A 0.5 g sample of an iron containing mineral mainly in the form of `CuFeS_2` was reduced suitable to convert all the ferric ions into the ferrous form and was obtained as a solution. In the absence of any interfering matter, the solution required 42 " mL of " 0.1 M `K_2Cr_2O_7` solution for titration calculate the percentage of `CuFeS_2` in the mineral `(Cu=63.5,Fe=55.8)`

A 0.5 g sample of an iron containing mineral mainly in the form of `Cu

1.	A 0.5 g sample of an iron containing mineral mainly in the form of `CuFeS_2` was reduced suitable to convert all the ferric ions into the ferrous form and was obtained as a solution. In the absence of any interfering matter, the solution required 42 " mL of " 0.1 M `K_2Cr_2O_7` solution for titration calculate the percentage of `CuFeS_2` in the mineral `(Cu=63.5,Fe=55.8)`
Answer» Equivalent weight of `CuFeS_2=` molecular weight `CuFeS_2toCuSdarr+FeS` `FeS+H_2SO_4toFeSO_4+H_2Suarr` `2FeSO_4+H_2SO_4+OtoFe(SO_4)_3+H_2O` Molecular weight of `CuFeS_2=63.5+55.8+64=183.3g` `42 " mL of " 0.01 M K_2Cr_2O_7=42xx0.01xx6NK_2Cr_2O_7` `=2.52 m" Eq of "K_2Cr_2O_7` `=2.52m" Eq of "Fe^(2+)` `=2.52 m" Eq of "CuFeS_2` `=2.52 m" Eq of "CuFeS_2` `=2.52xx10^(-3)xx18.3g of CuFeS_2` `=0.4619 g of CuFeS_2` `%` of `CuFeS_2=(0.4619xx100)/(0.5)=92.38%`

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