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InterviewSolution
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A `10g` mixture of `Cu_(2)S` and `CuS` was treated with `200 mL` of `075M MnO_(4)^(-)` in acid solution producing `SO_(2), CU^(2+)` and `Mn^(2+)`. The `SO_(2)` was boiled off and the excess of `MnO_(4)^(-)`. The `SO_(2)` was boiled off and the excess of `MnO_(4)^(-)` was titrated with `175 mL` of `1M Fe^(2+)` solution. Caculaate `%` fo `CuS` in original mixture. |
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Answer» Correct Answer - `57.94%` |
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