0.45 g of a dibasic acid is completely neutralized with 100 ml $\frac{\text{N}}{\text{10}}$  NaOH. The molecular weight of acid is 

milliequivalents of acid = milliequivalents of NaOH
$\frac{0.45}{E}\times 1000=100\times \frac{1}{10}$    $E=45$

Molecular weight =  $\text{E}\times \text{Basicity}$ 
=  45 x  2 = 90.