Question

If a gas effuses `1.89` times as fast as `"COCl"_2(g)`, what is its molar mass?

Answer 1

`M_r=27.7`

Graham's law states that :

`Rprop(1)/(sqrtM_r)`

For 2 gases this becomes:

`R_1/R_2=sqrt((M_2)/(M_1))`

`M_r[COCl_2]=99.0=M_1`

`R_2=1.89xxR_1`

So:

`(R_1)/(1.89R_1)=sqrt((M_2)/(99.0))`

So:

`1/1.89=sqrt((M_2)/(99.0))`

`0.53=sqrt((M_2)/(99.0))`

`0.28=(M_2)/(99.0)`

`M_2=0.28xx99.0=27.7`

Answer 2

The compound that effuses faster must have lighter molecules, if they are both at the same temperature.
Take the ratio that is `189 : 100 = 1.89` that means the smaller molecules are in the average moving 1.89 times faster than `COCl_2` molecules.

Square that value, `(V/v)^2 = V^2/v^2 = 1.89^2 = 3.57`.

This is the mass ratio between heavier and lighter molecules, because these have the same mean kinetic energies and root mean squared speeds, as consequence of the equipartition of energy :

`1/2Mv^2 = 1/2mV^2` implies: `V^2/v^2 = M/m = 3.57.`

So the lighter molecules have mass `m = M/3.57 = 98.9/3.57 = 27.7`

where 98.9 = C + O + 2Cl = 12.01 + 16.00 + 35.45·2 = 98.9.