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 ?

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

Tardigrade Admin · Accepted Answer

Change in oxidation number =(+7)-(+2)=5 ∴ Normality of solution =5 × 0.04=0.20 N Volume =250 mL ∴ Number of equivalents of KMnO 4 =0.20 × (250/1000)=0.05

Q. In an oxidation-reduction reaction, $M n O_{4}^{-}$ ion is converted to $M n^{2 +}$ . What is the number of equivalents of $K M n O_{4}$ (mol. wt. $= 158$ ) present in $250 m L$ of $0.04 M K M n O_{4}$ solution ?

0.02

0.05

0.04

0.07

Change in oxidation number $= (+ 7) - (+ 2) = 5$
$∴$ Normality of solution $= 5 \times 0.04 = 0.20 N$
Volume $= 250 m L$
$∴$ Number of equivalents of $K M n O_{4}$
$= 0.20 \times \frac{250}{1000} = 0.05$

Q. In an oxidation-reduction reaction, MnO4−​ion is converted to Mn2+. What is the number of equivalents of KMnO4​ (mol. wt. =158 ) present in 250mL of 0.04MKMnO4​ solution ?