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The rate at which a gas escapes through a tiny hole, or effuses, is an intriguing aspect of its behavior under certain conditions. Graham's Law of Effusion elegantly describes this rate as being inversely related to the square root of the gas's molar mass. This can be represented by the equation:
\[\begin{equation}Rate \(\mathrm{CH}_{4}\) = Rate \(\mathrm{HBr}\) \propto \frac{1}{\sqrt{M}}\end{equation}\]When comparing two gases under identical conditions of temperature and pressure, they effuse at rates that are inversely proportional to the square roots of their molar masses. If the effusion rates of methane (\(\mathrm{CH}_{4}\)) and hydrogen bromide (\(\mathrm{HBr}\)) are equal, as in the exercise provided, the equation simplifies to a ratio of 1. This forms the foundation of solving for the mole fraction of methane in the mixture. The real-world implication of this law can be seen in applications such as the separation of isotopes or assessing the purity of a gas.

The molar mass of a substance is a fundamental property, reflecting the mass per mole of its molecules, typically expressed in grams per mole (g/mol). It's integral to various chemical calculations, including the computation of effusion rates as per Graham's Law. For gases like methane (\(\mathrm{CH}_{4}\)) and hydrogen bromide (\(\mathrm{HBr}\)), with molar masses of 16 g/mol and 80.91 g/mol respectively, the molar mass is a critical factor in determining how fast they effuse.When we equate their effusion rates, as the exercise indicates, we apply the idea that heaviness (higher molar mass) restricts the speed at which a gas exits a container. By using the known molar masses, we can relate these to the mole fraction of methane. In the context of effusion, lower molar mass gases escape faster, illustrating why a balloon filled with helium deflates quicker than one with plain air.

Mole fraction is a measure of concentration, indicating the proportion of a component in a mixture. Denoted by \(x_{i}\) for a substance \(i\), it's defined as the number of moles of that substance divided by the total number of moles in the mixture. In the context of the given exercise, the mole fraction of methane (\(x_{\mathrm{CH}_{4}}\)) in a mixture with hydrogen bromide can showcase the methane's relative abundance.To elucidate the mole fraction from effusion rates, we calculate the square of the ratio obtained by Graham's Law. Since the combined mole fractions of methane and hydrogen bromide equal one, we can deduce the mole fraction of methane as\[\begin{equation}x_{\mathrm{CH}_{4}} = \frac{1}{1+(\sqrt{\frac{M_{\mathrm{HBr}}}{M_{\mathrm{CH}_{4}}}})^2}\end{equation}\]Through this expression, mole fraction becomes a simple calculation, interweaving the elements of effusion rates and molar mass, leading to a quantitative representation of the gas composition in the mixture.