normality of FeSo₄.7H₂O containing 5.56g/200ml which converts of ferric form in a reaction is?

Dear Student


Mass of FeSO₄.7H₂O = 5.56 g

Volume of solution = 200 mL


So, No. of moles = $\frac{Mass}{Molar Mass}$

Now, Molar Mass of   FeSO₄.7H₂O = (55.845 + 32 + (4 x 16) + (14 x 1) + (7 x 16)
                                                        = 55.845 + 32 + 64 + 14 + 112 =  277.845 g/mol or 278 g/mol


Thus, No. of moles of FeSO₄.7H₂O = $\frac{5.56 g}{278 g/mol}$ =  0.02 mol

Also, Normality = $\frac{Equivalent of Solution}{Volume of solution in L}$

In FeSO₄.7H₂O:

Fe²⁺ ----> Fe³⁺ + e^-

Here the oxidation state changes from 2 to 3, so, n factor = (3 - 2) = 1


Equivalent of FeSO₄.7H₂O = 0.02 mol x 1 eq/mol = 0.02

Volume of solution = 200 mL or 0.2 L

Hence, Normality = $\frac{0.02}{0.2}$ = 0.1 N


Regards