The calcium contained in a solution of 1.048 gm of the substance being analysed was precipitated with 25ml of `H_(2) C_(2) O_(4)`. The excess of `C_(2) O_(4)^(2-)` in one fourth of the filtrate was back titrated with 5ml of 0.1025 N `KMnO_(4)` solution. To determine the concentration of the `H_(2) C_(2) O_(4)` solution taken, it was diluted four fold, the titration of 25ml of the dilute solution used up 24.1 ml of `KMnO_(4)` solution. What is percentage of CaO in the substance being analysed.

Explanation

Equivalents of `KMnO_(4)` used up with one fourth of the filtrate containing excess `H_(2)C_(2) O_(4) = ( 5 xx 0.1025)/( 1000) = 5.125 xx 10^(-4)`
`:.` Excess equivalent of `H_(2) C_(2) O_(4) = 5.125 xx 10^(-4) xx 4 = 2.05 xx 10^(-3)`
Equivalent of `KMnO_(4)` reacting in the second titration `= ( 24.1 )/( 1000) xx 0.1025 = 2.47 xx 10^(-3)`
`:.` Equivalent `H_(2) C_(2) O_(4)` consumed by `CaO = 9.88 xx 10^(-3) - 2.05 xx 10^(-3) = 7.83 xx 10^(-3)`
`:.` Equivalents of `CaO = 7.83 xx 10^(-3)`
Moles of `CaO = ( 7.83 xx 10^(-3))/(2) = 3.915 xx 10^(-3)`
Mass of CaO `= 3.915 xx 10^(-3) xx 56 = 0.22 ` gm
% CaO = `( 0.22 )/( 1.048) xx 100 = 20.99 %`