If $50 \mathrm{mL}$ of $0.5\mathrm{M}$ oxalic acid is required to neutralise $25 \mathrm{mL}$ of $\mathrm{NaOH}$ solution, the amount of $\mathrm{NaOH}$ in $50 \mathrm{mL}$ of given $\mathrm{NaOH}$ solution is_______g.

$50$ml of $0.5 \mathrm{M}$ oxalic acid is completely neutralised by $25$ml of $\mathrm{NaOH}$ solution. 

For neutralisation reactions, ${\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2$

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

$\Rightarrow {\mathrm{N}}_{\mathrm{oxalic} \mathrm{acid}}=0.5\times 2=1\mathrm{N}$

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

For sodium hydroxide, molarity is the same as normality.

Molarity of sodium hydroxide = $2$M

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

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