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Hint: Start this question by writing the reaction between Mohr’s salt and \[KMn{{O}_{4}}\]. Balance the reaction to find the n-factor and find the equivalent mass of Mohr’s salt.

Complete step by step answer:

Mohr’s salt is an alternative name for ammonium iron (II) sulphate. It is a double salt of ferrous sulphate and ammonium sulphate. Its formula is – \[{{(N{{H}_{4}})}_{2}}Fe{{(S{{O}_{4}})}_{2}}.6{{H}_{2}}O\].

In the given question, let us consider the acidic medium to be sulphuric acid.

The molecular equation of the given reaction can be given as –

\[\begin{align}

& 2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+3{{H}_{2}}O+5[O]\cdots \cdots (i) \\

& 2FeS{{O}_{4}}{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O+{{H}_{2}}S{{O}_{4}}+[O]\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+2{{(N{{H}_{4}})}_{2}}S{{O}_{4}}+13{{H}_{2}}O\cdots \cdots (ii) \\

\end{align}\]

Now, \[(i)+5(ii)\] gives us –

\[2KMn{{O}_{4}}+8{{H}_{2}}S{{O}_{4}}+10FeS{{O}_{4}}{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+5F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+10{{(N{{H}_{4}})}_{2}}S{{O}_{4}}+68{{H}_{2}}O\]

The ionic equation of the given reaction can be given as –

\[\begin{align}

& MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{+2}}+4{{H}_{2}}O\cdots \cdots (iii) \\

& F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}\cdots \cdots (iv) \\

\end{align}\]

Now, \[(iii)+5(iv)\]gives us –

\[MnO_{4}^{-}+8{{H}^{+}}+5F{{e}^{2+}}\to 5F{{e}^{3+}}+M{{n}^{+2}}+4{{H}_{2}}O\]

So, From the above equation, we can see that the change in oxidation number of Iron

\[\begin{align}

& =\left( +3 \right)-\left( +2 \right) \\

& =+1 \\

\end{align}\]

Hence, n-factor = 1.

Also, the molecular mass of Mohr’s salt is 392g.

As we know, Equivalent weight of a compound is calculated by the formula –

\[Eq.wt.=\dfrac{Mol.wt.}{n-factor}\]\[=\dfrac{392}{1}\]\[=392g/mol\]

Therefore, the equivalent mass of Mohr’s salt is 392g. According to the question, the equivalent mass of Mohr’s salt is ABC. So, A=3, B=9, C=2.

Therefore, the answer is – B corresponds to 9.

Note: Equivalent weight is also known as gram equivalent). It is defined as, “the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance”. In other words, “the equivalent weight of an element is defined as the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine. These values correspond to the atomic weight divided by the usual valence”.

Complete step by step answer:

Mohr’s salt is an alternative name for ammonium iron (II) sulphate. It is a double salt of ferrous sulphate and ammonium sulphate. Its formula is – \[{{(N{{H}_{4}})}_{2}}Fe{{(S{{O}_{4}})}_{2}}.6{{H}_{2}}O\].

In the given question, let us consider the acidic medium to be sulphuric acid.

The molecular equation of the given reaction can be given as –

\[\begin{align}

& 2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+3{{H}_{2}}O+5[O]\cdots \cdots (i) \\

& 2FeS{{O}_{4}}{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O+{{H}_{2}}S{{O}_{4}}+[O]\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+2{{(N{{H}_{4}})}_{2}}S{{O}_{4}}+13{{H}_{2}}O\cdots \cdots (ii) \\

\end{align}\]

Now, \[(i)+5(ii)\] gives us –

\[2KMn{{O}_{4}}+8{{H}_{2}}S{{O}_{4}}+10FeS{{O}_{4}}{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+5F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+10{{(N{{H}_{4}})}_{2}}S{{O}_{4}}+68{{H}_{2}}O\]

The ionic equation of the given reaction can be given as –

\[\begin{align}

& MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{+2}}+4{{H}_{2}}O\cdots \cdots (iii) \\

& F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}\cdots \cdots (iv) \\

\end{align}\]

Now, \[(iii)+5(iv)\]gives us –

\[MnO_{4}^{-}+8{{H}^{+}}+5F{{e}^{2+}}\to 5F{{e}^{3+}}+M{{n}^{+2}}+4{{H}_{2}}O\]

So, From the above equation, we can see that the change in oxidation number of Iron

\[\begin{align}

& =\left( +3 \right)-\left( +2 \right) \\

& =+1 \\

\end{align}\]

Hence, n-factor = 1.

Also, the molecular mass of Mohr’s salt is 392g.

As we know, Equivalent weight of a compound is calculated by the formula –

\[Eq.wt.=\dfrac{Mol.wt.}{n-factor}\]\[=\dfrac{392}{1}\]\[=392g/mol\]

Therefore, the equivalent mass of Mohr’s salt is 392g. According to the question, the equivalent mass of Mohr’s salt is ABC. So, A=3, B=9, C=2.

Therefore, the answer is – B corresponds to 9.

Note: Equivalent weight is also known as gram equivalent). It is defined as, “the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance”. In other words, “the equivalent weight of an element is defined as the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine. These values correspond to the atomic weight divided by the usual valence”.

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