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Chapter 5: Problem 150

An equimolar mixture of Ne and Xe is accidentally placed in a container that has a tiny leak. After a short while, a very small proportion of the mixture has escaped. What is the mole fraction of Ne in the effusing gas?

Short Answer

Expert verified
The mole fraction of Ne in the effusing gas is approximately 0.717.

Step by step solution

01

- Understand Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion of gases is inversely proportional to the square root of their molar masses. Mathematically, it can be expressed as \( \frac{r_1}{r_2} = \frac{\text{MW}_2}{\text{MW}_1} \), where \( r_1 \) and \( r_2 \) are the rates of effusion of gases 1 and 2, respectively, and \( \text{MW}_1 \) and \( \text{MW}_2 \) are their respective molar masses.
02

- Identify the Gases and Their Molar Masses

For Ne (Neon), the molar mass \( \text{MW}_{\text{Ne}} \) is approximately 20.18 g/mol. For Xe (Xenon), the molar mass \( \text{MW}_{\text{Xe}} \) is approximately 131.29 g/mol.
03

- Apply Graham's Law to Calculate Relative Effusion Rates

Using Graham's Law \( \frac{r_{\text{Ne}}}{r_{\text{Xe}}} = \frac{\text{MW}_{\text{Xe}}}{\text{MW}_{\text{Ne}}} = \frac{131.29}{20.18} \). Simplifying this gives \( \frac{r_{\text{Ne}}}{r_{\text{Xe}}} \) approximately equal to 2.54.
04

- Find the Mole Fraction of Ne

Let \( r_{\text{total}} \) be the total rate of effusion for the mixture. The mole fraction of Ne in the effusing gas is then \( \frac{r_{\text{Ne}}}{r_{\text{Ne}} + r_{\text{Xe}}} \). Substituting \( r_{\text{Ne}} \) approximately equal to 2.54 \( r_{\text{Xe}} \), we get \( \frac{2.54 r_{\text{Xe}}}{2.54 r_{\text{Xe}} + r_{\text{Xe}}} = \frac{2.54}{3.54} \).
05

- Simplify the Fraction

Simplifying \( \frac{2.54}{3.54} \) gives a value of approximately 0.717. Therefore, the mole fraction of Ne in the effusing gas is approximately 0.717.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Effusion Rates
Graham's Law of Effusion helps us understand how gases escape through a small opening. The law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This means lighter gases effuse faster than heavier gases. The rate of effusion for two gases, given as \(\frac{r_1}{r_2} = \frac{\text{MW}_2}{\text{MW}_1}\), tells us how quickly one gas effuses compared to another. By knowing the molar masses of Neon (Ne) and Xenon (Xe), we can calculate their relative effusion rates.
Mole Fraction
The mole fraction represents the ratio of moles of one component to the total moles in a mixture. It is a way to express the concentration of a particular gas within a mixture. When a gas mixture effuses, gases with lower molar masses effuse more quickly, altering the composition over time. The mole fraction of Ne in the effusing gas can be found using the effusion rates calculated from Graham's Law. The formula for mole fraction of Ne is: \(\frac{r_{\text{Ne}}}{r_{\text{Ne}} + r_{\text{Xe}}}\). This gives the proportion of Ne gas escaping compared to the total gas escaping.
Molar Mass
Molar mass is the weight of one mole of a substance, expressed in grams per mole (g/mol). It is a critical value used in various chemical calculations, including effusion rates. For example, Ne has a molar mass of approximately 20.18 g/mol, while Xe has a molar mass of about 131.29 g/mol. These differences in molar masses explain why Ne, being lighter, effuses faster than Xe. Comparing molar masses allows us to apply Graham's Law effectively to determine how the gas mixture's composition changes over time.
Gas Mixtures
A gas mixture consists of two or more different gases combined together. In our example, Ne and Xe form an equimolar mixture, meaning they initially have equal quantities in moles. When this mixture experiences effusion, the rates at which Ne and Xe escape differ due to their molar masses. Over time, the lighter gas (Ne) will become more prevalent in the effusing gas compared to the heavier gas (Xe). Understanding how different gases in a mixture behave when they effuse is crucial in predicting changes in composition and behaviors of gas mixtures in various conditions.

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Most popular questions from this chapter

5.53 How many grams of phosphorus react with \(35.5 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at STP to form tetraphosphorus decoxide? $$\mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)$$

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