Question

Equivalent weight of $N_2$ in the change $N_2\to {NH}_3$ is:

• A. $\frac{28}{6}$
• B. $28$
• C. $\frac{28}{2}$
• D. $\frac{28}{3}$

Answer 1

Answer: (A)

$\frac{28}{6}$

$N_2⟶N{H}_3$

Oxidation no. of $N$ in $N_2=0$

Oxidation no. of $N$ in $N{H}_3=-3$

Valence factor $=$ no. of atoms $\times$ change in oxidation no. $=2\times |0-(-3)|=6$

Molecular weight of $N_2=28g$

As we know that,

$\text{eq. wt.}=\frac{\text{mol. wt.}}{\text{valence factor}}$

$\therefore$ Equivalent weight of $N_2$ for the given change $=\frac{28}{6}$

Hence for the given change, the equivalent weigt of $N_2$ will be $\frac{28}{6}$.