What is the average accurate titre volume in a titration experiment, and how is it calculated?

The blank should be filled with the average titre volume from your accurate titrations. This value is obtained by averaging the two (or more) concordant titre values recorded in the "Accurate" columns of the table.

For example, if your two accurate titre values were $20.10 \text{ cm}^3$ and $20.05 \text{ cm}^3$, the average would be:

$$\text{Average titre} = \frac{20.10 \text{ cm}^3 + 20.05 \text{ cm}^3}{2} = 20.075 \text{ cm}^3$$

You would then round this to an appropriate number of significant figures, typically two decimal places for burette readings, e.g., $20.08 \text{ cm}^3$.

Since the table is empty, I cannot provide a numerical answer. You need to perform the experiment and record your results.

Hence $\boxed{\text{Average accurate titre}}$ $\text{cm}^3$ of $\text{Na}_2\text{S}_2\text{O}_3$ reacted with $25 \text{ cm}^3$ of $\text{CuSO}_4$.