Question

What is the equivalent mass of $A{s}_2{S}_3$ in the following reactions? (Molar mass of $A{s}_2{S}_3$=M)
$A{s}_2{S}_3+N{O}_3^{-}+H^{+}\to A{s}_2{O}_5+N{O}_2^{-}+S{O}_3+H_{2}O$
A. M/4
B. M/24
C. M/28
D. M/12

Answer 1

Hint:The equivalent mass of the chemical compound gives the relation between the molecular weight or molar mass and the n- factor. For acid compounds, the n- factor is defined as the number of hydrogen ions which are displaced by the 1 mole of acid during a chemical reaction. For basic compounds, the n-factor is defined as the number of hydroxide ions which are displaced by the 1 mole of base during the chemical reaction.

Complete step by step answer:
$A{s}_2{S}_3+N{O}_3^{-}+H^{+}\to A{s}_2{O}_5+N{O}_2^{-}+S{O}_3+H_{2}O$
In this reaction oxidation of $A{s}_2{S}_3$ takes place, the oxidation reaction is shown below.
$A{s}_2{S}_3\to A{s}_2{O}_5+S{O}_3$
Balance the equation, the number of moles on both sides of the reaction should be the same.
$A{s}_2{S}_3\to A{s}_2{O}_5+3S{O}_3$
$A{s}_2^{+3}{S}_3^{-2}\to A{s}_2^{+5}{O}_5^{-3}+3S^{+6}{O}_3^{-2}$
The oxidation state of As changes from + 3 to +5, the oxidation state of S changes to -2 to +6.
Valence factor $=2\times (+2)+3\times (8)$
$\Rightarrow$Valence factor $=28$
There are a total 28 electrons gained by one molecule.
The equivalent weight is calculated by dividing molecular weight or molar mass by the valency factor.
The formula for calculating equivalent weight is shown below.
$Eq.Wt={\displaystyle \frac{M}{ValencyFactor}}$
Where,
M is the molecular weight.
To calculate the equivalent weight, substitute the value in the above equation.
$Eq.wt={\displaystyle \frac{M}{28}}$
Thus the equivalent mass of $A{s}_2{S}_3$ in this reaction is M/28.
Therefore, the correct option is C.

Note:
There is a relation between the equivalent weight, moles and valency factor when moles of the compound are given. The equivalent weight is thus calculated by multiplying moles by valency factor.