Question

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.

Answer 1

Answer: Argon

Explanation:

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:

$\frac{Rate_1}{Rate_2}=\sqrt{\frac{MolarMass_2}{MolarMass_1}}$

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:

$\frac{Rate_1}{Rate_2}=\frac{Time_2}{Time_1}$

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

$\sqrt{\frac{MolarMass_2}{MolarMass_1}}=\frac{Time_2}{Time_1}$

• Clear the unknown molar mass:

$\frac{MolarMass_2}{MolarMass_1}={(\frac{Time_2}{Time_1})}^2\\ \\ \\MolarMass_2=(\frac{Times_2}{Time_1} )^2 .MolarMass_1$

• 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.