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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» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Equivalent <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of `CuFeS_2=` molecular weight <br/> `CuFeS_2toCuSdarr+FeS` <br/> `FeS+H_2SO_4toFeSO_4+H_2Suarr` <br/> `2FeSO_4+H_2SO_4+OtoFe(SO_4)_3+H_2O` <br/> Molecular weight of `CuFeS_2=63.5+55.8+64=183.3g` <br/> `42 " <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> of " 0.01 M K_2Cr_2O_7=42xx0.01xx6NK_2Cr_2O_7` <br/> `=<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.52 m" Eq of "K_2Cr_2O_7` <br/> `=2.52m" Eq of "<a href="https://interviewquestions.tuteehub.com/tag/fe-460008" style="font-weight:bold;" target="_blank" title="Click to know more about FE">FE</a>^(2+)` <br/> `=2.52 m" Eq of "CuFeS_2` <br/> `=2.52 m" Eq of "CuFeS_2` <br/> `=2.52xx10^(-3)xx18.3g of CuFeS_2` <br/> `=0.4619 g of CuFeS_2` <br/> `%` of `CuFeS_2=(0.4619xx100)/(0.5)=92.38%`</body></html>


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