0.3 g of an oxalate salt was dissolved in 100 mL solution. The solution required 90 mL of N/20 KMnO4 for complete oxidation. The % of oxalate ion in salt is

No. of eq. of KMnO4 = No. of eq. of  C2O4-2NV(L) of KMnO4 =WEC2O4-2=nn-factorC2O4-2 90×10-3 × 120 = n2 (∵n-factor=change in oxidation state of C) n=no.of moles of oxalate = 9×10-32× 2 = 9×10-34Weight of oxalate = 94 × 88 × 10-3 = 22 × 9 × 10-3=198×10-3g% C2O4-2 = 0.1980.3 × 100 = 66%