0.45 g of acid (mol. wt. = 90) was exactly neutralized by 20 ml of 0.5 (M) NaOH.

The basicity of the given acid is

milliequivalents of acid = milliequivalents of base

$\frac{0.45}{90} \times \mathrm{basicity} = 20 \times 0.5 \times {10}^{-3}$

$\mathrm{basicity} = \frac{90 \times 20 \times 0.5 \times {10}^{-3}}{0.45}$

             $= 2000 \times {10}^{-3} = 2$