So solve the titration back into a clear blank format.

The consistent readings are $23.70\text{ cm}^3$ and $18.80\text{ cm}^3$.

$$\text{Average volume of HCl} = \frac{23.70\text{ cm}^3 + 18.80\text{ cm}^3}{2}$$ $$\text{Average volume of HCl} = \frac{42.50\text{ cm}^3}{2}$$ $$\text{Average volume of HCl} = \boxed{21.25\text{ cm}^3}$$

Given concentration of HCl = $0.100\text{ mol/dm}^3$.
Volume of HCl = $21.25\text{ cm}^3 = 0.02125\text{ dm}^3$.

$$\text{Moles of HCl} = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of HCl} = 0.100\text{ mol/dm}^3 \times 0.02125\text{ dm}^3$$ $$\text{Moles of HCl} = \boxed{0.002125\text{ mol}}$$

Assuming solution A is sodium carbonate ($\text{Na}_2\text{CO}_3$):

$$\text{Na}_2\text{CO}_3\text{(aq)} + 2\text{HCl(aq)} \rightarrow 2\text{NaCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$$

From the balanced equation, the mole ratio of $\text{Na}_2\text{CO}_3$ to HCl is 1:2.

$$\text{Moles of Na}_2\text{CO}_3 = \frac{\text{Moles of HCl}}{2}$$ $$\text{Moles of Na}_2\text{CO}_3 = \frac{0.002125\text{ mol}}{2}$$ $$\text{Moles of Na}_2\text{CO}_3 = \boxed{0.0010625\text{ mol}}$$

Volume of solution A = $25.0\text{ cm}^3 = 0.0250\text{ dm}^3$.

$$\text{Concentration of Na}_2\text{CO}_3 = \frac{\text{Moles}}{\text{Volume}}$$ $$\text{Concentration of Na}_2\text{CO}_3 = \frac{0.0010625\text{ mol}}{0.0250\text{ dm}^3}$$ $$\text{Concentration of Na}_2\text{CO}_3 = \boxed{0.0425\text{ mol/dm}^3}$$

Atomic masses: Na = 22.99, C = 12.01, O = 16.00.

$$\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 22.99) + 12.01 + (3 \times 16.00)$$ $$\text{Molar mass of Na}_2\text{CO}_3 = 45.98 + 12.01 + 48.00$$ $$\text{Molar mass of Na}_2\text{CO}_3 = \boxed{105.99\text{ g/mol}}$$

$$\text{Concentration (g/dm}^3) = \text{Concentration (mol/dm}^3) \times \text{Molar mass}$$ $$\text{Concentration (g/dm}^3) = 0.0425\text{ mol/dm}^3 \times 105.99\text{ g/mol}$$ $$\text{Concentration (g/dm}^3) = \boxed{4.50\text{ g/dm}^3} \text{ (to 3 significant figures)}$$

Original concentration of sample A = $5.00\text{ g/dm}^3$.

$$\text{Percentage purity} = \frac{\text{Concentration of pure Na}_2\text{CO}_3}{\text{Original concentration of sample A}} \times 100\%$$ $$\text{Percentage purity} = \frac{4.50\text{ g/dm}^3}{5.00\text{ g/dm}^3} \times 100\%$$ $$\text{Percentage purity} = 0.900 \times 100\%$$ $$\text{Percentage purity} = \boxed{90.0\%}$$