Question

Equivalent weight of $N{H}_3$ in the change $N_2\to N{H}_3$is:
A. ${\displaystyle \frac{17}{6}}$
B. 17
C. ${\displaystyle \frac{17}{2}}$
D. ${\displaystyle \frac{17}{3}}$

Answer 1

Hint: The equivalent weight is calculated by dividing the molecular weight of the compound by the valency factor. The charge of nitrogen in dinitrogen is zero as it is a divalent molecule and the charge of each nitrogen cancels out each other.

Complete step by step answer:
The equivalent weight is defined as the mass of the substance present in grams which is chemically equivalent to the eight grams of oxygen atom and one gram of hydrogen atom.
The equivalent weight is calculated by dividing the molecular weight by the valency factor.
The formula to calculate the equivalent weight is shown below.
$$E={\displaystyle \frac{M}{V}}$$
Where,
E is the equivalent weight
M is the molecular weight
V is the valency factor
In the given reaction, dinitrogen is converted to ammonia. Dinitrogen is a divalent molecule which contains two nitrogen atoms. There the charge on the nitrogen is zero.
The molecular weight of $N_2$ is 28.014 g/mol.
The equivalent weight of dinitrogen is also 28.014.
The molecular weight of ammonia is 17.031 g/mol.
The charge of nitrogen in ammonia $N{H}_3$is 3 and charge of hydrogen is 1.
To calculate the equivalent weight of nitrogen in ammonia, substitute the values of molecular weight and charge on the above equation.
$$\Rightarrow E={\displaystyle \frac{17.031}{3}}$$
Thus, the equivalent weight of $N{H}_3$ in the change $N_2\to N{H}_3$ is ${\displaystyle \frac{17}{3}}$.
Therefore, the correct option is D.

Note: Make sure that in calculating the equivalent weight of ammonia only charge of ammonia is divided as the charge of nitrogen changes from zero to three.