The balanced equation governing this reaction is:
3NaOH+FeCl3→3NaCl+Fe(OH)3
Molecular wt. of Fe(OH)3 is 56+(16+1)×3=107 g/mol
Thus no. of moles of Fe(OH)3 precipitated=1.425/107=0.01332 moles
No. of moles of FeCl3 initially present (a/c to balanced eqn.) =0.01332 moles
No. of equivalents of FeCl3 initailly present =no.ofmoles×valency=0.01332×3=0.0399
Normality of FeCl3 solution=No. of equivalentsVol. of sol. in litres
=0.03990.1
=0.399N≅0.40N
Thus the correct option is (d).