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What weight of `FeSO_(4)` ( mol.wt. `=152`) will be oxidised by `200 mL` of normal `KMnO_(4)` solution in acid solution?A. `30.4g`B. `60.8 g`C. `121.6g`D. `15.8 g` |
Answer» Correct Answer - A Meq.of `FeSO_(4)="Meq.of" KMnO_(4)=200xx1` `:. (w)/(152//1)xx1000=200` `w=30.4 g` |
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