0 . 45   g acid of molecular mass 90 was neutralise by 20 ml of 0 . 5   N   KOH . The…

Given that $0.45 \mathrm{g}$ of the acid of molecular mass $90$ was neutralized by $20 \mathrm{ml}$ of $0.5 \left(\mathrm{N}\right)$ caustic potash.

We know that,

$\mathrm{Normality} \left(\mathrm{N}\right) = \frac{\mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{base} \times 1000}{\mathrm{Eq}. \mathrm{weight}\times \mathrm{volume}}$

$\because$ Gram equivalents of an acid

Gram equivalents of $\mathrm{KOH}$ i.e. $\frac{0.45}{\mathrm{E}}=\frac{20\times 0.5}{1000}$

where, $\mathrm{E}=$ Equivalent mass $\mathrm{E}=\frac{0.45\times 1000}{20\times 0.5}=45$

Thus, basicity $=\frac{90}{45}=2$ $\left[\because \mathrm{Basicity}=\frac{\mathrm{Molecular}mass}{\mathrm{Equivalent}mass}\right]$