$K M n O_{4}$ acts as an oxidising agent in the neutral medium and gets reduced to $M n O_{2}$ .The equivalent weight of $K M n O_{4}$ in neutral medium

mol. wt/2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

mol. wt/3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

mol. wt/4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

mol .wt/7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B mol. wt/3
In neutral medium $M n^{+ 7}$ oxidation state changes into +4 oxidation state, hence equivalent weight of $K M n O_{4} = \frac{M}{3}$

Suggest Corrections

Similar questions

$K M n O_{4}$ acts as an oxidising agent in the neutral medium and gets reduced to $M n O_{2}$ . The equivalent weight of $K M n O_{4}$ in neutral medium is ______ .

Q. Assertion :

K M n O_{4}

acts as an oxidising agent in acidic, basic or neutral medium. Reason:

K M n O_{4}

oxidises ferrous sulphate to ferric sulphate.

Q. Why is

H C l

not used to make the medium acidic in oxidation reactions of

K M n O_{4}

in acidic medium?

Q. The equivalent weight of

K M n O_{4}

(formula weight

=

M), when it is used as an oxidant in neutral medium, is:

K M n O_{4}

acts as an oxidising agent in alkaline medium. When alkaline

K M n O_{4}

is treated with KI, iodide ion is oxidised to:

Join BYJU'S Learning Program

KMnO4 acts as an oxidising agent in the neutral medium and gets reduced to MnO2.The equivalent weight of KMnO4 in neutral medium

The correct option is B mol. wt/3In neutral medium Mn+7 oxidation state changes into +4 oxidation state, hence equivalent weight of KMnO4=M3

$K M n O_{4}$ acts as an oxidising agent in the neutral medium and gets reduced to $M n O_{2}$ .The equivalent weight of $K M n O_{4}$ in neutral medium

The correct option is B mol. wt/3
In neutral medium $M n^{+ 7}$ oxidation state changes into +4 oxidation state, hence equivalent weight of $K M n O_{4} = \frac{M}{3}$