IF 150 mL carbon dioxide effused in 25 seconds, what volume of methane would effuse in same time.

Explanation

The rate of effusion of carbon monoxide ,
`r_1=(150mL)/(25s)`
The rate of effusion of methane , `r_2=(x mL)/(25s)`
According to Graham's law of diffusion,
`r_(CO)/r_(CH_4)=sqrt((M_(CH_4))/(M_(CO)))`
`150/25times25/x=sqrt(15/28)`
x=198.5 mL
`therefore` In 25 seconds the volume of methane diffused =198.5 mL.