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Atomic Valency

The equivalen...

Question

The equivalent weight of $K M n O_{4}$ when it changes to $K_{2} M n O_{4}$ is ______.

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Solution

One mole of $K M n O_{4}$ accepts 1 mole of electrons to form one mole of $K_{2} M n O_{4}$ .

$E q u i v a l e n t w e i g h t = \frac{M o l e c u l a r w e i g h t}{N u m b e r o f e l e c t r o n s t r a n s f e r e d} = \frac{158}{1} = 158$

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Similar questions

M

is the molecular mass of

K M n O_{4}

. What will be its equivalent weight when it is converted into

K_{2} M n O_{4}

?

Q. If the molecular weight of

{K M n O}_{4}

is taken as M, its equivalent weight in the reaction where it is converted to

{K_{2} M n O}_{4}

Q. Molecular weight of

K M n O_{4}

158

K M n O_{4}

can be reduced to

M n {S O}_{4}, K_{2} M n O_{4}, M n O_{2}

and equivalent weight of

K M n O_{4}

comes out to be

158, 52.66, 31.6

. Match the equivalent weights with compounds formed by reduction.

Equivalent weight of $K M n O_{4}$	Compound formed
$158$	.......................
$52.66$	........................
$31.66$	........................

Q. The Molecular weight of Mohr's salt is 392. Its equivalent weight when it is oxidized by

K M n O_{4}

KMnO4, in acidic medium is:

Q. The equivalent weight of

F e C_{2} O_{4}

when it reacts with acidified

K M n O_{4}

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