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Recall that ** Graham's Law of Effusion** allows us to compare the rate of effusion of two gases. Graham's Law states that the rate of effusion of a gas is inversely proportional to its molar mass.

$\mathbf{rate}\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{MM}}_{\mathbf{gas}}}}$

This means that when comparing two gases:

$\overline{)\frac{{\mathbf{rate}}_{\mathbf{gas}\mathbf{}\mathbf{1}}}{{\mathbf{rate}}_{\mathbf{gas}\mathbf{}\mathbf{2}}}{\mathbf{=}}\sqrt{\frac{{\mathbf{MM}}_{\mathbf{gas}\mathbf{}\mathbf{2}}}{{\mathbf{MM}}_{\mathbf{gas}\mathbf{}\mathbf{1}}}}}$

The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusion of pure methane (CH_{4}) gas is 47.8 mL/min. What is the molar mass of the unknown gas?

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Our tutors have indicated that to solve this problem you will need to apply the Effusion concept. You can view video lessons to learn Effusion. Or if you need more Effusion practice, you can also practice Effusion practice problems.

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Based on our data, we think this problem is relevant for Professor Anliker's class at IUPUI.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl 2nd Edition practice problems.