An impure sample of sodium oxalate `(Na_(2) C_(2)0=O_(4))` weighing 0.20 g is dissolved in an aqueous solution of `H_(2)SO_(4)` and the solution is titrated at `70^(@)C`, requiring 45 ml `0.02 M KMnO_(4)` solution. The endpoint is overrun, and the back titration is carried out with 10 ml of 0.1 M oxalic acid solution. Find the per cent purity of `Na_(2)C_(2)O_(4)` in the sample.

Options

A. 75
B. 83.75
C. 90.25
D. None of these

Answer

B

Explanation

The number of milli equivalent of `Na_(2)C_(2)O_(4)` = milli equivalent of `KMnO_(4)` reacted
Total milli equivalent of `KMnO_(4)`- excess milli equivalent of `KmnO_(4)` reacts with `H_(2)C_(2)O_(4)`
`=45xx0.02xx5-10xx0.1xx2=2.5`
`1000xx(W)/(134)xx2=2.5`
`W_(Na_(2)C_(2)O_(4))=0.1675g`
Percent purity of `Na_(2)C_(2)O_(4)` in sample