A mixture of `H_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` weighing `2.02 g` was dissolved in water and the solution made uptp one litre. `10 mL` of this solution required `3.0 mL` of `0.1 N NaOH` solution for complete neutralization. In another experiment `10 mL` of same solution in hot dilute `H_(2)SO_(4)` medium required `4 mL` of `0.1N KMnO_(4) KMnO_(4)` for compltete neutralization. Calculate the amount of `H_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` in mixture.

Options

A. `0.1,0.1`
B. `0.1,0.05`
C. `0.05,0.1`
D. `0.05,0.05`

Answer

D

Explanation

Let molarity of `H_(2)SO_(4) = M_(1)` and `H_(2)C_(2)O_(4) =M`
No. of milli equivalents of `(H_(2)SO_(4) +H_(2)C_(2)O_(4))`
No. of milli equivalent of `NaOH`
`20 (2M_(1) +2M_(2)) = 40 xx (1)/(10), 40 (M_(1)+M_(2)) = 4`
`M_(1)+M_(2) = 0.1` -----(1)
No. of milli equivalents of `H_(2)C_(2)O_(4) =`
No. of milli equivalents `KMnO_(4)`
`20 (2M_(2)) = 20 xx (1)/(10)`
`M_(2) = (1)/(20) = 0.05M` ....(2)
`M_(1) = 0.1 - 0.05 = 0.05M`