Question

Find the equivalent mass of $N{O_3}^ -$ in the following reaction.
N{O_3}^ - \to ,N{H_4}^ + ]
A. $7.75$
B. $31$
C. $62$
D. $15$

Answer 1

We all know that there are two types of weight used. One is molecular weight which is based on mole concept and the other one is gram equivalent weight based on equivalent weight. Unlike molecular weight, equivalent weight is proportional to mass of chemical entities which combines or displaces other chemical entities.

Complete answer:
Equivalent weight is the measure of mass proportion of an element, compound or ion in which it combines with the mass of other chemical entities.
Generally we have to determine the equivalent weights of elements and few compounds by using this definition of equivalent weights.
A definition of equivalent weight is given as it is the ratio of the molecular weight of the chemical entities to the valence factor.
General formula for equivalent weight (E) as follows:
$Equivalent\,weight\,(E)\, = \dfrac{{Molecular\,\,weight}}{{Valence\,\,factor}}$
This valence factor changes w.r.t chemical entities. Let’s see the formula of equivalent weight of an element, acid, base, compound, ion and oxidizing and reducing agent.
Equivalent weight of an element
$Equivalent\,weight\,(E)\, = \dfrac{{Atomic\,\,weight}}{{Valency}}$
Equivalent weight of an acids and bases
$Equivalent\,weight\,of\,acids\,(E)\, = \dfrac{{Molecular\,\,weight}}{{Basicity}}$
$Equivalent\,weight\,of\,Bases\,(E)\, = \dfrac{{Molecular\,\,weight}}{{Acidity}}$
Equivalent weight of a compound/an ion
$Equivalent\,weight\,(E)\, = \dfrac{{Molecular\,\,weight\,of\,compound/ion}}{{Number\,of\,electronic\,\,ch\arg e}}$
Equivalent weight of an oxidizing or reducing agent
$Equivalent\,weight\,(E)\, = \dfrac{{Molecular\,\,weight\,of\,compound/ion}}{{Number\,of\,electrons\,transfer}}$
In this question, we have to determine the equivalent mass of $N{O_3}^ -$
We use this formula,
$Equivalent\,weight\,(E)\, = \dfrac{{Molecular\,\,weight\,of\,compound/ion}}{{Number\,of\,electronic\,\,ch\arg e}}$
Number of electronic charge of nitrogen is $+ 5$ and of oxygen is $3 \times \,( - 1)\, = - 3$
Total number of electronic charge = $8$
Molecular weight of $N{O_3}^ -$= $14\, + \,\,(16\, \times 3)$
Molecular weight of $N{O_3}^ -$= $62$
Equivalent weight (E) of $N{O_3}^ -$ = $\dfrac{{62}}{8}$
Equivalent weight (E) of $N{O_3}^ -$= $7.75$
Hence, the correct answer is option A.

Note:
Do not consider the positive and negative sign while calculating the total number of electronic charges. Just take the whole number value and add them. Do not forget to multiply the charge on one atom in an ion with the total number of atoms present in that particular ion.