Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A glass vessel fitted with a stopcock valve has a mass of 337.428 g when evacuated. When filled with Ar, it has a mass of 339.854 g. When evacuated and refilled with a mixture of Ne and Ar, under the same conditions of temperature and pressure, it has a mass of 339.076 g. What is the mole percent of Ne in the gas mixture?

Short Answer

Expert verified
The mole percent of Ne in the gas mixture is approximately 63.5%.

Step by step solution

01

Determine the mass of each gas

First, we need to determine the mass of the Ar gas and the Ne and Ar gas mixture separately. We will do this by subtracting the mass of the empty glass vessel from the mass of the vessel with each filling condition. Mass of Ar Gas = Mass of Glass Vessel with Ar - Mass of Empty Glass Vessel Mass of Ne and Ar Mixture = Mass of Glass Vessel with Mixture - Mass of Empty Glass Vessel
02

Calculate the moles of each gas using the molar mass

Next, we will calculate the moles of each gas using the molar mass of each element. The molar mass of Ar is 39.948 g/mol, and the molar mass of Ne is 20.180 g/mol. To find the amount (moles) of each gas, we can use the formula: Moles = Mass/Molar Mass
03

Determine the mole percent of Ne in the gas mixture

We now need to find the mole percent of Ne in the gas mixture. Let x be the mole fraction of Ne in the mixture. Then the mole fraction of Ar would be (1-x). The total mass of the mixture can be written as: Mass of Ne and Ar Mixture = (x * moles of Ne) + [(1-x) * moles of Ar] We can solve for x and then find the mole percent of Ne in the mixture by multiplying x by 100. Mole percent of Ne = x * 100
04

Calculate the moles of each gas using the mass of each gas

Calculate the mass and moles of Ar gas and Ne, Ar gas mixture using the values we have: Mass of Ar Gas = 339.854 g - 337.428 g = 2.426 g Mass of Ne and Ar Mixture = 339.076 g - 337.428 g = 1.648 g Moles of Ar Gas = 2.426 g / 39.948 g/mol = 0.0607 mol
05

Create an equation to find the mole percent of Ne in the mixture and solve for x

We can write an equation with x as the mole fraction of Ne in the mixture: 1.648 g = x * (0.0607 mol of Ne * 20.180 g/mol) + (1-x) * (0.0607 mol * 39.948 g/mol) Now, we will solve this equation to find the mole fraction x of Ne in the mixture. 1.648 g = x * (1.225 g) + (1-x) * (2.426 g) Solving for x: x ≈ 0.635 Thus, the mole percent of Ne in the mixture is: Mole percent of Ne = 0.635 * 100 ≈ 63.5%

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Stoichiometry
Stoichiometry is a section of chemistry that involves the calculation of the quantities of reactants and products involved in a chemical reaction. It is a powerful tool that allows chemists to predict the outcomes of reactions, design experiments, and make conversions between different chemical entities such as moles, atoms, or molecules.

One of the key concepts in stoichiometry is the mole, which is a standard unit of measurement that represents a large quantity of very small entities such as atoms, molecules, or other particles. One mole is equivalent to Avogadro's number, which is approximately 6.022 \( \times \) 10^23 entities.

In the context of our problem, stoichiometry is used to calculate the mole percent of neon (Ne) in a mixture with argon (Ar). To do so, we must first determine the molar masses of the gases involved and use them to calculate the number of moles from the mass of gas in the vessel. With the number of moles of each gas, we can then find the composition of the gas mixture – a prime example of stoichiometry in action.
Molar Mass
The molar mass of a substance is the mass of one mole of that substance. It is a critical concept in chemistry, particularly in stoichiometry, because it allows for the conversion between moles and grams. The molar mass is usually reported in grams per mole (g/mol), and it is numerically equivalent to the average mass of one molecule of the substance, expressed in unified atomic mass units (u or amu).

For instance, the molar mass of argon (Ar) is 39.948 g/mol, while that of neon (Ne) is 20.180 g/mol. These values are pivotal in the exercise at hand because they help to calculate the number of moles of each gas in the mixture by using the formula \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \). Through this formula, the mass of the gas mixture obtained from the vessel can be used to determine the moles of each constituent gas.
Gas Mixture Composition
Understanding the composition of a gas mixture is fundamental in various fields, including chemistry, environmental science, and industrial processes. It involves determining the percentage by volume or by mole of each gas within a mixture. In practical scenarios, this would allow scientists and engineers to control reactions, optimize processes, and monitor environmental conditions.

The mole percent is a way of expressing the concentration of a component in a mixture. It is calculated by taking the number of moles of one component and dividing it by the total number of moles of all components in the mixture, then multiplying by 100 to get a percentage.

In the given exercise, once the masses of the individual gases are known, the students can use molar masses to find the moles of Ne and Ar. Then, by creating the equation mentioned in the solution, which involves mass and mole calculations, they can solve for the mole fraction of Ne in the mixture. Multiplying the mole fraction by 100 yields the mole percent of Ne, giving insight into the composition of the gas mixture.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Indicate which of the following statements regarding the kinetic-molecular theory of gases are correct. (a) The average kinetic energy of a collection of gas molecules at a given temperature is proportional to \(m^{1 / 2}\) . (b) The gas molecules are assumed to exert no forces on each other. (c) All the molecules of a gas at a given temperature have the same kinetic energy. (d) The volume of the gas molecules is negligible in comparison to the total volume in which the gas is contained. (e) All gas molecules move with the same speed if they are at the same temperature.

(a) Are you more likely to see the density of a gas reported in \(\mathrm{g} / \mathrm{mL}, \mathrm{g} / \mathrm{L},\) or \(\mathrm{kg} / \mathrm{cm}^{3} ?(\mathbf{b})\) Which units are appropriate for expressing atmospheric pressures, \(\mathrm{N}, \mathrm{Pa},\) atm, kg/m \(^{2} ?\) (c) Which is most likely to be a gas at room temperature and ordinary atmospheric pressure, \(\mathrm{F}_{2}, \mathrm{Br}_{2}, \mathrm{K}_{2} \mathrm{O} .\)

The planet Jupiter has a surface temperature of 140 \(\mathrm{K}\) and a mass 318 times that of Earth. Mercury (the planet) has a surface temperature between 600 \(\mathrm{K}\) and 700 \(\mathrm{K}\) and a mass 0.05 times that of Earth. On which planet is the atmosphere more likely to obey the ideal-gas law? Explain.

A set of bookshelves rests on a hard floor surface on four legs, each having a cross-sectional dimension of \(3.0 \times 4.1 \mathrm{cm}\) in contact with the floor. The total mass of the shelves plus the books stacked on them is 262 kg. Calculate the pressure in pascals exerted by the shelf footings on the surface.

Gaseous iodine pentafluoride,\(\mathrm{IF}_{5}\) can be prepared by the reaction of solid iodine and gaseous fluorine: $$\mathrm{I}_{2}(s)+5 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{5}(g)$$ A 5.00 -L flask containing 10.0 g of \(\mathrm{I}_{2}\) is charged with 10.0 \(\mathrm{g}\) of \(\mathrm{F}_{2},\) and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is \(125^{\circ} \mathrm{C}\) . (a) What is the partial pressure of \(\mathrm{IF}_{5}\) in the flask? (b) What is the mole fraction of \(\mathrm{IF}_{5}\) in the flask (c) Draw the Lewis structure of \(\mathrm{IF}_{5}\). (d) What is the total mass of reactants and products in the flask?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free