Diffusion of `CH_4 (g) and O_2 (g) ` occurs under similar conditions, then the ratio of their rates of diffusion is

Options

A. 1.414
B. 0.707
C. 2.312
D. 1.732

Answer

A

Explanation

From Graham's law of diffusion :
`r_(CH_4)/r_(O_2)=sqrt(M_(O_2)/M_(CH_4))`
`"here"" "r_(CH_4)="rate of diffusion of "CH_4" gas"`
`" "r_(O_2)="rate of diffusion of "O_2" gas"`
`" "M_(CH_4)="molecular mass of "CH_4=16`
`"and"" "M_(O_2)="molecular mass of "O_2" gas"=32=16`
`"Hence, "r_(CH_4)/r_(O_2)=sqrt(32/16)=sqrt(2)=1.414`