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

Options

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

Answer

B

Explanation

`(t_(H_(2)))/(t_(x)) = sqrt((M_(H_(2)))/(M_(x)))`
`(5)/(t_(x)) = sqrt((2)/(M_(x)))`
`(25)/(t_(x)^(2)) = (2)/(M_(x))`
`(t_(x)^(2))/(M_(x)) = (25)/(2)`
Mol. Mass of `He, O_(2), CO " and " CO_(2)` are 4, 32, 28 and 44 respectively.
`(t_(He)^(2))/(M_(He)) = (100)/(4) = (50)/(2), (t_(O_(2))^(2))/(M_(O_(2))) = (400)/(32) = (25)/(2)`
`(t_(CO)^(2))/(M_(CO)) = (625)/(28), (t_(CO_(2))^(2))/(M_(CO_(2))) = (3025)/(44)`
`[t_(He) = 10 s, t_(O_(2)) = 20 s, t_(CO) = 25 s' t_(CO_(2)) = 550]`