Calculate the moles of a dibasic acid that reacts with 25 cm extsuperscript3 of 8 g/L NaOH solution.

$$\text{Molar mass of NaOH} = \text{Atomic mass of Na} + \text{Atomic mass of O} + \text{Atomic mass of H}$$ $$\text{Molar mass of NaOH} = 23.0 + 16.0 + 1.0 = 40.0 \text{ g/mol}$$

The concentration is given as 8 g per litre.

$$\text{Concentration of NaOH (mol/L)} = \frac{\text{Mass of NaOH per litre}}{\text{Molar mass of NaOH}}$$ $$\text{Concentration of NaOH (mol/L)} = \frac{8 \text{ g/L}}{40.0 \text{ g/mol}} = 0.2 \text{ mol/L}$$

The volume of $\text{NaOH}$ solution used is $25 \text{ cm}^3$, which is $0.025 \text{ L}$.

$$\text{Moles of NaOH} = \text{Concentration} \times \text{Volume (in L)}$$ $$\text{Moles of NaOH} = 0.2 \text{ mol/L} \times 0.025 \text{ L}$$ $$\text{Moles of NaOH} = 0.005 \text{ mol}$$

A dibasic acid reacts with $\text{NaOH}$ in a 1:2 molar ratio. Let the dibasic acid be $\text{H}_2\text{A}$.
The neutralization reaction is:

$$\text{H}_2\text{A} + 2\text{NaOH} \rightarrow \text{Na}_2\text{A} + 2\text{H}_2\text{O}$$

From the equation, 1 mole of dibasic acid reacts with 2 moles of $\text{NaOH}$.

$$\text{Moles of acid} = \text{Moles of NaOH} \times \frac{1 \text{ mol acid}}{2 \text{ mol NaOH}}$$ $$\text{Moles of acid} = 0.005 \text{ mol} \times \frac{1}{2}$$ $$\text{Moles of acid} = 0.0025 \text{ mol}$$

The mass of the acid is $0.245 \text{ g}$.

$$\text{Relative molecular mass} = \frac{\text{Mass of acid}}{\text{Moles of acid}}$$ $$\text{Relative molecular mass} = \frac{0.245 \text{ g}}{0.0025 \text{ mol}}$$ $$\text{Relative molecular mass} = 98 \text{ g/mol}$$

The relative molecular mass of the acid is $\text{98}$.