0.5 g sample of copper ore is converted into `CuSO_4` solution. The resulting solution is acidified with dilute `CH_3COOH` (acetic acid) and excess KI added. The liberated `I_2` required 0.248 g `Na_2S_2O_3.5H_2O` for complete reaction. Calculate the percantage of Cu in the ore.

0.5 g sample of copper ore is converted into `CuSO_4` solution. The re

1.	0.5 g sample of copper ore is converted into `CuSO_4` solution. The resulting solution is acidified with dilute `CH_3COOH` (acetic acid) and excess KI added. The liberated `I_2` required 0.248 g `Na_2S_2O_3.5H_2O` for complete reaction. Calculate the percantage of Cu in the ore.
Answer» `Mw(Na_2S_2O_3.5H_2O)=248g` `Ew=(248)/(1)(n=1)` `Ew(CuSO_4 or Cu)=(M)/(1)` `Cu-=CuSO_4-=KI-=I_2-=Na_2S_O_3` `-=(0.248)/(248)=10^(-3)Eq` " Eq of "`Cu=10^(-3)Eq` "Weight of" of `Cu=EqxxEq=10^(-3)xx(63.5)/(1)g(n=1)` `%` of `Cu=(63.5xx10^(-3)gxx100)/(0.5g)=(635xx100)/(1000xx5)=12.7%`

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0.5 g sample of copper ore is converted into `CuSO_4` solution. The resulting solution is acidified with dilute `CH_3COOH` (acetic acid) and excess KI added. The liberated `I_2` required 0.248 g `Na_2S_2O_3.5H_2O` for complete reaction. Calculate the percantage of Cu in the ore.

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