Calculate the molar concentration of MOH given the reaction with sulfuric acid.

Given atomic masses: H = 1.0, S = 32.0, O = 16.0.

$$\text{Molar mass of } H_2SO_4 = (2 \times 1.0) + 32.0 + (4 \times 16.0) = 2.0 + 32.0 + 64.0 = 98.0 \text{ g/mol}$$

Given concentration of $H_2SO_4 = 4.9 \text{ g/dm}^3$.

$$\text{Molar concentration of } H_2SO_4 = \frac{\text{Mass concentration}}{\text{Molar mass}} = \frac{4.9 \text{ g/dm}^3}{98.0 \text{ g/mol}} = 0.05 \text{ mol/dm}^3$$

Volume of $H_2SO_4$ used = $12 \text{ cm}^3 = \frac{12}{1000} \text{ dm}^3 = 0.012 \text{ dm}^3$.

$$\text{Moles of } H_2SO_4 = \text{Molar concentration} \times \text{Volume} = 0.05 \text{ mol/dm}^3 \times 0.012 \text{ dm}^3 = 0.0006 \text{ mol}$$

The balanced chemical equation is: $2MOH_{aq} + H_2SO_4 \rightarrow M_2SO_4 + 2H_2O$.
From the equation, 2 moles of MOH react with 1 mole of $H_2SO_4$.

$$\text{Moles of MOH} = 2 \times \text{Moles of } H_2SO_4 = 2 \times 0.0006 \text{ mol} = 0.0012 \text{ mol}$$

Volume of MOH solution used = $6 \text{ cm}^3 = \frac{6}{1000} \text{ dm}^3 = 0.006 \text{ dm}^3$.

$$\text{Molar concentration of MOH} = \frac{\text{Moles of MOH}}{\text{Volume of MOH solution}} = \frac{0.0012 \text{ mol}}{0.006 \text{ dm}^3} = 0.2 \text{ mol/dm}^3$$

The initial MOH solution was prepared by dissolving 11.2 g of MOH in enough solvent to make one litre ($1 \text{ dm}^3$) of solution.
Since the concentration is $0.2 \text{ mol/dm}^3$, one litre of solution contains 0.2 moles of MOH.

$$\text{Molar mass of MOH} = \frac{\text{Mass of MOH}}{\text{Moles of MOH}} = \frac{11.2 \text{ g}}{0.2 \text{ mol}} = 56 \text{ g/mol}$$

The formula of the hydroxide is MOH.
Molar mass of MOH = Relative atomic mass of M + Relative atomic mass of O + Relative atomic mass of H.

$$56 = M + 16.0 + 1.0$$ $$56 = M + 17.0$$ $$M = 56 - 17.0$$ $$M = 39.0$$

The relative atomic mass of M is $\text{39.0}$.