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 data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.