A student is performing a titration experiment involving hydrochloric acid (solution A) and sodium bicarbonate. The provided image contains experimental data, including burette readings. The student n

Solution A was prepared by dissolving $0.825 \text{ g}$ of $\text{HCl}$ in $250 \text{ cm}^3$ of water.

$$\text{Moles of HCl} = \frac{\text{Mass of HCl}}{\text{Molar mass of HCl}} = \frac{0.825 \text{ g}}{36.5 \text{ g/mol}} \approx 0.02260 \text{ mol}$$

The volume of solution A is $250 \text{ cm}^3 = 0.250 \text{ dm}^3$.

$$\text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of solution}} = \frac{0.02260 \text{ mol}}{0.250 \text{ dm}^3} \approx 0.0904 \text{ mol/dm}^3$$

The concentration of solution A is $\boxed{\text{0.0904 mol/dm}^3}$.

$$\text{Molar mass of NaHCO}_3 = \text{Na} + \text{H} + \text{C} + (3 \times \text{O}) = 23.0 + 1.0 + 12.0 + (3 \times 16.0) = 23.0 + 1.0 + 12.0 + 48.0 = 84.0 \text{ g/mol}$$

The molar mass of $\text{NaHCO}_3$ is $\boxed{\text{84.0 g/mol}}$.

To complete the problem, you would need to provide the burette readings (initial and final volumes of solution A) from the titration experiment. Once those are available, the average volume of A used can be calculated, and then further calculations regarding solution B's purity can be performed.