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# $1$ mole of $FeS{O_4}$ (atomic weight of Fe is $55.84gmo{l^{ - 1}}$ ) is oxidized $F{e_2}{(S{O_4})_3}$ . Calculate the equivalent weight of ferrous ions. $(A)55.84$  $(B)27.92$  $(C)18.61$  $(D)111.68$  $(E)83.76$

Last updated date: 14th Jul 2024
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Hint :Oxidation number of an element is the measure of the loss of electrons when they are in a particular state. It is basically a number which denotes whether the atom has oxidized or reduced when reaction has occurred. In oxidation the number increases and if the atom reduces oxidation number decreases.
Equivalent mass $= \dfrac{{{\text{molar mass }}}}{{{\text{no of electrons gained or lost}}}}$

To calculate the equivalent weight of ferrous ion, we need to first calculate the change in oxidation state of Fe when one mole of $FeS{O_4}$ oxidizes to $F{e_2}{(S{O_4})_3}$ .
Let us write the chemical equation involves with the above case:
$FeS{O_4} \to F{e_2}{(S{O_4})_3} \\$
$FeS{O_4} \to F{e^{ + 2}} + S{O_4}^{ - 2}$
$F{e_2}{(S{O_4})_3} \to F{e^{ + 3}} + 3S{O_4}^{ - 2}$
$F{e^{ + 2}} \to F{e^{ + 3}} + {e^{ - 1}}$
Thus the change of electrons in oxidation of $F{e^{ + 2}}$ to $F{e^{ + 3}}$ involves loss of one electron.
Given the molecular mass of iron is $55.84gmo{l^{ - 1}}$ and the value of n is $1$ . By applying the formula for equivalent weight:
Equivalent mass $= \dfrac{{{\text{molecular weight of Iron}}}}{n}$
Equivalent mass $= \dfrac{{55.84}}{1}$
Thus, the equivalent mass of ferrous ion when it oxidizes from $FeS{O_4}$ to $F{e_2}{(S{O_4})_3}$ is $55.84$ .
Hence option (A) $55.84$ is the correct choice.

Note :
Equivalent weight is helpful in determining the normality of a solution. Normality is defined as the gram equivalent of solute present in solution. To calculate equivalent weight of acid and base we divide their molar mass by basicity and acidity respectively. For salts we take molar mass to the ratio of positive valency of salt.