Question

The equivalent mass of nitrogen in nitrogen pentoxide is :-
(A). 14
(B). 28
(C). 2.8
(D). 4.66

Answer 1

Equivalent mass of any substance can be calculated by keeping in mind the atomic mass of that particular element and the valency .

Complete step by step answer:
Volume of nitrogen is to be calculated first of all. So, we will begin with finding out the valency of Nitrogen. The atomic number of nitrogen is $7$ . So, its electronic configuration will be $2,5.$
Valency is the combining capacity of an element. In order to calculate its valency , we need to see the electrons in this outermost shell which are $5$ .
So, its requires $3$ more electrons to complete this octet, which gives this valency to be $3.$
The atomic mass of nitrogen is $14{\text{ amu}}$ Now , we how to calculate the valency of nitrogen in ${N_2}{O_{5.}}$ .
Let we can assume that the valency of $N$ in ${N_2}{O_5}$ is $x$ and we also know that the valency of $0$ is $2$ or the oxidation state of oxygen is $- 2$.
We will calculate it like this –
${N_2}{O_5} = 2 \times x + 5 \times \left( { - 2} \right) = 0$
$= 2x - 10 = 0$
$x = 5$
This gives is the valency of nitrogen in ${N_2}{O_5}$ as $5.$
Now as per the formula of equivalent mass,
$\operatorname{equivalent} mass = \dfrac{{Atomic mass}}{{Valency}} - - - - - - - - - - - - - - \left( 1 \right)$
We will substitute the values of Atomic mass and valency in this formula $\left( 1 \right)$
$\operatorname{equivalent} mass {\text{ = }}\dfrac{{14}}{5} = 2.8$
So, the equivalent mass of nitrogen in ${N_2}{O_5}$ is $2.8$. $\left( C \right)$ is the correct option.

Note:
Equivalent mass is different from molecular mass,that is, it is the mass of chemical or compound or ion which displaces or combines 1 part of hydrogen or 8 parts of oxygen or $35.5$ parts of chlorine by mass .Sometimes, when the valency of an element is not given it becomes difficult to calculate the equivalent mass of the element but in that case there are some alternative methods to do so.
1. Hydrogen displacement method
$equivalent{\text{ }}mass = \;\;\dfrac{{\left( {mass{\text{ }}of{\text{ }}metal{\text{ }} \times {\text{ }}11200} \right)}}{{volume{\text{ }}of\;hydrogen{\text{ }}liberated{\text{ }}at{\text{ }}stp}}$
2. Oxide method
$equivalent{\text{ }}mass{\text{ }} = \dfrac{{\left( {mass{\text{ }}of{\text{ }}metal{\text{ }} \times 8} \right)}}{{mass{\text{ }}of{\text{ }}oxygen{\text{ }}in\;\;the{\text{ }}oxide}}$
3. Chloride method
$equivalent{\text{ }}mass = \dfrac{{\left( {mass{\text{ }}of{\text{ }}metal{\text{ }} \times 35.5} \right)}}{{mass{\text{ }}of{\text{ }}chlorine{\text{ }}in{\text{ }}the{\text{ }}chloride}}$