50ml of 0.5M oxalic acid is completely neutralised by 25ml of NaOH solution.
For neutralisation reactions, N1V1=N2V2
Normality of oxalic acid = Molarity×2
⇒Noxalicacid=0.5×2=1N
50×1=25×NNaOH
For sodium hydroxide, molarity is the same as normality.
Molarity of sodium hydroxide = 2M
The number of moles of sodium hydroxide= 50×2×10−3mol.
Hence, the mass of sodium hydroxide =50×2×10−3×40=4g