An impure sample of sodium oxalate `(Na_(2)C_(2)O_(4)` weighing 0.20 g is dissolved in aqueous solution of `H_(2)SO_94)` and solution is titrated at `70^(@)C`,requiring 45 mL of 0.02 M `KMnO_(4)` solution. The end point is overrun, and back titration in carried out with 10 mL of 0.1 M oxalic acid solution.Find the purity of `Na_(2)C_(2)O_(4)` in sample:

Answer

83.75

Explanation

m-eq of `Na_(2)CO_(3)` = m-eq of `KMnO_(4)` reacted total m-eq of `KMnO_(4)` excess m-eq of `KMnO_(4)` reacted with `H_(2)C_(2)O_(4)`
`=45 xx 0.02 xx 5 -10 xx 0.1 xx 2 = 2.5`
`1000 xx W/134 xx 2 = 2.5`
`W_(Na_(2)C_(2)O_(4)) = 0.1675`g
% purity of `Na_(2)CO_(3)` in sample
`=0.1675/0.2 xx 100 = 83.75`