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Concentration Terms

10. In alkali...

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10. In alkaline medium KMnO4 reacts as follows: 2KMnO4 + 2KOH - 2K2MnO4+H2O + O2. therefore it's equivalent mass will be : (1)31.6 (2)52.7(3)79 (4)158

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Q. In alkaline conditions,

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, its equivalent mass will be:

Q. In an 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

Equivalent weight of

K M n O_{4}

Q. In an alkaline medium,

K M n O_{4}

reacts as follows (At wt. of

K = 39.09

M n = 54.94

O = 16.0

2 K M n O_{4} + 2 K O H \to 2 K_{2} {M n O}_{4} + H_{2} O + [O]

i) The equivalent weight of

K M n O_{4}

in alkaline medium is 158.
ii) The number of electrons gained by one molecule of

K M n O_{4}

in the alkaline medium is 1.
iii) In the alkaline medium,

K M n O_{4}

acts as a reducing agent.

Select the correct option.

Q. In neutral medium,

K M n O_{4}

acts as a powerful oxidizing agent according to the reaction:

2 K M n O_{4} + H_{2} O \to 2 M n O_{2} + 2 K O H + 3 (O)

. The change involves

Q. Calculate the equivalent mass of

K M n O_{4}

in the below reaction.

M n O_{4}^{-} (+ 7) \to M n O_{4}^{- 2} (+ 6)

(alkaline medium)

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