Equivalent weight of ${NH}_{3}$ in the change
$N_{2} ⟶ {NH}_{3}$

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$\frac{17}{6}$

$\frac{17}{2}$

$\frac{17}{3}$

answer is D.

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Detailed Solution

Molecular weight of NH 3 =17g

Oxidation number of nitrogen in $N_{2} = 0$

Oxidation no. of N in ${NH}_{3}$ = -3

We already know that

eq. weight = mole weight / valence factor.

NH3 equivalent weight for the specified change Consequently, the equivalent weight of NH with the specified change $= \frac{17}{3}$ shall be

Equivalent weight of $N H_{3}$ in the change $N_{2} ⟶ N H_{3}$ is : 17/3

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