`200 mL` of `SO_(2)` diffuse through a porous plug in 600 seconds. What volume of methane `(CH_(4))` will diffuse in the same time ?

Explanation

Rate of diffusion of `SO_(2) = (200)/(600) mL s^(-1)`
Rate of diffusion of `CH_(4)(r_(CH_(4))) = (V)/(600) mL s^(-1)`
Molar mass of `SO_(2)(M_(SO_(2))) = 64`
Molar mass of `CH_(4)(M_(CH_(4))) = 16`
According to Graham's law of emission :
`(r_(SO_(2)))/(r_(CH_(4))) = sqrt((M_(CH_(4)))/(M_(SO_(2))))` or `(200)/(600) xx (600)/(V) = sqrt((16)/(64)) = 1/2`
`200= 1/2` or `V = 200 xx 2 = 400 mL`