1.80 g of impure sample of oxalate was dissolved in water and the solution made to 250 mL. On titration 20 mL of this solution required 30 mL of M/50 KMnO_4 solution . Calculated the percentage purity of the sample.

1.80 g of impure sample of oxalate was dissolved in water and the solu

1.	1.80 g of impure sample of oxalate was dissolved in water and the solution made to 250 mL. On titration 20 mL of this solution required 30 mL of M/50 KMnO_4 solution . Calculated the percentage purity of the sample.
Answer» Solution :The BALANCED chemical equation is : `2MnO_4^(-)+5C_2O_4^(2-)+16H^(+)rarr2Mn^(2+)+10CO_2+8H_2O` Applying morality equation , `((M_1V_1)/n_1)_(KMnO_4)-=((M_2V_2)/n_1)_(C_2O_4^(2-))` `1/50xx30/2-=(M_2xx20)/5` `M_2=(30xx5)/(50xx2xx20)=0.075M` MOLECULAR wt. of `C_2O_4^(2-) =88` Wt of `C_2O_4^(2-) = 0.075 xx88=6.6g` AMOUNT of `C_2O_4^(2-) ` in 250 mL `=(6.6)/1000xx250=1.65g` % Purity `=(1.65)/(1.80)xx100=91.7%`