# A certain volume of argon gas ( Mol.wt = 40) requires 45 s to diffuse through a hole at a certain pressure and temperature. The same volume of another gas of unknown molecular weight requires 60 s to pass through the same hole under same conditions?

Sep 1, 2017

71 g/ mol

#### Explanation:

since kinetic energy of every gases dipends only from the absolute temperature (Ek = KT) and then at a certain temerature is equal for every gasses, and kinetic energy is also given from ${E}_{k} = \frac{1}{2} m {v}^{2}$ you have that v^2 is proportional to the inverse of the molecular mass.
For that, the Graham's law says $\frac{M b}{M a} = {V}_{a}^{2} / {V}_{b}^{2}$
but speed is proportional to the inverse of the time occurred to diffuse so
$M b = M a \times {t}_{b}^{2} / {t}_{a}^{2} = 40 \frac{g}{m o l} \times {60}^{2} / {45}^{2} = 71 \frac{g}{m o l}$