Solution:
Solution
Meq. Of KMnO4=M eq. of C2O4−2
90×1 / 2=100×NC2O4−2
Mmole of oxalate = $$\displaystyle=\frac{{{9}}}{{{2}\times{2}}}=\frac{{{9}}}{{{4}}}$$
Weight of oxalate =9 ×88×10−3/ 4=22×9×10−3
198×10−3
= $$\displaystyle\%{C}_{{{2}}}{{O}_{{{4}}}^{{-{2}}}}=\frac{{{.198}}}{{{.300}}}\times{100}={66}\%$$
