10 mL of 0.02 M $KMn{O}_4$ is required to oxidize 20 mL of oxalic acid of certain strength. 25 mL of the same oxalic acid is required to neutralize 20 mL of NaOH of unknown strength. Find the amount (gm) of NaOH in one litre of solution.

In acidic medium  $1MKMn{O}_4=5NKMn{O}_4$
 $0.02MKMn{O}_4=0.1NKMn{O}_4$
According to normality equation,
 $$\begin{gathered} N_1{V}_1\left(KMn{O}_4\right)=N_2{V}_2\left(Oxalicacid\right) \\ 0.1\times 10=N_2\times 20\Rightarrow N_2=\frac{10\times 0.1}{20}0.05N \end{gathered}$$
 Again, $N_1{V}_1$ (Oxalic acid)=$N_2{V}_2$  (NaOH)
 $$\begin{gathered} 0.05\times 25=N_2\times 20 \\ {N}_2=\frac{0.05\times 25}{20}=0.0625N \end{gathered}$$
 Strength = Normality × equiv. wt.
= 0.0625 × 40 = 2.5 $g{L}^{-1}$