Question

The equivalent weight of Mohr’s salt is

Answer 1

Explanation

• Mohr’s salt:
• Mohr’s Salt is an inorganic compound named after the German scientist Karl Friedrich MohrIt is also referred to as ammonium iron(II) sulfate.
• The chemical formula of anhydrous Mohr’s salt is $\mathrm{Fe}\left({\mathrm{SO}}_4\right){\left({\mathrm{NH}}_4\right)}_2\left({\mathrm{SO}}_4\right)$. The hexahydrate form can be represented by the chemical formula $\mathrm{Fe}\left({\mathrm{SO}}_4\right){\left({\mathrm{NH}}_4\right)}_2\left({\mathrm{SO}}_4\right).6{\mathrm{H}}_2\mathrm{O}$ .
• The hexahydrate of Mohr’s salt is known to have a molecular weight of 392.13 grams per mole.

Equivalent weight can be calculated by

• Equivalent weight can be calculated by
• .$\mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{Molar}\mathrm{mass}}{\mathrm{number}\mathrm{of}\mathrm{electrons}\mathrm{lost}\mathrm{or}\mathrm{gained}}$

The equivalent weight of Mohr’s salt :

• ${\mathrm{FeSO}}_4$ in the Mohr’s salt,
• In a redox reaction gets oxidized to form ${\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3$
• Fe⁺²⟶Fe⁺³
• The oxidation of Fe varies from +2 to +3.
• The n-factor has described the change in the oxidation state
• The n-factor of Mohr’s salt = change in the oxidation state of Fe.
• Thus, n-factor = 1.
• $\mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{molecular}\mathrm{weight}}{\mathrm{n}-\mathrm{factor}}$
• On substituting the known values we get
  Equivalent weight$=\frac{392}{1}=392$
• Thus, we get the equivalent weight = Molecular weight. which is 392.