U haven't solved the questions under titration.!!!

The consistent readings are $23.70\text{ cm}^3$ and $18.80\text{ cm}^3$. The average volume of hydrochloric acid used is $(23.70 + 18.80) / 2 = 21.25\text{ cm}^3$.

Formula: moles = concentration × volume (in dm³)
Volume of HCl = $21.25\text{ cm}^3 = 0.02125\text{ dm}^3$
Concentration of HCl = $0.100\text{ mol/dm}^3$
Moles of HCl = $0.100\text{ mol/dm}^3 \times 0.02125\text{ dm}^3 = 0.002125\text{ mol}$

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.
Moles of $\text{Na}_2\text{CO}_3$ = (Moles of HCl) / 2
Moles of $\text{Na}_2\text{CO}_3$ = $0.002125\text{ mol} / 2 = 0.0010625\text{ mol}$

Concentration = moles / volume (in dm³)
Volume of solution A = $25.0\text{ cm}^3 = 0.0250\text{ dm}^3$
Concentration of $\text{Na}_2\text{CO}_3$ = $0.0010625\text{ mol} / 0.0250\text{ dm}^3 = 0.0425\text{ mol/dm}^3$

Molar mass of $\text{Na}_2\text{CO}_3 = (2 \times 22.99) + 12.01 + (3 \times 16.00) = 45.98 + 12.01 + 48.00 = 105.99\text{ g/mol}$
Concentration in g/dm³ = concentration in mol/dm³ × molar mass
Concentration of $\text{Na}_2\text{CO}_3$ = $0.0425\text{ mol/dm}^3 \times 105.99\text{ g/mol} = 4.50\text{ g/dm}^3$ (to 3 significant figures)

Percentage purity = (Concentration of $\text{Na}_2\text{CO}_3$ in g/dm³ / Original concentration of sample A in g/dm³) × 100%
Percentage purity = $(4.50\text{ g/dm}^3 / 5.00\text{ g/dm}^3) \times 100\% = 90.0\%$

The concentration of solution A is $0.0425\text{ mol/dm}^3$ or $4.50\text{ g/dm}^3$.
The percentage purity of sample A is $90.0\%$.