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During electrolysis of `H_(2)O` , the molar ratio of `H_(2)` and `O_(2)` formed isA. ` 2 : 1`B. `1 : 2`C. `1 : 3`D. `1 : 1` |
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Answer» Correct Answer - A `H_(2)O to H_(2) + 1//2 O_(2)` `therefore` The ratio of `H_(2)` to `O_(2)` `1 : 0.5 therefore 2 : 1` |
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