Question

A \(455 \mathrm{cm}^{3}\) vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is \(2.245 \mathrm{g}\) at \(25.0^{\circ} \mathrm{C}\) and the pressure is \(760 .\) Torr, calculate the mole fraction of Xe in the mixture.

Short answer

The mole fraction of Xe in the mixture is approximately 0.7388 or 73.88%.

Step-by-step solution

Convert the given data into appropriate units

First, let's convert the given data into appropriate units. We have the following data: - Volume: 455 cm³ -> 0.455 L (1 L = 1000 cm³) - Mass of the gas mixture: 2.245 g - Temperature: 25.0°C -> 298.15 K (K = °C + 273.15) - Pressure: 760 Torr -> 1 atm (1 atm = 760 Torr) Now we have: - Volume: 0.455 L - Mass of the gas mixture: 2.245 g - Temperature: 298.15 K - Pressure: 1 atm

Calculate the moles of the gas mixture using the Ideal Gas Law

The Ideal Gas Law can be expressed as: PV = nRT Where: P = pressure (atm) V = volume (L) n = number of moles R = ideal gas constant (0.08206 L⋅atm/mol⋅K) T = temperature (K) We can now solve for the total number of moles (n) in the mixture using the given data: \(1 \cdot 0.455 = n \cdot 0.08206 \cdot 298.15\) n = \(\frac{0.455}{0.08206 \cdot 298.15}\) n ≈ 0.01842 moles

Calculate the weighted average molar mass of the gas mixture

We can calculate the weighted average molar mass using the total mass of the mixture and the total moles: Weighted Average Molar Mass = \(\frac{Total Mass}{Total Moles}\) Weighted Average Molar Mass = \(\frac{2.245 \mathrm{g}}{0.01842 \mathrm{moles}}\) Weighted Average Molar Mass ≈ 121.863 g/mol

Set up the equations for the mole fractions of Ar and Xe

Let x be the mole fraction of Xe, then the mole fraction of Ar is (1 - x). The moles of each gas can be expressed as: Moles of Ar = (1 - x) * 0.01842 Moles of Xe = x * 0.01842 The molar masses of Ar and Xe are: - Ar: 39.948 g/mol - Xe: 131.293 g/mol Using the weighted average molar mass of the gas mixture, we can write the equation: Weighted Average Molar Mass = (1 - x)(39.948) + x(131.293)

Solve for the mole fraction of Xe (x)

Solve the weighted average molar mass equation for x: 121.863 = (1 - x)(39.948) + x(131.293) Rearrange and solve for x: x = \(\frac{121.863 - 39.948}{131.293 - 39.948}\) x ≈ 0.7388

Calculate the mole fraction of Xe

The calculated mole fraction of Xe is approximately 0.7388 or 73.88%.

Key concepts

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture. It's especially useful in chemistry for comparing gases in a mixture. The mole fraction of a substance is the ratio of the moles of that substance to the total moles of all substances present. This can be expressed mathematically as:

• Mole Fraction (X) = \( \frac{n_i}{n_{total}} \)

where \( n_i \) is the moles of the gas in interest, and \( n_{total} \) is the total moles of all gases in the mixture.
In this exercise, we calculated the mole fraction of Xenon (Xe) in a mixture with Argon (Ar). If the mole fraction of Xe is known, it is easy to evaluate the mole fraction of Ar, as the sum of mole fractions in a mixture always equals 1. For example, if the mole fraction of Xe is approximately 0.7388, the mole fraction of Ar would be 1 - 0.7388 = 0.2612.

Weighted Average Molar Mass

The weighted average molar mass of a gas mixture is important for understanding the collective behavior of the gases involved. It's calculated using the formula:

• Weighted Average Molar Mass = \( \frac{Total Mass}{Total Moles} \)

This tells us the average molar mass based on the mass and moles of the composite gases. For this particular problem, by using the total mass of the mixture (2.245 g) and the total moles derived from the Ideal Gas Law (approximately 0.01842 moles), we calculated the weighted average molar mass to be about 121.863 g/mol.
This can then be used to set up equations involving the mole fractions and molar masses of individual gases to solve for unknown quantities like the mole fraction.

Gas Mixtures

Gas mixtures involve more than one gas and their properties are influenced by the partial pressures and molar masses of each component. Each gas in a mixture doesn't interact and behaves independently due to the principles of the Ideal Gas Law (PV = nRT).
For gas mixtures, the concept of partial pressure becomes relevant. The partial pressure of a gas is the hypothetical pressure the gas would exert if it alone occupied the entire volume. The total pressure of the gas mixture is the sum of all partial pressures of the individual gases.
In our exercise, Argon and Xenon coexist in a vessel. Both gases contribute to the total pressure of 1 atm, but our main interest was to find how much each gas contributes to the moal results. We calculated the moles and mole fractions to understand their individual roles in the mixture.

Pressure Conversion

Pressure conversion is often necessary when dealing with gas laws and varying units of measurement. Pressure can be measured in several different units, such as torr, atm, or Pa, and it is crucial to use consistent units when applying formulas.
In our problem, we began with a pressure of 760 Torr, which is a common measurement in chemistry. To use the Ideal Gas Law effectively, which utilizes atmospheres (atm), we converted this pressure to 1 atm using the conversion factor: 1 atm = 760 Torr.
Being comfortable with these conversions is vital and allows you to seamlessly integrate and apply formulas under standardized conditions. By doing so, solving the gas law problems becomes more intuitive and accurate.