Calculate the concentration (mol dm⁻³) of your sodium oxalate solution (Na₂C₂O₄) given the mass of sodium oxalate used and the volume of the solution prepared.

$$M(\text{Na}_2\text{C}_2\text{O}_4) = (2 \times 23.0) + (2 \times 12.0) + (4 \times 16.0)$$ $$M(\text{Na}_2\text{C}_2\text{O}_4) = 46.0 + 24.0 + 64.0 = 134.0 \text{ g mol}^{-1}$$

$$\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1.250 \text{ g}}{134.0 \text{ g mol}^{-1}}$$ $$\text{Moles} = 0.009328 \text{ mol}$$

$$\text{Concentration} = \frac{\text{Moles}}{\text{Volume (dm}^3)} = \frac{0.009328 \text{ mol}}{0.250 \text{ dm}^3}$$ $$\text{Concentration} = \boxed{0.0373 \text{ mol dm}^{-3}}$$

1 (b) The concentration (mol dm⁻³) of the potassium permanganate in solution G.

The balanced equation for the reaction between $\text{KMnO}_4$ and $\text{Na}_2\text{C}_2\text{O}_4$ is:
$2\text{MnO}_4^-(aq) + 5\text{C}_2\text{O}_4^{2-}(aq) + 16\text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 10\text{CO}_2(g) + 8\text{H}_2\text{O}(l)$
The mole ratio of $\text{MnO}_4^-$ to $\text{C}_2\text{O}_4^{2-}$ is 2:5.

$$\text{Moles of C}_2\text{O}_4^{2-} = \text{Concentration} \times \text{Volume (dm}^3)$$ $$\text{Moles of C}_2\text{O}_4^{2-} = 0.0373 \text{ mol dm}^{-3} \times \frac{25.0}{1000} \text{ dm}^3$$ $$\text{Moles of C}_2\text{O}_4^{2-} = 0.0009325 \text{ mol}$$

$$\text{Moles of MnO}_4^- = \text{Moles of C}_2\text{O}_4^{2-} \times \frac{2}{5}$$ $$\text{Moles of MnO}_4^- = 0.0009325 \text{ mol} \times \frac{2}{5} = 0.000373 \text{ mol}$$

$$\text{Concentration of KMnO}_4 = \frac{\text{Moles of MnO}_4^-}{\text{Volume of KMnO}_4 \text{ (dm}^3)}$$ $$\text{Concentration of KMnO}_4 = \frac{0.000373 \text{ mol}}{20.00/1000 \text{ dm}^3}$$ $$\text{Concentration of KMnO}_4 = \boxed{0.01865 \text{ mol dm}^{-3}}$$

2 Find the concentration (mol dm⁻³) of the sodium nitrite in solution H.

The balanced equation for the reaction between $\text{KMnO}_4$ and $\text{NaNO}_2$ is given:
$2\text{MnO}_4^-(aq) + 5\text{NO}_2^-(aq) + 6\text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 5\text{NO}_3^-(aq) + 3\text{H}_2\text{O}(l)$
The mole ratio of $\text{MnO}_4^-$ to $\text{NO}_2^-$ is 2:5.

$$\text{Moles of MnO}_4^- = \text{Concentration of KMnO}_4 \times \text{Volume of KMnO}_4 \text{ (dm}^3)$$ $$\text{Moles of MnO}_4^- = 0.01865 \text{ mol dm}^{-3} \times \frac{22.50}{1000} \text{ dm}^3$$ $$\text{Moles of MnO}_4^- = 0.0004196 \text{ mol}$$

$$\text{Moles of NO}_2^- = \text{Moles of MnO}_4^- \times \frac{5}{2}$$ $$\text{Moles of NO}_2^- = 0.0004196 \text{ mol} \times \frac{5}{2} = 0.001049 \text{ mol}$$

$$\text{Concentration of NaNO}_2 = \frac{\text{Moles of NO}_2^-}{\text{Volume of NaNO}_2 \text{ (dm}^3)}$$ $$\text{Concentration of NaNO}_2 = \frac{0.001049 \text{ mol}}{25.0/1000 \text{ dm}^3}$$ $$\text{Concentration of NaNO}_2 = \boxed{0.04196 \text{ mol dm}^{-3}}$$

3 Find the mass of sodium nitrite actually present in 1 dm³ of solution H (O = 16.0).

$$M(\text{NaNO}_2) = 23.0 + 14.0 + (2 \times 16.0)$$ $$M(\text{NaNO}_2) = 23.0 + 14.0 + 32.0 = 69.0 \text{ g mol}^{-1}$$

$$\text{Mass} = \text{Concentration} \times \text{Molar mass} \times \text{Volume}$$ $$\text{Mass} = 0.04196 \text{ mol dm}^{-3} \times 69.0 \text{ g mol}^{-1} \times 1 \text{ dm}^3$$ $$\text{Mass} = \boxed{2.895 \text{ g}}$$

4 Calculate the percentage purity of the sodium nitrite used.

To calculate the percentage purity, the initial mass of the impure sodium nitrite used to prepare solution H is required. This information is not provided in the problem.

If, for example, 3.00 g of impure sodium nitrite was dissolved to make 1 dm³ of solution H, then the calculation would be:

$$\text{Percentage purity} = \frac{\text{Mass of pure NaNO}_2 \text{ in 1 dm}^3}{\text{Mass of impure NaNO}_2 \text{ used to make 1 dm}^3} \times 100\%$$ $$\text{Percentage purity} = \frac{2.895 \text{ g}}{3.00 \text{ g}} \times 100\% = \boxed{96.5\%}$$

(This is an example calculation assuming an initial impure mass of 3.00 g.)