The correct option is B 0.4 N
Molecular weight of Na2CO3=(23×2)+12+(3×16)=106 g mol−1Volume=2501000 L=0.25 Lnf=2
Since,
Normality=Number of g-equivalents of soluteVolume of solution (L)Number of g-equivalents of solute=Weight of soluteEquivalent weight of soluteEquivalent weight of solute=Molecular Mass of soluten-factor
Putting the values we get,
Normality=Weight of solute×n-factorMolecular Mass of solute×Volume of solution (L)Normality=5.3 g×2106 g mol−1×0.25 LNormality=0.4 N