X mL of `H_(2)` gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical condition is:

Options

A. `10 s`, `He`
B. `20 s`, `O_(2)`
C. `25 s`, `CO`
D. `55 s`, `CO_(2)`

Explanation

For the same volume diffused,
`t prop M` (Graham' s law)
`(t_(1))/(t_(2))=sqrt((M_(1))/(M_(2))`
(`a`) `(5)/(10)!=sqrt((2)/(4))`
(`b`) `(5)/(20)=sqrt((2)/(32))`
(`c`) `(5)/(25)!=sqrt((2)/(28))`
(`d`) `(5)/(55)!=sqrt((2)/(44))`