0.5 g mixture of oxalic acid `(H_(2)C_(2)O_4)` and some sodium oxalate `(Na_(2)C_(2)O_4)` with some impurities requires 40 ml of 0.1M NaOH for complete neutralization and 6ml of 0.2 M `KMnO_4` for complete oxidation. Calculate the % of `Na_2C_(2)O_4` in the mixture

Options

A. `90%`
B. ` 26.8%`
C. `40%`
D. `50%`

Answer

B

Explanation

No. of milli equivalents of `H_(2)C_(2)O_4` = No .of milli equivalents of NaOH = 4 No.of milli
equivalents of `H_(2)C_(4)O_(4)+Na_(20)C_(2)O_(4)` = No. of
milli equivalents of `KMnO_(4)= 6 xx 0.2 xx 5 = 6`
Milli equivalent of `Na_(2)C_(2)O_(4)= 2`
Weight of `Na_(2)C_(2)O_(4)=2xx10^(-3)xx67 =0.134` g
% `Na_(2)C_2O_4=(0.134)/(0.5)xx100=26.8`