Please resolve the calculations under titration for better visibility. I can't see what u've done

The consistent readings are $23.70\text{ cm}^3$ and $18.80\text{ cm}^3$.
Average volume of HCl used = $(23.70\text{ cm}^3 + 18.80\text{ cm}^3) / 2$
Average volume of HCl used = $42.50\text{ cm}^3 / 2$
Average volume of HCl used = $21.25\text{ cm}^3$

Formula: Moles = Concentration × Volume (in dm³)
Volume of HCl = $21.25\text{ cm}^3 = 21.25 / 1000\text{ dm}^3 = 0.02125\text{ dm}^3$
Concentration of HCl = $0.100\text{ mol/dm}^3$

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

Assuming solution A is sodium carbonate ($\text{Na}_2\text{CO}_3$), the reaction with hydrochloric acid is:
$\text{Na}_2\text{CO}_3\text{(aq)} + 2\text{HCl(aq)} \rightarrow 2\text{NaCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$

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

Formula: Concentration = Moles / Volume (in dm³)
Volume of solution A = $25.0\text{ cm}^3 = 25.0 / 1000\text{ dm}^3 = 0.0250\text{ dm}^3$

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

First, calculate the molar mass of $\text{Na}_2\text{CO}_3$:
Molar mass of $\text{Na}_2\text{CO}_3 = (2 \times \text{Na}) + (1 \times \text{C}) + (3 \times \text{O})$
Molar mass of $\text{Na}_2\text{CO}_3 = (2 \times 22.99) + 12.01 + (3 \times 16.00)$
Molar mass of $\text{Na}_2\text{CO}_3 = 45.98 + 12.01 + 48.00 = 105.99\text{ g/mol}$

Formula: Concentration (g/dm³) = Concentration (mol/dm³) × Molar mass
Concentration of $\text{Na}_2\text{CO}_3$ = $0.0425\text{ mol/dm}^3 \times 105.99\text{ g/mol}$
Concentration of $\text{Na}_2\text{CO}_3$ = $4.50\text{ g/dm}^3$ (to 3 significant figures)

The original concentration of sample A was $5.00\text{ g/dm}^3$.
Percentage purity = (Concentration of pure $\text{Na}_2\text{CO}_3$ in g/dm³ / Original concentration of sample A in g/dm³) × 100%

Percentage purity = $(4.50\text{ g/dm}^3 / 5.00\text{ g/dm}^3) \times 100\%$
Percentage purity = $0.900 \times 100\%$
Percentage purity = $90.0\%$