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

My apologies for the visibility issue. Let's re-do the titration calculations with clearer formatting.

Titration of $25.0 cm^3$ of solution A with $0.100 mol/dm^3$ hydrochloric acid.

Step 1: Record the burette readings and calculate the average volume of HCl used.

| Titration | 1st | 2nd | 3rd | | :--------------- | :-------- | :-------- | :-------- | | Final reading | $24.50 cm^3$ | $48.20 cm^3$ | $47.00 cm^3$ | | Initial reading | $0.00 cm^3$ | $24.50 cm^3$ | $48.20 cm^3$ | | Volume used | $24.50 cm^3$ | $23.70 cm^3$ | $18.80 cm^3$ |

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

Step 2: Calculate the number of moles of hydrochloric acid (HCl) used. Formula: Moles = Concentration × Volume (in dm³) Volume of HCl = $21.25 cm^3 = 21.25 / 1000 dm^3 = 0.02125 dm^3$ Concentration of HCl = $0.100 mol/dm^3$

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 ($\text{Na}_2CO_3$), the reaction with hydrochloric acid is: $\text{Na}_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2\text{(g)}$

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

Moles of $\text{Na}_2CO_3$ = $0.002125 mol / 2$ Moles of $\text{Na}_2CO_3$ = $0.0010625\text{ mol}$

Step 5: Calculate the concentration of solution A in $\text{mol/dm}^3$. Formula: Concentration = Moles / Volume (in dm³) Volume of solution A = $25.0 cm^3 = 25.0 / 1000 dm^3 = 0.0250 dm^3$

Concentration of $\text{Na}_2CO_3$ = $0.0010625 mol / 0.0250 dm^3$ Concentration of $\text{Na}_2CO_3$ = $0.0425 mol/dm^3$

Step 6: Calculate the concentration of solution A in $\text{g/dm}^3$. First, calculate the molar mass of $\text{Na}_2CO_3$: Molar mass of $\text{Na}_2CO_3 = (2 \times Na) + (1 \times C) + (3 \times O)$ Molar mass of $\text{Na}_2CO_3 = (2 \times 22.99) + 12.01 + (3 \times 16.00)$ Molar mass of $\text{Na}_2CO_3 = 45.98 + 12.01 + 48.00 = 105.99\text{ g/mol}$

Formula: Concentration (g/dm³) = Concentration (mol/dm³) × Molar mass Concentration of $\text{Na}_2CO_3$ = $0.0425 mol/dm^3 \times 105.99\text{ g/mol}$ Concentration of $\text{Na}_2CO_3$ = $4.50 g/dm^3$ (to 3 significant figures)

Step 7: Calculate the percentage purity of sample A. The original concentration of sample A was $5.00 g/dm^3$. Percentage purity = (Concentration of pure $\text{Na}_2CO_3$ in g/dm³ / Original concentration of sample A in g/dm³) × 100%

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