A) Give a balanced equation for the reaction.

$$\text{Moles of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of H}_2\text{SO}_4 = 1\text{mol/dm}^3 \times \frac{25}{1000}\text{dm}^3 = 1\text{mol/dm}^3 \times 0.025\text{dm}^3 = 0.025\text{mol}$$

From the equation, $1\text{mol}$ of $\text{H}_2\text{SO}_4$ reacts with $2\text{mol}$ of $\text{XOH}$.

$$\text{Moles of XOH} = 2 \times \text{Moles of H}_2\text{SO}_4 = 2 \times 0.025\text{mol} = 0.05\text{mol}$$

$$\text{Molar mass of XOH} = \frac{\text{Mass of XOH}}{\text{Moles of XOH}}$$ $$\text{Molar mass of XOH} = \frac{2.8\text{g}}{0.05\text{mol}} = 56\text{g/mol}$$

The molar mass of $\text{XOH}$ is the sum of the relative atomic masses of X, O, and H.

$$\text{Molar mass of XOH} = \text{Ar}(\text{X}) + \text{Ar}(\text{O}) + \text{Ar}(\text{H})$$ $$56 = \text{Ar}(\text{X}) + 16 + 1$$ $$56 = \text{Ar}(\text{X}) + 17$$ $$\text{Ar}(\text{X}) = 56 - 17 = 39$$

The relative atomic mass of X is $\boxed{39}$.