During electrolysis of water, the volume of oxygen liberate is `2.24 dm^(3).` The volume of hydrogen liberated, under same conditions will be

Options

A. `2.24 dm^(3)`
B. `1.12 dm^(3)`
C. `4.48 dm^(3)`
D. `0.56 dm^(3)`

Answer

C

Explanation

During electrolysis volume of `O_(2)` and `H_(2)` liberated are in the ratio of 1 : 2, hence volume of `H_(2)` liberated will be 4.48 `dm^(3)`.