$0.115 \mathrm{g}$ of a dibasic acid required $25 \mathrm{mL}$ of $0.1 \mathrm{N} \mathrm{NaOH}$ solution for complete neutralization. Find the molecular weight of the acid.

For complete neutralization, number of equivalent of NaOH is equal to number of equivalent of dibasic acid. 

$\begin{array}{ccc}\mathrm{No}. \mathrm{of} \mathrm{equivalents} \mathrm{of} \mathrm{NaOH}& =& \mathrm{No}. \mathrm{of} \mathrm{equivalents} \mathrm{of} \mathrm{dibasic} \mathrm{acid}\\ \mathrm{Normality}\times \mathrm{Volume} (\mathrm{in} \mathrm{L})& =& \frac{\mathrm{Mass}}{{\displaystyle \frac{\mathrm{Molecular} \mathrm{weight} }{2}}}\\ 0.1 \mathrm{N}\times 25 \mathrm{mL}\times \frac{1 \mathrm{L}}{1000 \mathrm{mL}}& =& \frac{0.115 \mathrm{g}\times 2}{\mathrm{Molecular} \mathrm{weight}}\\ \mathrm{Molecular} \mathrm{weight}& =& 92\end{array}$