The equivalent weights of oxidising and reducing agents can be calculated by the number of electrons gained or lost. The equivalent weight of an oxidising agent is the number of parts by weight of the substance which gains one electron. Thus, it is equal to the molecular weight of the substance divided by the number of electrons gained in the balanced chemical equation. Similarly, equivalent weight of a reducing agent is equal to the molecular weight divided by the number of electrons lost as represented in the balanced chemical equation
The equivalent weght of `As_(2),S_3` in the following reaction
`As_(2)S_(3) + H^(+) + NO_(3)^(-) rarr NO + H_(2)O + AsO_(4)^(3-) + SO_(4)^(2-)` is related to its molecular weight as

Options

A. M/5
B. M /7
C. M / 14
D. M/28

Answer

D

Explanation

Calculate the total change in oxidation number of one molecule of `As_(2)O_(3)`
`As_(2)O_(3) =2As^(3+) +3S^(2-)`
`2As^(3+) rarr 2(AsO_4)^(3-) `
Increase in O.N = 4 `(6 rarr 10)`
`3S^(-2) rarr 3 overset(+6)(SO_4)^(2-)`
Increase in O.N. = 24 (`-6 rarr +18) `
Total increasein O.N. = 4 + 24 =28
Hence Eq. wt. = Mol . wt /28.