Question

Find out the $\%$ of oxalate ion in given sample of oxalate salt of which $0.3g$ is present in $100mL$ of solution required $90mL$ of $N/20KMn{O}_4$ for complete oxidation.

Answer 1

Answer: 66.00

The redox changes are $M{n}^{+7}+5e^{-}\to M{n}^{2+}$

${\left(C^{3+}\right)}_2\to 2C^{4+}+2e^{-}$

$\because$ meq. of oxalate ion $=$ meq. of $KMn{O}_4$ $(w/E)\times 1000=90\times (1/20)$

Or $\left(\frac{w}{88/2}\right)\times 1000=\frac{9}{2}\left(\because E_{C_2{O}_4^{-2}}=\frac{88}{2}\right)$

or $w_{C_2{O}_4^{-2}}=0.198g$

$\therefore \%$ of oxalate in sample $=(0.198\times 100)/0.3=66\%$