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

No. of eq. of ${\mathrm{KMnO}}_4$ = No. of eq. of  ${\mathrm{C}}_2{\mathrm{O}}_4^{-2}$

$$\begin{gathered} \mathrm{NV}\left(\mathrm{L}\right) \mathrm{of} {\mathrm{KMnO}}_4 ={\left(\frac{\mathrm{W}}{\mathrm{E}}\right)}_{{\mathrm{C}}_2{\mathrm{O}}_4^{-2}}={\left(\frac{\mathrm{n}}{\mathrm{n}-\mathrm{factor}}\right)}_{{\mathrm{C}}_2{\mathrm{O}}_4^{-2}} \\ 90\times {10}^{-3} \times \frac{1}{20} = \frac{\mathrm{n}}{2} (\because \mathrm{n}-\mathrm{factor}=\mathrm{change} \mathrm{in} \mathrm{oxidation} \mathrm{state} \mathrm{of} \mathrm{C}) \\ \mathrm{n}=\mathrm{no}.\mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{oxalate} = \frac{9\times {10}^{-3}}{2\times 2} = \frac{9\times {10}^{-3}}{4} \end{gathered}$$

Weight of oxalate = $\frac{9}{4} \times 88 \times {10}^{-3} = 22 \times 9 \times {10}^{-3}=198\times {10}^{-3}\mathrm{g}$

$\% {\mathrm{C}}_2{\mathrm{O}}_4^{-2} = \frac{0.198}{0.3} \times 100 = 66\%$