Question

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

• A. $\frac{17}{6}$
• B. $17$
• C. $\frac{17}{2}$
• D. $\frac{17}{3}$

Answer 1

Answer: (D)

$\frac{17}{3}$

Solution:- (D) $\frac{17}{3}$

$N_2⟶N{H}_3$

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.

$\therefore$ Valence factor for $N{H}_3=1\times |0-(-3)|=3$

Molecular weight of $N{H}_3=17g$

As we know that,

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

$\therefore$ Equivalent weight of $N{H}_3$ for the given change $=\frac{17}{3}$

Hence for the given change, the equivalent weigt of $N{H}_3$ will be $\frac{17}{3}$.