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

The percentage of oxalate ion in salt is 66%.

Number of mili equivalent of  $KMn{O}_4$$=90\times 1/20=4.5 m$ equivalents

Number of mili equivalents of oxalate ions = number of mili equivalents of $KMn{O}_4$$=4.5 m$equivalents
$\because$Number of equivalents  $=\frac{ Mass }{ equivalent mass }$

& equivalent mass of oxalate ion $=88/2=44$

$\therefore$ Mass of oxalate $=44\times 4.5\times {10}^{-3}grams$

$=198\times {10}^{-3} g$ 
$=0.198 g$

$\therefore$ percentage in salt oxalate $=\frac{0.198}{0.3}\times 100=66\%$

Hence, option B is correct.