Calculate the concentration of the oxalate solution.

From Table 1, the mass of oxalate used is $1.73 \text{ g}$.

The molar mass of $\text{Na}_2\text{C}_2\text{O}_4$ is $134.0 \text{ g/mol}$.

From Procedure 1, the sodium oxalate was dissolved in a $250 \text{ cm}^3$ volumetric flask.
Volume of solution = $250 \text{ cm}^3 = 0.250 \text{ dm}^3$.

Rounding to three significant figures:

The accurate titres are $26.7 \text{ cm}^3$ and $26.8 \text{ cm}^3$. The approximate titre ($28.1 \text{ cm}^3$) is usually discarded for mean calculations.

This matches the value written on the paper.

The reaction between permanganate ions ($\text{MnO}_4^-$) and oxalate ions ($\text{C}_2\text{O}_4^{2-}$), in acidic medium, is:

From the balanced equation, the mole ratio of $\text{KMnO}_4$ ($\text{MnO}_4^-$) to $\text{Na}_2\text{C}_2\text{O}_4$ ($\text{C}_2\text{O}_4^{2-}$) is $2:5$.
So, $n_{\text{KMnO}_4} = 2$ and $n_{\text{Na}_2\text{C}_2\text{O}_4} = 5$.

• Concentration of $\text{Na}_2\text{C}_2\text{O}_4$ ($M_{\text{Na}_2\text{C}_2\text{O}_4}$) = $0.0516416 \text{ mol/dm}^3$ (from Question 1)
• Volume of $\text{Na}_2\text{C}_2\text{O}_4$ ($V_{\text{Na}_2\text{C}_2\text{O}_4}$) = $25.0 \text{ cm}^3$ (from Procedure 2)
• Volume of $\text{KMnO}_4$ ($V_{\text{KMnO}_4}$) = $26.75 \text{ cm}^3$ (mean titre from Question 2)

The titration formula is:

Rearranging to solve for $M_{\text{KMnO}_4}$:

Substitute the known values:

Rounding to three significant figures: