An excess of NaOH was added to 100 mL of a FeCl3 solution which gives 2.14 of Fe(OH)3. Calculate the normality of FeCl3 solution.

3NaOH+FeCl3      ⟶Fe(OH)3+3NaCl(excess)   100 mL                                   1mmol≡1mmol   FeCl3≡Fe(OH)3 M×100≡2.14107×1000 MFeCl3=0.2 NFeCl3=0.2×3( valency factor =3) =0.6N