Question

The equivalent weight of $FeSO_{ 4 }.(NH_{ 4 })_{ 2 }SO_{ 4 }.24H_{ 2 }O$ is:
(At. wt of Fe = 56, S = 32, O = 16, N = 14, H = 1)
A.) 58
B.) 358
C.) 179
D.) 89.5

Answer 1

Hint: The gram equivalent weight of any given compound is equal to its molecular weight (sum of the weight of atoms present in this compound) divided by the valency factor of the compound. You will surely get your answer after applying these steps.

Complete step by step answer:
First, let’s calculate the molecular mass of $FeSO_{ 4 }.(NH_{ 4 })_{ 2 }SO_{ 4 }.24H_{ 2 }O$:
At. wt of Fe = 56, S = 32, O = 16, N = 14, H = 1

56+32+(4x16)+(2x18)+32+(4x16)+(24x18) = 716 g/mol
Valency factor = total positive or total negative charge on ions in the solution (here we are considering positive charge)
= 2(from $Fe^{ 2+ }$) + 2(from $NH_{ 4 }^{ + }$) = 4 ( it means there is a transfer of 4 electrons from cations to anions )
$FeSO_{ 4 }.(NH_{ 4 })_{ 2 }SO_{ 4 }.24H_{ 2 }O\quad \rightarrow \quad Fe^{ 2+ }\quad +\quad 2NH_{ 4 }^{ + }\quad +\quad 2SO_{ 4 }^{ 2- }\quad +\quad 24H_{ 2 }O$
Now we have to calculate the gram equivalent weight of Mohr's salt.

Formula used :
$Equivalent\quad weight\quad =\quad \dfrac { Molecular\quad weight }{ Valency\quad factor }$
$Equivalent\quad weight\quad =\quad \dfrac { 716 }{ 4 }$
Equivalent weight = 179 g/eq

Hence, the equivalent weight of $FeSO_{ 4 }.(NH_{ 4 })_{ 2 }SO_{ 4 }.24H_{ 2 }O$ is 179 g/eq. Therefore, the correct answer to this question is option C.

Note: This formula is somewhat like Mohr’s salt. Ammonium iron(II) sulfate, or Mohr's salt, is the inorganic compound with the formula $(NH_{ 4 })_{ 2 }Fe(SO_{ 4 })_{ 2 }(H_{ 2 }O)_{ 6 }$. Containing two different cations, $Fe^{ 2+ }$ and $NH_{ 4 }^{ + }$, it is classified as a double salt of ferrous sulfate and ammonium sulfate.