Neon Volume In Gas Mixture: A Chemistry Problem

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Hey there, chemistry enthusiasts! Let's dive into a cool problem involving gas mixtures, specifically nitrogen (N₂) and neon (Ne). We're going to figure out the volume fraction of neon when we're given some nifty information about the mixture's density. Sounds fun, right? Buckle up, because we're about to explore the fascinating world of gas behavior!

Understanding the Problem: Density and Volume Fractions

Okay, so the problem tells us that the density of a gas mixture of nitrogen and neon relative to hydrogen is 13. What does that even mean, you ask? Well, it means that if you compare the mass of a certain volume of this gas mixture to the mass of the same volume of hydrogen, the gas mixture is 13 times heavier. This concept is super important in understanding how gases behave and interact.

First, let's break down the main goal: to determine the volume fraction of neon. The volume fraction essentially tells us what proportion of the total volume of the gas mixture is occupied by neon. It's like asking, "If we have a balloon filled with this gas mixture, what percentage of the balloon is filled with neon?" To solve this, we'll need to use some clever chemistry tricks involving molar masses and the ideal gas law (even though we won't explicitly use the ideal gas law formula here).

To begin, let's denote the volume fraction of neon as 'x'. This means that the volume fraction of nitrogen will be (1 - x), since the total volume fraction of all components in a mixture must add up to 1. This is your first crucial step in solving these problems! Now, let's look at the density of the mixture. Density, in this case, is relative to hydrogen (H₂). This means we're comparing the mixture's density to hydrogen's density under the same conditions (temperature and pressure). Remember that the relative density is 13. This information is key to the solution.

Now, let's relate this relative density to the molar masses of the gases involved. The molar mass of a substance is the mass of one mole of that substance (measured in grams per mole). For hydrogen (H₂), the molar mass is approximately 2 g/mol. For nitrogen (N₂), it's about 28 g/mol, and for neon (Ne), it's about 20 g/mol. We'll use these values to establish a connection between the volume fractions and the relative density.

Setting up the Equation: The Key to the Solution

Alright, let's get into the nitty-gritty and create an equation that connects all the pieces. We know the relative density of the gas mixture compared to hydrogen is 13. This relative density can be expressed as the ratio of the molar mass of the mixture to the molar mass of hydrogen. The molar mass of the mixture can, in turn, be expressed as a weighted average of the molar masses of its components, weighted by their volume fractions. Trust me, it’s not as scary as it sounds!

So, let’s consider the molar mass of the mixture (M_mixture). It will be a combination of nitrogen and neon. Using the volume fractions, we can write:

  • M_mixture = x * M_Ne + (1 - x) * M_N₂*

where:

  • x is the volume fraction of neon,
  • M_Ne is the molar mass of neon (20 g/mol),
  • M_N₂ is the molar mass of nitrogen (28 g/mol).

Next, the relative density (D_relative) is given as 13. This relative density is defined as the ratio of the molar mass of the mixture (M_mixture) to the molar mass of hydrogen (M_H₂), which is 2 g/mol. Therefore:

  • D_relative = M_mixture / M_H₂*

Since D_relative = 13 and M_H₂ = 2 g/mol, we can calculate the molar mass of the mixture:

  • M_mixture = D_relative * M_H₂ = 13 * 2 = 26 g/mol*

Now we have two equations:

  1. M_mixture = x * 20 + (1 - x) * 28
  2. M_mixture = 26

We can substitute equation (2) into equation (1):

26 = x * 20 + (1 - x) * 28

Now, all that's left is to solve this equation for 'x'. It's all about algebra from here on out. It looks a bit complex, but you guys have got this!

Solving for the Volume Fraction of Neon

Let’s roll up our sleeves and solve the equation we just created. We've got: 26 = x * 20 + (1 - x) * 28. Our goal is to isolate 'x' to find the volume fraction of neon.

First, let's expand the equation:

26 = 20x + 28 - 28x

Now, combine like terms:

26 = 28 - 8x

Next, let’s rearrange the equation to isolate the term with 'x'. Subtract 28 from both sides:

26 - 28 = -8x

This simplifies to:

-2 = -8x

Finally, divide both sides by -8 to solve for 'x':

x = -2 / -8

x = 1/4

So, the volume fraction of neon (x) in the mixture is 1/4. This means that 25% of the mixture's volume is occupied by neon, and the remaining 75% is nitrogen.

Now, look back at the original problem. We were given four options:

A) 1/3 B) 1/4 C) 3/4 D) 2/3

The correct answer is B) 1/4. Congrats, we solved it! This is the kind of problem that, with a little bit of practice, becomes much easier.

Tips and Tricks: Mastering Gas Mixture Problems

Want to become a gas mixture guru? Here are some extra tips and tricks to help you ace these types of problems:

  • Memorize Molar Masses: Knowing the molar masses of common gases like nitrogen, oxygen, hydrogen, and neon is super helpful. It will save you time during exams and practice sessions.

  • Understand Relative Density: Make sure you grasp the concept of relative density. Remember it is always relative to a reference gas (often hydrogen or air).

  • Practice, Practice, Practice: The more problems you solve, the better you'll become at recognizing patterns and applying the correct formulas. Do a lot of exercises!

  • Visualize the Mixture: Imagine the gases mixing together in a container. This mental picture can help you understand the relationship between volume fractions and partial pressures.

  • Use the Ideal Gas Law (Sometimes): While we didn't explicitly use it here, the Ideal Gas Law (PV = nRT) can be useful in many gas mixture problems. Be prepared to use it if the problem involves temperature, pressure, or the number of moles.

  • Check Your Units: Always double-check your units to make sure they are consistent throughout your calculations. This is a common source of errors. For example, make sure you use consistent units for volume (liters, cubic meters, etc.), pressure (atmospheres, Pascals, etc.), and temperature (Kelvin).

Final Thoughts: You've Got This!

Solving gas mixture problems, like this one, is all about understanding the relationships between volume fractions, molar masses, and density. By breaking down the problem step by step, setting up the right equations, and practicing consistently, you can master these types of problems!

Keep exploring the fascinating world of chemistry, and never stop asking questions. Keep in mind that chemistry can be challenging, but it's also incredibly rewarding. Keep up the awesome work, and keep those curious minds engaged!