0.5 g of an oxalate was dissolved in water and the solution made to 100 mL. On titration 10 mL of this solution required 15 mL of `(N)/(20)KMnO_(4)`. Calculate the percentage of oxalate in the sample .

Explanation

The reduction half reaction is `C_(2)O_(4)^(2-) rarr 2CO_(2) +2e^(-)`
`therefore` Eq we of `C_(2)O_(4)^(2-)=(2xx12+4xx16)/(2)=44`
Let `N_(1)` be the normality of the oxalte solution applying normality equation we have
`15xx1//20(KmnO_(4))=N_(1)xx10(C_(2) O_(4))^(2-) therefore N_(1)=(15)/(10xx20)=3/40` N
Amount of pure `C_(2)O_(4)^(2-)` present in 100 mL =`(33/1000)xx100=0.33 g`
But amount of `C_(2)O_(4)^(2-)` present in impure sample =0.5 g (given )
Percentage of pure oxalate `=(0.33)/(0.50)xx100=66`