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Q.
0.3 g of an oxalate salt was dissolved in $100$ mL solution. The solution required 90 mL of $N/20 KMnO_{4}$ for complete oxidation. The % of oxalate ion in salt is:
Redox Reactions
Solution:
Meq. of $KMnO_{4}$ = M eq. of $C_{2}O^{-2}_{4}$
$90\times\frac{1}{2}=100 \times N_{C_{ 2}O_{4}^{-2}}$
Mmole of oxalate $=\frac{9}{2\times2}=\frac{9}{4}$
Weight of oxalate $=\frac{9}{4} \times88\times10^{-3}=22 \times9\times10^{-3}$
$=198 \times10^{-3} $
$\% C_{2} O_{4}^{-2} \times100=66\%$