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It takes $161 \mathrm{~s}$ for an unknown gas to effuse through a porous wall and $97 \mathrm{~s}$ for the same volume of $\mathrm{N}_{2}$ gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas?
$\square$ $\mathrm{g} / \mathrm{mol}$

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∻Solution∻ ‖ 1 ⋮ **Use Graham's Law of Effusion**: Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. ‖ ‖ 2 ⋮ **Set up the ratio**: According to Graham's Law, $\frac{ ext{Rate of effusion of unknown gas}}{ ext{Rate of effusion of } N_2} = \sqrt{\frac{ ext{Molar mass of } N_2}{ ext{Molar mass of unknown gas}}}$. Since the rates are inversely proportional to the times taken to effuse, we can write $\frac{97}{161} = \sqrt{\frac{28.02}{M_{ ext{unknown}}}}$, where $M_{ ext{unknown}}$ is the molar mass of the unknown gas. ‖ ‖ 3 ⋮ **Solve for the molar mass of the unknown gas**: Squaring both sides of the equation gives us $\left(\frac{97}{161} ight)^2 = \frac{28.02}{M_{ ext{unknown}}}$. Solving for $M_{ ext{unknown}}$ gives us $M_{ ext{unknown}} = \frac{28.02}{\left(\frac{97}{161} ight)^2}$. ‖ ‖ 4 ⋮ **Calculate the molar mass**: After performing the calculation, we find the molar mass of the unknown gas. ‖ ∻Answer∻ ⚹ $M_{ ext{unknown}} = \frac{28.02}{\left(\frac{97}{161} ight)^2} \approx 44.01 ext{ g/mol}$ ⚹ ∻Key Concept∻ ⚹ Graham's Law of Effusion ⚹ ∻Explanation∻ ⚹ Graham's Law relates the rates of effusion of two gases to their molar masses, allowing us to calculate the molar mass of an unknown gas when compared to a known gas. ⚹◊What is the formula to calculate the molar mass of a gas based on effusion rates?⍭ Generate me a similar question◊

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