So solve the titration back into a clear blank format.

Okay, Ruddy, let's present the titration calculations in a clear, step-by-step format.

Titration Calculations

Step 1: Calculate the average volume of hydrochloric acid (HCl) used. The consistent readings are $23.70 cm^3$ and $18.80 cm^3$. $Average volume of HCl = \frac{23.70 cm^3 + 18.80 cm^3}{2}$ $Average volume of HCl = \frac{42.50 cm^3}{2}$ Average volume of HCl = 21.25\text{ cm^3}

Step 2: Calculate the number of moles of hydrochloric acid (HCl) used. Given concentration of HCl = $0.100 mol/dm^3$. Volume of HCl = $21.25 cm^3 = 0.02125 dm^3$. $Moles of HCl = Concentration \times Volume$ $Moles of HCl = 0.100 mol/dm^3 \times 0.02125 dm^3$ $Moles of HCl = 0.002125\text{ mol}$

Step 3: Write the balanced chemical equation for the reaction. Assuming solution A is sodium carbonate ($\text{Na}_2CO_3$): $Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)$

Step 4: Calculate the number of moles of solution A ($\text{Na}_2CO_3$) in $25.0 cm^3$. From the balanced equation, the mole ratio of $\text{Na}_2CO_3$ to HCl is 1:2. $Moles of Na_2CO_3 = \frac{Moles of HCl}{2}$ $Moles of Na_2CO_3 = \frac{0.002125 mol}{2}$ $Moles of Na_2CO_3 = 0.0010625\text{ mol}$

Step 5: Calculate the concentration of solution A ($\text{Na}_2CO_3$) in $\text{mol/dm}^3$. Volume of solution A = $25.0 cm^3 = 0.0250 dm^3$. $Concentration of Na_2CO_3 = \frac{Moles}{Volume}$ $Concentration of Na_2CO_3 = \frac{0.0010625 mol}{0.0250 dm^3}$ Concentration of Na_2CO_3 = 0.0425\text{ mol/dm^3}

Step 6: Calculate the molar mass of $\text{Na}_2CO_3$. Atomic masses: Na = 22.99, C = 12.01, O = 16.00. $Molar mass of Na_2CO_3 = (2 \times 22.99) + 12.01 + (3 \times 16.00)$ $Molar mass of Na_2CO_3 = 45.98 + 12.01 + 48.00$ $Molar mass of Na_2CO_3 = 105.99\text{ g/mol}$

Step 7: Calculate the concentration of solution A ($\text{Na}_2CO_3$) in $\text{g/dm}^3$. $Concentration (g/dm^3) = Concentration (mol/dm^3) \times Molar mass$ $Concentration (g/dm^3) = 0.0425 mol/dm^3 \times 105.99 g/mol$ Concentration (g/dm^3) = 4.50\text{ g/dm^3} (to 3 significant figures)

Step 8: Calculate the percentage purity of sample A. Original concentration of sample A = $5.00 g/dm^3$. $Percentage purity = \frac{Concentration of pure Na_2CO_3}{Original concentration of sample A} \times 100\%$ $Percentage purity = \frac{4.50 g/dm^3}{5.00 g/dm^3} \times 100\%$ $Percentage purity = 0.900 \times 100\%$ $Percentage purity = 90.0\%$