An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is

Equivalent weight of oxalic acid COOH|COOH 2H2O=Molecular wt.2=1262=63Normality of oxalic acid = Wt.E×1000V=6.363×1000250 N=0.4  NSuppose, x mL volume of 0.1 N NaOH is required to complete neutralization of 10 mL of this solution.On complete neutralization, one M. eq. of acid is neutralized with one M. eq. of base.So, M. eq. of NaOH = M. eq. of oxalic acid∴  0.1  ×  x  =  0.4  ×  10x=0.4  ×  100.1  =  40  mL