In the Haber's process the equivalent weight of ammonia is:

[Molecular weight of $N H_{3}$ = M]

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M/3

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2M/3

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Solution

The correct option is C M/3
According to Haber's process,

$3 H_{2} + N_{2} ⇋ 2 N H_{3}$

Change in oxidation number of $N = 0 - (- 3) = 3$

Equivalent weight of $N H_{3} = \frac{M}{3}$

Hence, option B is correct.

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