0 . 63 g of Oxalic acid is dissolved in order to obtain 250 cm 3 of its solution. Find the normality of this solution. [Oxalic acid: ( COOH ) 2 · 2 H 2 O

Step-by-step solution

Molar mass of oxalic acid = 126   g / mol Mole = Given   mass Molar   mass = 0 . 63 126 = 0 . 005   mol Molarity = Mole   of   solute Volume   of   solution   in   ml × 1000 = 0 . 005 250 × 1000 = 0 . 02 M Equivalent mass of oxalic acid = 126 2 = 63 Gram equivalent of the solute = 0 . 63 63 = 0 . 01 Normality = Gram   equivalent   of   solute Volume   of   solution   in   ml × 1000 = 0 . 01 250 × 1000 = 0 . 04 N