Question

The equivalent weight of Mohr's salt, $FeS{O}_{4} . {(N{H}_{4})}_{2}S{O}_{4} . 6{H}_{2}O$ is equal to:
a.) its molecular weight
b.) atomic weight
c.) half-its molecular weight
d.) one-third its molecular weight

Answer 1

Hint: Equivalent is the quantity that combines with or replaces 1 g of hydrogen or 8 g of oxygen. Molecular weight is the sum of the weight of the individual atoms of the molecule.

Complete step by step solution:
Equivalent weight can be defined as the one equivalent which is the mass of a given substance which combines with or replaces a fixed quantity of another substance. It is the mass of a substance that combines with or is chemically equivalent to one gram of hydrogen or eight grams of oxygen.
$ FeS{O}_{4}$ in the Mohr's salt, when used in a redox reaction gets oxidized to form ${Fe}_{2}{(S{O}_{4})}_{3}$.
$ { Fe }^{ +2 }\quad \longrightarrow \quad { Fe }^{ +3 }$
The oxidation of Fe changes from +2 to +3.
The n-factor is the change in the oxidation state of any compound.
Therefore, n-factor of Mohr's salt = change in oxidation state of Fe.
Thus, n-factor = 1.
The equivalent weight is molecular weight divided by the n-factor.
$Equivalent\quad weight\quad =\quad \cfrac { Molecular\quad Weight }{ n-factor }$
Substituting the values, we get,
$Equivalent\quad weight\quad =\quad \cfrac { Molecular\quad Weight }{ 1 }$
Thus, we get the equivalent weight = Molecular weight.
Hence, option (a) is the correct option.

Note: Students tend to understand the molecular weight and the equivalent weight of a compound as equal. But they are different from each other. Molecular weight is the sum of the weights of the individual atoms in the molecule whereas, the equivalent weight is molecular weight divided by the n-factor.