A sample of neon effuses from a container in 77 seconds. The same amount of an unknown noble gas requires 157 seconds. Identify the gas.

To find the amount of an unknown noble gas that requires 157 seconds fo effuse from a container from which a same amount of sample of neon effuses in 77 seconds, you must use Graham's Law of effusion.

Graham's law of effusion states the rate of effusion of a gas is inversely proportional to the square root of the masses of its particles. Mathematically, that is:

[tex]\frac{Rate_1}{Rate_2}=\sqrt{\frac{MolarMass_2}{MolarMass_1}}[/tex]

Since, the other conditions (amount of gas and container) are the same for both gases, the rates and the times are inversely proportional; this is:

[tex]\frac{Rate_1}{Rate_2}=\frac{Time_2}{Time_1}[/tex]

By using the previous equations, you can relate the times and the molar masses of the gases:

[tex]\sqrt{\frac{MolarMass_2}{MolarMass_1}}=\frac{Time_2}{Time_1}[/tex]

Clear the unknown molar mass:

[tex]\frac{MolarMass_2}{MolarMass_1}={(\frac{Time_2}{Time_1})}^2\\ \\ \\MolarMass_2=(\frac{Times_2}{Time_1} )^2 .MolarMass_1[/tex]

Substitute the data:

MolarMass₂ = (157 s / 77 s) × MolarMass₁

From a periodic table, obtain the molar mass of neon gas:

Molar Mass₁ = 20.180 g/mol

∴ MolarMass₂ = (157 s / 77 s) × 20.180 g/mol = 41 g/mol

Again from the periodic table, find the noble gas with the molar mass closest to 41 g/mol.

It is argon. The molar mass of argon is 30.948 g/mol. So, argon (Ar) is the answer.