Question

Equivalent mass of $A{s_2}{S_3}$ in the following reaction is
 \[A{s_2}{S_3} + HN{O_3} \to {H_3}As{O_4} + {H_2}S{O_4} + NO\]
A. $E = \dfrac{M}{{10}}$
B. $E = \dfrac{M}{{28}}$
C. $E = \dfrac{M}{{20}}$
D. $E = \dfrac{M}{{24}}$

Answer 1

Hint: Since the given equation in the question is a redox reaction, so in order to determine the equivalent weight in this type of reactions, first of all, we have to obtain the molecular weight of Arsenic sulphide. And secondly, we have to follow the change in oxidation number of ‘As’ before the reaction and after it to determine the valency. Also, we know that Equivalent weight is $\dfrac{M}{{{N_f}}}$.

Complete step by step answer:
The given equation is
 \[A{s_2}{S_3} + HN{O_3} \to {H_3}As{O_4} + {H_2}S{O_4} + NO\]
This reaction is a redox involving multiple oxidation
Both as and S has been oxidised
 \[A{s_2}{S_3}\] is oxidised to \[{H_3}As{O_4}\]
We know that the,
Equivalent weight $ = \dfrac{M}{{{N_f}}}$
Also
 \[{N_f} = \sum (\Delta OS)\]

Which actually means that we need to add all the changes in oxidation states
Now, N factor or \[{N_f} = (2 \times 2)\,for\,\;A{S^ + }(8 \times 3)\,for\,\;S = 28\]

Thus, equivalent mass of \[A{s_2}{S_3} = \dfrac{{\left( {Molar\,mass} \right)}}{{\left( {change\,in\,oxidation} \right)}} = \dfrac{M}{{28}}\]​

Therefore, the correct answer is option (B).

Note: N factor \[\left( {{N_f}} \right)\] means a conversion factor by which we divide molar mass of substance to get equivalent mass and it depends on nature of substance which varies from one condition to another condition. We can divide n-factor calculation into two categories. In case of non-redox reaction, n factor of acid is equal to the basicity of the acid and n factor of base is equal to the acidity of the base. The salts that react in such a way that only one atom undergoes change in oxidation state and goes only in one product, so the n-factor of such salts is defined as the number of moles of electrons exchanged (lost or gained) by one mole of the salt.