In the oxidation of $N_{2} H_{4}$ to $N_{2}$ , Equivalent weight of $N_{2} H_{4}$ would be________

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Solution

Step 1: Oxidation number of the molecule
The oxidation state of $N$ in Hydrazine ( $N_{2} H_{4}$ ) is -2
The oxidation state of $N$ in Nitrogen molecule ( $N_{2}$ ) is 0
Hence, the total change in oxidation number per molecule of Hydrazine ( $N_{2} H_{4}$ ) is 4.
Step 2: Equivalent mass of Hydrazine.
Formula mass of Hydrazine = $2 (14) + 4 = 32$ .
$Equivalent mass = \frac{molecular mass}{total change in the oxidation state}$
So,
$Equivalent mass of Hydrazine = \frac{Formula mass of Hydrazine}{total change in the oxidation number of nitrogen permolecule of Hydrazine} = \frac{32}{4} = 8$
Therefore, the Equivalent weight of $N_{2} H_{4}$ would be 8.

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