When $N_{2}$ is converted into ${NH}_{3},$ the equivalent weight of nitrogen will be

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4.67

1.67

3.67

2.67

answer is D.

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Detailed Solution

$\begin{array}{l} {\overset{0}{N}}_{2} \overset{}{\to} \overset{- 3}{N} H_{3} \\ Change in oxidation state = n - factor = 3 \\ Eq . wt of' N' = mol . wt . / n factor = \frac{14}{3} = 4 .67 \end{array}$

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