30 mL 0.05 M `KMnO_(4)` is required for complete oxidation of 0.5 g oxalate in acidic medium .Calculate the percent amount of oxalate in that salt sample

Explanation

`2MnO_(4)^(-) +5C_(2)O_(4)^(2-) +16H^(+)to2Mn^(2+)+10CO_(2)+8H_(2)O`
Accoding to the equation ,`2molMnO_(4)^(-)-=5molC_(2)O_(4)^(2-)`
`therefore 1mol MnO_(4)^(-)-=(5)/(2)mol C_(2)O_(4)^(2-)`
Again ,1000 ml 0.5(M) `KMnO(4)implies 0.05mol KMnO_(4)`
`therefore 30mL 0.05(M) MnO_(4)implies(0.05xx30)/(1000)molKMnO_(4)`
`1.5xx10^(-3) mol KMnO_(4)`Mbr> Now 1 mol `MnO(4)^(-)-=(5)/(2)mol C_(2)O_(4)^(2-)`
`therefore1.5xx10^(-)molMnO(4)^(-)-=(5)/(2)xx1.5xx10^(-3)molC_(2)O_(4-)^(2-)`
`=(5)/(2)xx1.5xx10^(-3)xx88gC_(2)O_(4)^(2-)=0.33g C_(2)O_(4)^(2-)`
Percentage of `C_(2)O_(4)^(2-)`in the sample =`(0.33xx100)/(0.5)=66%`