$0.3 \mathrm{g}$ of an oxalate salt was dissolved in $100 \mathrm{mL}$ solution. The solution required $90 \mathrm{mL}$ of $\frac{\mathrm{N}}{20}{\mathrm{KMnO}}_4$ for complete oxidation. The $\%$ of oxalate ion in salt is :-

A $0.518 \mathrm{g}$ sample of limestone is dissolved in $\mathrm{HCl}$ and then the calcium is precipitated as ${\mathrm{CaC}}_2{\mathrm{O}}_4.$ After filtering and washing the precipitate, it requires $40.0 \mathrm{mL}$ of $0.250 {\mathrm{NKMnO}}_4$ solution acidified with ${\mathrm{H}}_2{\mathrm{SO}}_4$ to titrate it as. The percentage of $\mathrm{CaO}$ in the sample is :-

${\mathrm{MnO}}_4^{-}+{\mathrm{H}}^{+}+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}⟶{\mathrm{Mn}}^{2+}+{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$

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?

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