Calculate the molecular weight of a gas `X` which diffuses four times as fast as another gas `Y`, which in turn diffuses twice as fast as another `Z`. Molecular weight of the gas `Z` is `128`.

Answer

`2 g mol^(-1)`

Explanation

Step I. Molecular mass of gas y
`(r_((y)))/(r_((z))) = 2` and `M_((z)) = 128, M_((y)) = ?`
Applying Graham's Law of diffusion
`(r_((y)))/(r_((z))) = sqrt((M_((z)))/(M_((x)))) = 2` or `(M_((z)))/(M_((y)))=4` or `M_((y)) = (M_((z)))/(4)=128/4=32`
Step II. Molecular mass of gas y
`(r_((y)))/(r_((z))) = 4` and `M_((y)) = 32, M_((x)) = ?`
Applying Graham's Law of diffusion,
`(r_((x)))/(r_((y))) = sqrt((M_((y)))/(M_((x))))=4` or `(M_((y)))/(M_(x))=16` or `M_((x)) = (M_((y)))/(16)=32/16=2`.