0.900 g of an acyclic saturated dicarboxylic acid is exactly neutralised by 20.0 mL of 1.00 M aqueous $\mathrm{NaOH}$. The acid is (take C = 12, H = 1, O = 16)

Moles of NaOH = 0.0200. A dicarboxylic acid is dibasic, so moles of acid = 0.0100. Molar mass = 0.900 / 0.0100 = 90 g/mol, which equals ethanedioic acid $\mathrm{HOOC}-\mathrm{COOH}$ (2 x 12 + 2 x 16 + 2 x 16 + 2 = 90). The next members are 104, 118 and 132.