InterviewSolution
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A `1.10 g` sample of copper ore is dissolved and the `Cu^(2+)` of is treated with excess `KI`. The liberated `I_(2)` requires `12.12 mL` of `0.10M Na_(2)S_(2)O_(3)` solution for titration. Find the `%` copper by mass in ore. |
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Answer» Correct Answer - 7 `{:(Cu^(2+),+e,rarr,Cu^(+),),(,2I^(-),rarr,I_(2)+2e,),(2S_(2)O_(3)^(2-),,rarr,S_(4)O_(6)^(2-),+2e):}` `"Meq. of" Cu^(2+) = "Meq. Of liberated" I_(2)` `= "Meq. of" Na_(2)S_(2)O_(3)`. `= 12.12 xx 0.1 xx 1 = 1.212` `:. (w_(Cu^(2+)))/(63.6//1) xx 1000 = 1.212` `:. w_(Cu) = w_(Cu^(2+)) = 0.077 (because Cu overset(H_(2)SO_(4))rarr CuSO_(4))` `:. %Cu = (0.077)/(1.10) xx 100 = 7%` |
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