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

Comparing rates of diffusion of gas B and Gas C.
`(r_(B))/(r_(C ))=sqrt((M_(C ))/(M_(B)))`
`M_(C )=128, =(r_(B))/(r_(C ))=2, M_(B)=?`
`2=sqrt((128)/(M_(B)))`
`"or "4=(128)/(M_(B))`
`therefore" "M_(B)=(128)/(4)=32`
Now, comparing rates of diffusion of gas B and A
`(r_(A))/(r_(B))=sqrt((M_(B))/(M_(A)))`
`(r_(A))/(r_(B))=4, M_(B)=32, M_(A)=?`
`4=sqrt((32)/(M_(A)))`
`"or "16=(32)/(M_(A))`
`therefore" "M_(A)=(32)/(16)="2 g mole"^(-1)`.