0.16 g of dibasic acid required 25 ml of decinormal NaOH solution for complete neutralisation. The molecular weight of the acid will be 

 $\begin{array}{l}{\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2; \mathrm{where} 1 \mathrm{represents} \mathrm{dibasicacid} \& 2 \mathrm{represents} \mathrm{base}\left(\mathrm{NaOH}\right)\\ \frac{\mathrm{W}}{\mathrm{E}}\times 1000=\frac{1}{10}\times 25 [\mathrm{since} \mathrm{NV}(\mathrm{ml})=\frac{\mathrm{W}}{\mathrm{E}}\times 1000]\\ \frac{0.16}{\mathrm{E}}\times 1000=\frac{25}{10}\Rightarrow \mathrm{E}=64\\ \mathrm{M}=2\times \mathrm{E}=2\times 64=128.\end{array}$