Question

\[HN{O_3}\] is converted to \[N{H_3}\], the equivalent weight of \[HN{O_3}\] will be:
(a) M/2
(b) M/1
(c) M/6
(d) M/8

Answer 1

Hint: The atomic weight of a compound is the product of its valence factor and equivalent weight. Thus, we can easily find the equivalent weight if we know the valence factor.

Complete step by step solution:
Valency or Valence factor refers to the valency in element, acidity of bases, basicity of acids and total charge of cation or anion in an ionic compound. The concept of equivalent weight allows us to explore the fact that atoms combine to form molecules in fixed number ratios, not mass ratios. That is, while element masses differ, when it comes to bonding with other atoms, the number of atoms, expressed in moles, is the determining factor in how much of a given element or compound will react with a given mass of another.
Valence factor denotes the change in oxidation number.
On conversion from nitric acid to ammonia,
The oxidation number for nitrogen in \[HN{O_3}\] on the reactant side is = +5
And the oxidation number for nitrogen in \[N{H_3}\] on the product side is = -3
Therefore,
Valence factor of \[HN{O_3}\]= Total change in oxidation number of nitrogen = 5-(-3) = 8.
We know,
\[Equivalent\,weight = \dfrac{{Molecular\,weight}}{{Valence\,factor}}\]
Now, we substitute the values in the equation:
\[Equivalent\,weight = \dfrac{M}{8}\]
Hence, the correct answer would be Option (D) \[\dfrac{M}{8}\]

Note: We must not confuse Valency and Oxidation number as there is a difference between them. Valency is the property of an isolated atom while the oxidation number is for an atom in a bonded state in a molecule.
We must remember Valency is combining capacity, it does not have +or - charge, while oxidation no. shows whether it can gain (+ve charge) or lose (-ve charge) electrons.