0.1 moles of diacidic base is dissolved in a 500 ml solution. Normality of the solution is-----N

N=Number of gram equivalents×1Vl; N=Number of gram equivalents×1000Vml; Number of gram equivalents=wsoluteGEW of solute; Number of gram equivalents=Number of moles×Valency; For a diacidic base, valency is 2 ; N=0.1×2×1000500=0.4 N ;