If $$50$$ mL of $$0.5$$ M oxalic acid is required to neutralise $$25$$ mL of NaOH solution, the amount of NaOH in $$50$$ mL of given NaOH solution is ______ g.

We need to find the amount of NaOH in 50 mL of NaOH solution, given that 50 mL of 0.5 M oxalic acid neutralises 25 mL of the NaOH solution.

The balanced neutralisation reaction is:

$$H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O$$

Oxalic acid ($$H_2C_2O_4$$) is diprotic, so 1 mole of oxalic acid reacts with 2 moles of NaOH.

$$\text{Moles of } H_2C_2O_4 = M \times V = 0.5 \times \frac{50}{1000} = 0.025 \text{ mol}$$

From the stoichiometry, moles of NaOH = 2 × moles of oxalic acid:

$$\text{Moles of NaOH in 25 mL} = 2 \times 0.025 = 0.05 \text{ mol}$$

Since the concentration is uniform, moles in 50 mL = 2 × moles in 25 mL:

$$\text{Moles of NaOH in 50 mL} = 2 \times 0.05 = 0.1 \text{ mol}$$

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

$$\text{Mass} = 0.1 \times 40 = 4 \text{ g}$$

Therefore, the amount of NaOH in 50 mL of the solution is 4 g.