Practice Problem: Graham's Law of Effusion

The video tutorial explains how to use Graham's law of effusion to identify an unknown gas. Methane effuses at 1.414 times the rate of the unknown gas. By using the formula relating effusion rates and molar masses, the tutorial guides viewers through the calculation process. The molar mass of methane is known, and by solving the equation, the molar mass of the unknown gas is found to be approximately 32 grams per mole, suggesting it is likely oxygen gas.

What is the main concept being practiced in this video tutorial?

• Charles's Law

• Graham's Law of Effusion

• Avogadro's Law

• Boyle's Law

According to Graham's Law, the rate of effusion of a gas is inversely proportional to what?

• The square root of its molar mass

• The pressure of the gas

• The temperature of the gas

• The volume of the gas

In the given problem, what is the known molar mass of methane used in the calculations?

• 20 grams per mole

• 16 grams per mole

• 12 grams per mole

• 18 grams per mole

What mathematical operation is performed to eliminate the square root in the equation?

• Multiplying by a constant

• Squaring both sides

• Dividing by the molar mass

• Taking the logarithm

Which gas is identified as the unknown gas based on the calculated molar mass?

• Carbon Dioxide (CO2)

• Nitrogen (N2)

• Helium (He)

• Oxygen (O2)