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

Question

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

0.02

0.05

0.04

0.07