Equivalent weight of potassium permanganate in alkaline medium is:

m o l e c u l a r w e i g h t

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\frac{m o l e c u l a r w e i g h t}{3}

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\frac{m o l e c u l a r w e i g h t}{5}

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\frac{m o l e c u l a r w e i g h t}{6}

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Solution

The correct option is A $m o l e c u l a r w e i g h t$
The reaction in alkaline medium is:

$K M n O_{4} + 1 e^{-} \to K_{2} M n O_{4}$

Thus, n factor is 1.

$E q u i v a l e n t w e i g h t = \frac{m o l e c u l a r w e i g h t}{n - f a c t o r}$ $\Rightarrow \frac{m o l e c u l a r w e i g h t}{1}$

Hence, the equivalent weight of potassium permanganate in alkaline medium is $\frac{m o l e c u l a r w e i g h t}{1}$ .

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Similar questions

H_{3} P O_{2} = \frac{m o l e c u l a r w e i g h t}{1}

H_{3} P O_{3} = \frac{m o l e c u l a r w e i g h t}{2}

H_{3} P O_{4} = \frac{m o l e c u l a r w e i g h t}{3}

N H_{3}

N_{2} \to N H_{3}

N_{2}

= \frac{Molecular weight}{Number of electrons lost or gained per molecule}

2 {N H}_{3} ⟶ N_{2} + 2 {N H}_{3}

{N H}_{3}

= \frac{M o l e c u l a r w e i g h t}{y}

y

= \frac{M o l e c u l a r w e i g h t}{A c i d i t y}

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