Question

In an oxidation-reduction reaction, $MnO _{4}^{-}$ion is converted to $Mn ^{2+}$. What is the number of equivalents of $KMnO _{4}$ (mol. wt. $=158$ ) present in $250\, mL$ of $0.04\, M\, KMnO _{4}$ solution ?

• A. 0.02
• B. 0.05
• C. 0.04
• D. 0.07

Answer 1

Answer: (B)

0.05

Answer 2

Change in oxidation number $=(+7)-(+2)=5$
$\therefore$ Normality of solution $=5 \times 0.04=0.20\, N$
Volume $=250\, mL$
$\therefore$ Number of equivalents of $KMnO _{4}$
$=0.20 \times \frac{250}{1000}=0.05$
So, the correct option is (B): 0.05.