The molecular mass of KMnO4 is M. The equivalent weight of KMnO4 , when it is converted into KMnO4 is:

Question

The molecular mass of KMnO4 is M. The equivalent weight of KMnO4 , when it is converted into KMnO4 is: (A)  M  (B)  (M/3)  (C)  (M/5)  (D)  (M/7)

Tardigrade Admin · Accepted Answer

In this conversion, the oxidation-state change of Mn is equal to 1. overset+ 7 mathopKMO4 xrightarrow overset+ 6 mathopK2MnO4 Hence, equivalent weight of KMnO4=( textmolecular textmass/ textchange textin textoxidation textstate) =(M/1)=M

Q. The molecular mass of $K M n O_{4}$ is M. The equivalent weight of $K M n O_{4}$ , when it is converted into $K M n O_{4}$ is:

$M$

$\frac{M}{3}$

$\frac{M}{5}$

$\frac{M}{7}$

In this conversion, the oxidation-state change of Mn is equal to 1. $K M O_{4} + 7 K_{2} M n O_{4} + 6$ Hence, equivalent weight of $K M n O_{4} = \frac{molecular mass}{change in oxidation state}$ $= \frac{M}{1} = M$

Q. The molecular mass of KMnO4​ is M. The equivalent weight of KMnO4​ , when it is converted into KMnO4​ is:

M

3M​

5M​

7M​

In this conversion, the oxidation-state change of Mn is equal to 1. KMO4​+7​​K2​MnO4​+6​ Hence, equivalent weight of KMnO4​=changeinoxidationstatemolecularmass​ =1M​=M

Q. The molecular mass of $K M n O_{4}$ is M. The equivalent weight of $K M n O_{4}$ , when it is converted into $K M n O_{4}$ is:

$M$

$\frac{M}{3}$

$\frac{M}{5}$

$\frac{M}{7}$

In this conversion, the oxidation-state change of Mn is equal to 1. $K M O_{4} + 7 K_{2} M n O_{4} + 6$ Hence, equivalent weight of $K M n O_{4} = \frac{molecular mass}{change in oxidation state}$ $= \frac{M}{1} = M$