Question

The equivalent weight of Mohr’s salt \[FeSO4.{{\left( N{{H}_{2}} \right)}_{2}}S{{O}_{4}}.6{{H}_{2}}O\] is equal to:
A. Atomic weight
B. Its molecular weight
C. Half its molecular weight
D. One third its molecular weight

Answer 1

Hint: We should here revise the concept of equivalent weight. We have to find n-factor for determination of equivalent weight.

Complete step by step answer:
Mohr’s salt is ferrous ammonium sulphate. It is classified as a double salt of ferrous sulphate and ammonium sulphate.
We use Ferrous Ammonium Sulphate in common labs for performing a qualitative chemical analysis which is used to identify the unknown concentration of a solution. Ferrous ions of Mohr’s salt go through the hydrolysis in aqueous solution.
As Mohr’s salt is a double salt. When we dissolve it in aqueous solution Mohr’s salt gives ammonium ion, ferrous ion and sulphate ions.
In aqueous medium it acts only as a reducing agent. It transforms from ferrous to ferric ion.
\[F{{e}^{2+}}\to F{{e}^{3+}}+\text{ }{{e}^{-}}\]
We can easily calculate the molecular weight of Mohr’s salt.
Molecular weight= 392.2 gram per mole
And after observing the above reaction we get that only one electron is lost during the reaction.
$Equivalent\,weight=\dfrac{Molecular\,weight\,of\,Mohr\,salt}{Number\,of\,electrons\,gained\,or\,lost}=\dfrac{392.2}{1}$= 392.2

So, above discussion we found our answer. Our correct answer is option B. this is because, in aqueous medium it only loses one electron and thus its valency factor becomes equal to one. By dividing, the molecular weight of Mohr’s salt with valency factor its molecular weight will remain the same.
So our correct option is B, which is equivalent to the weight of Mohr’s salt will be equal to molecular weight.

Additional information:
We use Mohr’s salt as double salt because a double salt is a crystalline salt having the composition of a mixture of two simple salts but with an unlike crystal structure from either. It contains two different cations of \[F{{e}^{2+}}\], \[N{{H}^{4+}}\].
We can prepare Mohr’s salt by dissolving an equal amount of mixture of hydrated ferrous sulphate and ammonium sulphate in water that contains sulphuric acid. Because of this crystallisation occurs and the solution looks green. The reaction looks like this
\[FeS{{O}_{4}}.7H_2O+{{\left( N{{H}_{4}} \right)}_{2}}SO_4\text{ }\to \,~FeSO_4.{{\left( N{{H}_{2}} \right)}_{2}}S{{O}_{4}}.6{{H}_{2}}O\]

Note: We should carefully follow precaution when we are handling Mohr’s salt, because inhalation of this salt causes irritations in the nose and throat. Intake of the chemical causes nuisance in the mouth and stomach. Toxic ammonia and oxides of nitrogen may form in fires.