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

Explanation

Rate of diffusion of `CO (r_1)`
` ( " volume")/( "Time") = ( 150)/(125)`
Molecular weight of CO (`M_1`) = 28
Rate of diffusion of methane (`r_2`)
` = ( " volume")/( "time") = (V )/(25 ) ml. Sec^(-1)`
Molecular weight of methane `(M_2) = 16`
According to Graham.s law of diffusion
` (r_1)/(r_2) sqrt(( M_2)/(M_1) ) = > ( 150)/(25) xx ( 25)/(v ) = sqrt((16 )/(28 )) = 0.756 `
` V= ( 150 )/( 0.756 )= 198 .4 ml `.