To tabulate results, but no titration data (burette readings) has been provided. Therefore, this part cannot be completed.

Here's the solution to the volumetric analysis problem:

• Part (a): Tabulate your results and calculate the average volume of A used. The question asks to tabulate results, but no titration data (burette readings) has been provided. Therefore, this part cannot be completed. To proceed with the calculations in part (b), I will assume an average titre (volume of A used) of $15.00 \text{ cm}^3$. I will also assume that $25.0 \text{ cm}^3$ of solution B was pipetted for each titration.

• Part (b): From your results and information provided, calculate:

  • Relative atomic masses: $\text{H}=1, \text{O}=16, \text{Na}=23, \text{S}=32$.
  • Molar mass of $\text{H}_2\text{SO}_4$ (Solution A): $$\text{Molar mass of } \text{H}_2\text{SO}_4 = (2 \times 1) + 32 + (4 \times 16) = 2 + 32 + 64 = 98 \text{ g/mol}$$
  • Molar mass of $\text{NaOH}$ (Solution B): $$\text{Molar mass of } \text{NaOH} = 23 + 16 + 1 = 40 \text{ g/mol}$$