Thank you for reporting, we will resolve it shortly
Q.
Equivalent weight of $NH_3$ in the change, $N_2 \rightarrow NH_3$ is
Redox Reactions
Solution:
$N _{2} \longrightarrow NH _{3}$
Oxidation no. of $N$ in $N _{2}=0$
Oxidation no. of $N$ in $NH _{3}=-3$
Valence factor $=$ no. of atoms $\times$ change in oxidation no.
$\therefore$ Valence factor for $NH _{3}=1 \times 10-(-3) I =3$
Molecular weight of $NH _{3}=17 g$
As we know that,
$
\text { eq. wt. }=\frac{\text { mol. wt. }}{\text { valence factor }}
$
$\therefore$ Equivalent weight of $NH _{3}$ for the given change
$=\frac{17}{3}$
Hence for the given change, the equivalent weigt of $NH _{3}$ will be
$\frac{17}{3}$.