$referrer_host = www.google.com. Graham's Law of Effusion Practice Problems - Chemistry Steps

General Chemistry

In the previous post, we talked about the Graham’s law which shows that the rate of effusion of a gas is inversely proportional to the square root of its molar mass:

 

\[{u_{{\rm{rms}}}}\;{\rm{ = }}\;\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \]

 

One application of Graham’s law is the formula obtained for two gases correlating their molar masses and effusion rates:

 

This allows for determining the molar mass of unknown gas by comparing its effusion rate to standard gas. 

Below are some practice examples of the Graham’s law of effusion:

 

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Practice

1.

What is the ratio of the effusion rates of hydrogen gas (H2) and carbon dioxide (CO2) at the same pressure and temperature?

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2.

The rate of effusion of an unknown gas is 9.20 mL/min. Under identical conditions, the rate of effusion of pure nitrogen (N2) gas is 14.65 mL/min. Identify the unknown gas using the Graham’s law.
a) O2 b) C3H8 c) C4H10  d) NO2  e) Cl2

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3.

A sample of krypton effuses from a container in 95 seconds. The same amount of an unknown gas requires 55 seconds. Identify the unknown gas.

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4.

If a sample of Br2 vapor can effuse from an opening in a heated vessel in 46 s, how long will it take the same amount of He to effuse under identical conditions?

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5.

It has been demonstrated that 3.56 mL of an unknown gas effuses through a hole in the same time that 8.64 mL of argon does under the same conditions. Determine the molecular mass of the unknown gas.

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