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

Let mass of each of the gases be wg. In the mixture,
`n_(H_(2))=(w)/(2) and , n_(O_(2))=(w)/(32) mol and n_(CH_(4)) =(w)/(16) mol`.
Total number of moles in the mixture`=(w)/(2)+(w)/(32)+(w)/(16)=(19)/(32)w`.
so, in the mixture, `x_(H_(2))=(w//2)/(19w//32)=(16)/(19),x_(O_(2))=(1)/(19)`
and `x_(CH_(4))=(2)/(19)`.
Volume fraction of a component in the mixture=molefraction of the component`xx`total volume of the mixture
`therefore V_(H_(2))=(16)/(19)xxV,V_(O_(2))=(1)/(19)xxV and V_(CH_(4)) =(2)/(19)xxV`
`therefore V_(H_(2)) : V_(O_(2)) : V_(CH_(4))=16:1:2`