The Graham's law states that ''at constant pressure and temperature the rate of diffusion or effusion of a gas is inversely proportional to the squar root of its density Rate of diffusion `prop (1)/(sqrt(d))`
If `r_(1)` and `r_(2)` represent the rates of diffusion of two gases and `d_(1)` and `d_(2)` are their respective densities, then
`r_(1)/(r_(2))=sqrt((d_(2))/(d_(1)))`
`r_(1)/(r_(2)) =sqrt((M_(2))/(M_(1))) xx P_(1)/(P_(2))`
`(V_(1)xxt_(2))/(V_(2)xxt_(1)) = sqrt((d_(2))/(d_(1))) = sqrt((M_(2))/(M_(1)))`
`V prop n` (where n is no of moles)
`V_(1) prop n_(1)` and `V_(2) prop n_(2)`
The time taken for a certain volume of gas `X` to diffuse through a small hole is 2 minutes It takes 5.65 minutes for oxygen to diffuse under the simillar conditions The molecular weight of `X` is .

Answer

4

Explanation

`(r_(A))/(ro_(2)) = sqrt(M_(O_(2))/(M_(A))) = (t_(O_(2)))/(t_(A))`
or `sqrt((32)/(M_(A))) = (5.65)/(2) or M_(A)=4`