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During electrolysis of `H_(2)O` , the molar ratio of `H_(2)` and `O_(2)` formed is
A. ` 2 : 1`
B. `1 : 2`
C. `1 : 3`
D. `1 : 1`

1 Answer

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Best 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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