Substituting we have 28 × 0.1 = 25 × N2
∴ N2 = (29 x 0.1)/25 = 0.116
∴ Normality of sodium hydroxide solution = 0.116
We know that W = N × E for 100 ml
i.e., Mass per dm3 = Normality × eq. nlass = 0.116 × 40 g
∴ Mass of sodium hydroxide in 500 cm3 of the solution = (0.116 x 40)/2 = 2.32 g.