0. 45g acid of molecular mass 90 was neutralise by 20ml of 0. 5NKOH...

Given that& $0 . 45g$ of the acid of molecular mass& $90$ was neutralized by& $20ml$ of& $0 . 5$\left(N\right)$caustic potash.</p><p>We know that,</p><p>Normality\left(N\right) = \dfrac{Weightofthebase \times 1000}{Eq . weight \times volume}$$</p><p>$∵$Gram equivalents of an acid</p><p>Gram equivalents of$KOH$i.e.$$\dfrac{0 . 45}{E} = \dfrac{20 \times 0 . 5}{1000}$

where, $E =$ Equivalent mass $E = \dfrac{0 . 45 \times 1000}{20 \times 0 . 5} =45$

Thus, basicity $= \dfrac{90}{45} =2 \left[∵ Basicity = \dfrac{Molecular mass}{Equivalent mass}\right]$