Calculate the mass of anhydrous oxalic acid, which can be oxidised to ${\mathrm{CO}}_2\left(\mathrm{g}\right)$ by $100 \mathrm{ml}$ of an ${\mathrm{MnO}}_4^{-}$ solution, $10 \mathrm{ml}$ of which is capable of oxidising $50 \mathrm{ml}$ of $1\mathrm{N} {\mathrm{I}}^{-}$ to ${\mathrm{I}}_2.$

To a $10 \mathrm{ml} 1 \mathrm{M}$ aqueous solution of ${\mathrm{Br}}_2,$ excess of $\mathrm{NaOH}$ is added so that all ${\mathrm{Br}}_2$ is disproportional to $\mathrm{Br}$ and${{\mathrm{BrO}}_3}^{-}.$ The resulting solution is freed from ${\mathrm{Br}}^{-},$ by extraction and excess of ${\mathrm{OH}}^{-}$ neutralised by acidifying the solution. The resulting solution is sufficient to react with $1.5 \mathrm{gm}$ of impure ${\mathrm{CaC}}_2{\mathrm{O}}_4\left(\mathrm{M}=128 \mathrm{gm} {\mathrm{mol}}^{-1}\right)$ample. The $\%$ purity of Oxalate sample is

The relevant reactions are  ${\mathrm{Br}}^2\left(\mathrm{aq}.\right)+{\mathrm{OH}}^{-} \to {\mathrm{Br}}^{-}\left(\mathrm{aq}.\right)+\mathrm{Br}{{\mathrm{O}}_{{}^3}}^{-}$

${\mathrm{BrO}}_3{}^{-}+{\mathrm{C}}_2{\mathrm{O}}_4{}^{2-}\to \mathrm{Br}+{\mathrm{CO}}_2$

What is the equivalent weight of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in the reaction?

${\mathrm{H}}_2{\mathrm{SO}}_4+\mathrm{Nal}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{I}}_2+{\mathrm{H}}_2 \mathrm{S}+{\mathrm{H}}_2\mathrm{O}$