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Hint: Reductant is also known as reducing agent means it helps in the reduction process by losing electrons from itself. Equivalent weight of any molecule is calculated as molecular weight is divided by valency of that molecule or acidity or basicity. But, in the given question for calculating equivalent weight molecular weight will be divided by number of electrons involved in the given process.
Complete step by step solution: Chemical formula of ferrous sulphate is \[{\text{FeS}}{{\text{O}}_{\text{4}}}\], in which iron (Fe) is present in +2 oxidation state and sulphur is also present in the -2 oxidation state. As it is mentioned in the question that ferrous sulphate is present as reductant where they remove one electron for helping in the reduction reaction & turns into the +3 oxidation state.
-For calculating molecular weight we have to add the atomic weight of all atoms present in the molecule.
\[\begin{gathered}
{\text{FeS}}{{\text{O}}_{\text{4}}} = 55.8 + 32 + (4 \times 16) \\
{\text{FeS}}{{\text{O}}_{\text{4}}} = 151.8 \\
\end{gathered} \]
So, the molecular weight of ferrous sulphate is $151.8$.
-Equivalent weight of ferrous sulphate = molecular weight of ferrous sulphate/ involved electron.
\[{\text{FeS}}{{\text{O}}_{\text{4}}} = \dfrac{{151.8}}{1} = 151.8\]
So, equivalent weight of ferrous sulphate which is present as reductant is $151.8$ which is equal to the molecular weight of ferrous sulphate.
Hence, option (B) is correct.
Note: Here, some of you may divide the molecular weight by 3 because by behaving as a reductant oxidation state of iron it will be as +3. But in this condition in the single molecule reductant & oxidant both are present that’s why we divide by the involved electron only.
Complete step by step solution: Chemical formula of ferrous sulphate is \[{\text{FeS}}{{\text{O}}_{\text{4}}}\], in which iron (Fe) is present in +2 oxidation state and sulphur is also present in the -2 oxidation state. As it is mentioned in the question that ferrous sulphate is present as reductant where they remove one electron for helping in the reduction reaction & turns into the +3 oxidation state.
-For calculating molecular weight we have to add the atomic weight of all atoms present in the molecule.
\[\begin{gathered}
{\text{FeS}}{{\text{O}}_{\text{4}}} = 55.8 + 32 + (4 \times 16) \\
{\text{FeS}}{{\text{O}}_{\text{4}}} = 151.8 \\
\end{gathered} \]
So, the molecular weight of ferrous sulphate is $151.8$.
-Equivalent weight of ferrous sulphate = molecular weight of ferrous sulphate/ involved electron.
\[{\text{FeS}}{{\text{O}}_{\text{4}}} = \dfrac{{151.8}}{1} = 151.8\]
So, equivalent weight of ferrous sulphate which is present as reductant is $151.8$ which is equal to the molecular weight of ferrous sulphate.
Hence, option (B) is correct.
Note: Here, some of you may divide the molecular weight by 3 because by behaving as a reductant oxidation state of iron it will be as +3. But in this condition in the single molecule reductant & oxidant both are present that’s why we divide by the involved electron only.
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