In alkaline medium, KMnO4 reacts as follows 2 KMnO 4+2 KOH arrow 2 K 2 MnO 4+ H 2 O +[O] Therefore, the equivalent weight of K MnO 4 will be

Question

In alkaline medium, KMnO4 reacts as follows 2 KMnO 4+2 KOH arrow 2 K 2 MnO 4+ H 2 O +[O] Therefore, the equivalent weight of K MnO 4 will be (A) 31.6 (B) 52.7 (C) 7.0 (D) 158.0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/e2a254ae5322362015d628947c57e3aa-.png" /> Equivalent wt. of KMnO 4=( text molecular weight /n) =(158/1)=158.0

Q. In alkaline medium, $K M n O_{4}$ reacts as follows
$2 K M n O_{4} + 2 K O H \to 2 K_{2} M n O_{4} + H_{2} O + [O]$
Therefore, the equivalent weight of $K M n O_{4}$ will be

31.6

52.7

7.0

158.0

Equivalent wt. of $K M n O_{4} = \frac{molecular weight}{n}$
$= \frac{158}{1} = 158.0$

Q. In alkaline medium, KMnO4​ reacts as follows 2KMnO4​+2KOH→2K2​MnO4​+H2​O+[O] Therefore, the equivalent weight of KMnO4​ will be

31.6

52.7

7.0

158.0

Equivalent wt. of KMnO4​=n molecular weight ​ =1158​=158.0

Q. In alkaline medium, $K M n O_{4}$ reacts as follows
$2 K M n O_{4} + 2 K O H \to 2 K_{2} M n O_{4} + H_{2} O + [O]$
Therefore, the equivalent weight of $K M n O_{4}$ will be

Equivalent wt. of $K M n O_{4} = \frac{molecular weight}{n}$
$= \frac{158}{1} = 158.0$