A 10 g samle of only CuS and `Cu_(2)S` was treated with 100 ml of `1.25 M K_(2) Cr_(2) O_(7) ` . The Products obtained were `Cr^(3+) , Cu^(2+) " and " SO_(2) `. The excess oxidant was reacted with 50 ml of `Fe^(2+)` Solution : 25 ml of the same `Fe^(2+)` solution required `0.875 M KMnO_(4)` under acidic condition , the volume of `KMnO_(4)` used was 20 ml . Find the % of CuS and `Cu_(2)S` in the sample .

Explanation

Equivalents of dichromate initially ` = (1.25 xx 6 xx 100)/1000 = 0.75`
Equivalents of `Fe^(2+) ` in 25 ml ` = (0.875 xx 5 xx 5 xx 20)/1000 = 0.0875`
Equivalents of excess dichromate = `0.175`
` :. ` Equivalents of dichromate consumed by `(CuS " and " Cu_(2)S)`
` = 0.75 - 0.175 = 0.575`
If x g is the mass of CuS, the mass of `Cu_(2)S " is " (10 - x) ` g
`x/(95.5) xx 6 + ((10 -x))/(159) xx 8 = 0.575`
` :. x = 5.74 ` gm
` % CuS = (5.74)/10 xx 100 = 57.4 %`
` % Cu_(2)S = 42.6 %`