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The rate of effusion is an essential concept when it comes to understanding how gases pass through small openings. Effusion occurs when gas molecules escape through a tiny hole into an evacuated space. The rate at which this happens depends on several factors, such as the gas's molar mass and the conditions of pressure and temperature.
According to Graham's Law of Effusion, the rate of effusion of gas is inversely proportional to the square root of its molar mass. In formula terms, it can be represented as:

• For gas A: \( Rate_A = \frac{1}{\sqrt{M_A}} \)
• For gas B: \( Rate_B = \frac{1}{\sqrt{M_B}} \)
• Ratio of their effusion rates: \( \frac{Rate_A}{Rate_B} = \frac{\sqrt{M_B}}{\sqrt{M_A}} \)

In our given problem, helium (He) and a mixture of carbon monoxide (CO) and carbon dioxide (CO2) are involved. Helium effuses at a known rate, and the task is to compare this with the effusion rate of the mixture.
The formula aids in calculating how much larger or smaller the volume of one gas is compared to the other. This insight is crucial for determining the proportion of each gas in a mixture.

Molar mass is a critical component in understanding Graham's Law of Effusion since the rate of effusion is strongly influenced by it. The molar mass of a substance refers to the mass of one mole of its chemical entities (atoms, molecules, etc.), typically expressed in grams per mole (g/mol).
In the context of this exercise, we examine helium, carbon monoxide, and carbon dioxide. Their molar masses are respectively:

• Helium (He): \(4.00\text{ g/mol} \)
• Carbon monoxide (CO): \(28.01\text{ g/mol} \)
• Carbon dioxide (CO2): \(44.01\text{ g/mol} \)

When applying Graham's Law, these values help determine how fast each gas effuses through the small hole. Because helium is much lighter than CO and CO\(_2\), it tends to effuse more quickly.
Understanding molar masses helps set up the equation necessary to solve for the volume percentages of gas mixtures, enabling us to substitute them appropriately into the ratio formula derived from effusion rates.

Volume percent composition is a useful metric to express concentrations of different components in a gas mixture, typically given in percentages. It indicates how much of the total volume each gas occupies in a mixture.
In problems involving effusion and gas mixtures, Graham's Law allows us to determine these percentages by comparing effusion rates and molar masses. An equation can be set up such that:
\[ \frac{1}{\sqrt{M}} = \text{effusion ratio} \]
Given that \(x\) is the volume percent of CO, the remaining percentage \((1-x)\) represents the volume percent of CO2. The formula derived from the effusion rates enables us to solve for \(x\):

• Equation: \( \frac{1}{\sqrt{(x \cdot 28.01) + ((1-x) \cdot 44.01)}} = 2.97 \)

Evaluating this, we find that CO contributes to about 61.6% of the mixture, demonstrating how Graham's Law provides insights into both effusion and the composition of gas mixtures in terms of volume.