The equivalent weight of KMnO4 in alkaline medium is equal to:

Question

The equivalent weight of KMnO4 in alkaline medium is equal to: (A)  (M/3)  (B)  (M/5)  (C)  (M/2)  (D)  M

Tardigrade Admin · Accepted Answer

In alkaline medium overset+72 KMnO 4+2 KOH longrightarrow overset+62 K 2 MnO 4+ H 2 O + O Change in oxidation number =7-6=1 Equivalent weight =( text molecular weight / text change in oxidation number ) =(M/1)=M

Q. The equivalent weight of $K M n O_{4}$ in alkaline medium is equal to:

$\frac{M}{3}$

$\frac{M}{5}$

$\frac{M}{2}$

$M$

In alkaline medium
$2 K M n O_{4} + 7 + 2 K O H ⟶ 2 K_{2} M n O_{4} + 6 + H_{2} O + O$
Change in oxidation number $= 7 - 6 = 1$
Equivalent weight
$= \frac{molecular weight}{change in oxidation number}$
$= \frac{M}{1} = M$

Q. The equivalent weight of KMnO4​ in alkaline medium is equal to:

3M​

5M​

2M​

M

In alkaline medium 2KMnO4​+7​+2KOH⟶2K2​MnO4​+6​+H2​O+O Change in oxidation number =7−6=1 Equivalent weight = change in oxidation number molecular weight ​ =1M​=M

Q. The equivalent weight of $K M n O_{4}$ in alkaline medium is equal to:

$\frac{M}{3}$

$\frac{M}{5}$

$\frac{M}{2}$

$M$

In alkaline medium
$2 K M n O_{4} + 7 + 2 K O H ⟶ 2 K_{2} M n O_{4} + 6 + H_{2} O + O$
Change in oxidation number $= 7 - 6 = 1$
Equivalent weight
$= \frac{molecular weight}{change in oxidation number}$
$= \frac{M}{1} = M$