If 50mL of 0 . 5M oxalic acid is required to neutralise 25mL of...

$50$ml of $0 . 5M$ oxalic acid is completely neutralised by $25$ml of $NaOH$ solution. 

For neutralisation reactions, ${N}_{1}{V}_{1} = {N}_{2}{V}_{2}$

Normality of oxalic acid = Molarity$\times$$2$

$⇒ {N}_{oxalicacid} = 0 . 5 \times 2 = 1N$

$50 \times 1 = 25 \times {N}_{NaOH}$

For sodium hydroxide, molarity is the same as normality.

Molarity of sodium hydroxide = $2$M

The number of moles of sodium hydroxide= $50 \times 2 \times {10}^{- 3}mol$.

Hence, the mass of sodium hydroxide $= 50 \times 2 \times {10}^{- 3} \times 40 = 4g$