. 60
• B. 20
• C. 40
• D. 6

Answer 1

Answer: (D)

6

Answer 2

Reaction of $KMnO _{4}$ with $I ^{-}$ converting $I _{2}$ is

$2 KMnO _{4}+10 I ^{-} \longrightarrow 5 I _{2}+2 Mn ^{+2}$
$n$ -factor of $KMnO _{4}=5$ (in acidic medium)

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

Number of equivalent of $KMnO _{4}$

= Number of equivalent of $I^{-}$
$0.02 \times V \times 5 =60 \times 0.01 \times 2$
$V=6 \,mL$