Equal masses of `H_(2)`, `O_(2)` and methane have been taken in a container of volume `V` at temperature `27^(@)C` in identical conditions. The ratio of the volume of gases `H_(2) : O_(2)` : methane would be

Options

A. `8:16:1`
B. `16:8:1`
C. `16:1:2`
D. `8:1:2`

Answer

C

Explanation

Suppose mass of each of `H_(2), O_(2)` and `CH_(4)` taken
=w g
`w g H_(2)=(w)/(2)"mole", w g O_(2)=(w)/(32)"mole"`,
`w g CH_(4)=(w)/(16)"mole"`
As under identical conditions, volume are in the ratio of moles. Hence,
volume of `H_(2) : O_(2): CH_(4)=(w)/(2) : (w)/(32) : (w)/(16)`
`=(1)/(2) : (1)/(32) : (1)/(16)=16 : 1 : 2`