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Standard XII

Chemistry

Valency and Oxidation States

In order to p...

Question

In order to prepare 1 litre normal solution of $K M n O_{4}$ , how many grams of $K M n O_{4}$ are required if the solution is to be used in acid medium for oxidation?

158 g

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31.6 g

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62 g

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790 g

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Solution

The correct option is B 31.6 g
In acid medium, Mn $^{7 +}$ +5e $\to$ Mn $^{2 +}$
Equivalent weight = $\frac{M}{5}$
Therefore, No. of equivalents
$\frac{ω}{M} \times 5$ =1N
$ω$ = $\frac{M}{5}$ = $\frac{158}{5}$ =31.6 g

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In order to prepare 1 litre normal solution of KMnO4, how many grams of KMnO4 are required if the solution is to be used in acid medium for oxidation?

The correct option is B 31.6 gIn acid medium, Mn7+ +5e → Mn2+ Equivalent weight = M5Therefore, No. of equivalents ωM×5 =1N ω = M5 = 1585 =31.6 g

In order to prepare 1 litre normal solution of $K M n O_{4}$ , how many grams of $K M n O_{4}$ are required if the solution is to be used in acid medium for oxidation?

The correct option is B 31.6 g
In acid medium, Mn $^{7 +}$ +5e $\to$ Mn $^{2 +}$
Equivalent weight = $\frac{M}{5}$
Therefore, No. of equivalents
$\frac{ω}{M} \times 5$ =1N
$ω$ = $\frac{M}{5}$ = $\frac{158}{5}$ =31.6 g