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Assertion: Equivalent weight of $\mathrm{NH}_3$ in the reaction, $\mathrm{N}_2 \rightarrow \mathrm{NH}_3$ is $17 / 3$ while that of $\mathrm{N}_2$ is 28/6.
Reason : Equivalent weight
$=\frac{\text { Molecular weight }}{\begin{array}{c}\text { Number of } e^{-} \text {lost or gained } \\ \text { per mole }\end{array}}$
Options:
Reason : Equivalent weight
$=\frac{\text { Molecular weight }}{\begin{array}{c}\text { Number of } e^{-} \text {lost or gained } \\ \text { per mole }\end{array}}$
Solution:
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Verified Answer
The correct answer is:
If both assertion and reason are true and reason is the correct explanation of assertion.
$: \stackrel{0}{\mathrm{~N}}_2+6 e^{-} \rightarrow 2 \mathrm{~N}^{3-}$
$\therefore \quad$ Equivalent weight of $\mathrm{NH}_3=\frac{14+3}{3}=\frac{17}{3}$
(Mol. wt. of $\mathrm{NH}_3=17$ )
While, equivalent weight of $\mathrm{N}_2=\frac{14 \times 2}{6}=\frac{28}{6}$
$\therefore \quad$ Equivalent weight of $\mathrm{NH}_3=\frac{14+3}{3}=\frac{17}{3}$
(Mol. wt. of $\mathrm{NH}_3=17$ )
While, equivalent weight of $\mathrm{N}_2=\frac{14 \times 2}{6}=\frac{28}{6}$
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