Question

A certain saturated hydrocarbon effuses about half as fast as methane, `CH_4`. What is the molar mass, in g/mol, of this hydrocarbon ?

Answer 1

`M_r=64`

Rate of effusion is`prop(I)/(sqrtM_r)`

Since Rate `=V/t`:

`tpropsqrtM_r`

So for 2 gases:

`t_1/t_2=sqrt((M_1)/(M_2))`

We are told that `2t_1=t_2`

So:

`(t_1)/(2t_1)=sqrt((M_1)/(M_2))=1/2`

Squaring both sides we get:

`1/4=(M_1)/(M_2)`

`M_1` is the `M_r` of methane`=(12+(4xx1)) = 16`

So `M_2=M_1xx4=16xx4=64`

Here is a video which shows how to solve a different problem using Graham's law.

Video from: Noel Pauller

Hope this helps!