Question 6.16: Relating Effusion Times and Molar Masses A sample of Kr(g) e...
Relating Effusion Times and Molar Masses
A sample of Kr(g) escapes through a tiny hole in 87.3 s. The same amount of an unknown gas escapes in 42.9 s under identical conditions. What is the molar mass of the unknown gas?
Analyze
Because the unknown gas effuses faster, it must have a smaller molar mass than Kr. Before we set the ratio
\frac{\text{effusion time for unknown}}{\text{effusion time for Kr}}
equal to \sqrt{\text{ratio of two molar masses}}, we must make sure the ratio of molar masses is smaller than one. Thus, the ratio of molar masses must be written with the molar mass of the lighter gas (the unknown gas) in the numerator.
\frac{\text{effusion time for unknown}}{\text{effusion time for Kr}} = \frac{42.9 s}{87.3 s} = \sqrt{\frac{M_{\text {unk }}}{M_{ Kr }}} = 0.491
M_{\text {unk }} = (0.491)^2 \times M_{Kr} = (0.491)^2 \times 83.80 = 20.2 g/mol
Assess
Use the final result and work backward. The molar mass of the unknown is about 4 times as small as that of Kr. Because effusion rate \alpha 1/\sqrt{M}, the unknown will effuse about \sqrt{4} = 2 times as fast as Kr. The effusion times show that the unknown does indeed effuse 2 times as fast as Kr.