2 mole, equimolar mixture of Na₂​C₂O₄ and H₂C₂O₄​ required V₁​L of 0.1M KMnO₄ in acidic medium.

As the mixture is equimolar, 1 mole of each, Na₂​C₂O₄ and H₂​C₂O₄​ are present.

eq. of Na₂C₂O₄+eq.of H₂C₂O₄  = eq. of KMnO₄

1x2+1x2= V₁x0.1x5 ; ($\because$n-factor of KMnO₄=5 , n-factor of Na₂C₂O₄=n-factor of H₂C₂O₄=2)

  V₁=8L
$\mathrm{eq}.\mathrm{of} {\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4= \mathrm{eq}.\mathrm{of} \mathrm{NaOH}$
1x2 =0.2 x V₂; V₂=10L
Hence, ${\mathrm{V}}_1:{\mathrm{V}}_2=8:10=4:5$