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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» <html><body><p></p>Solution :The reduction half reaction is `C_(2)O_(4)^(2-) rarr 2CO_(2) +2e^(-)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Eq we of `C_(2)O_(4)^(2-)=(2xx12+4xx16)/(2)=44` <br/> Let `N_(1)` be the normality of the oxalte solution applying normality equation we have <br/> `15xx1//20(KmnO_(4))=N_(1)xx10(C_(2) O_(4))^(2-) therefore N_(1)=(15)/(10xx20)=3/40` <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> <br/> Amount of pure `C_(2)O_(4)^(2-)` present in 100 mL =`(33/1000)xx100=0.33 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>` <br/> But amount of `C_(2)O_(4)^(2-)` present in impure sample =0.5 g (given ) <br/> <a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of pure <a href="https://interviewquestions.tuteehub.com/tag/oxalate-1144431" style="font-weight:bold;" target="_blank" title="Click to know more about OXALATE">OXALATE</a> `=(0.33)/(0.50)xx100=66`</body></html>


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