Question

Equivalent weight of $N{H}_3$ in the change, $N_2\to N{H}_3$ is

• A. 17/6
• B. 17
• C. 17/2
• D. 17/3

Answer 1

Answer: (D)

$N_2⟶N{H}_3$
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 $N{H}_3=1\times 10-(-3)I=3$
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
$\frac{17}{3}$.