0.5 g o fan impure sample of oxalate was dissolved in water and the solution made 100 mL on titration 10 mLof this solution requared 15 mL of N//20 KMnO_(4) solution . Calculate the percentage of pure oxalate in the sample

0.5 g o fan impure sample of oxalate was dissolved in water and the so

1.	0.5 g o fan impure sample of oxalate was dissolved in water and the solution made 100 mL on titration 10 mLof this solution requared 15 mL of N//20 KMnO_(4) solution . Calculate the percentage of pure oxalate in the sample
Answer» Solution :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`