Potassium bromide (KBr) has a density of 2.75 × 10 3 2.75\times10^3 kg/m 3 and the same crystal structure as NaCl. The mass of a potassium atom is 6.49 × 10 − 26 6.49\times10^{-26} kg, and the mass of a bromine atom is 1.33 × 10 − 25 1.33\times10^{-25} kg. Calculate the average spacing between adjacent atoms in a KBr crystal.

Hello, fellow physicist today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem in a KCL crystal determine the average spacing between potassium and chlorine atoms. If the density of KCL is 1.984 multiplied by 10 to the power of 3 kg per meters cubed. The individual mass of the potassium atom is 6.49 multiplied by 10 to the power of negative 26 kg. And the chlorine atom is 5.89 multiplied by 10 to the power of negative 26 kg. Awesome. So ultimately, our end goal, what we're trying to solve for for this problem is we're trying to figure out what the average spacing between this potassium and chlorine atoms will be if the specific density I of the KCL is 1.984 multiplied by 10 to the power of 3 kg per meters cubed. And also considering that we're provided the mass of both the potassium atom and the mass of the chlorine atom. Awesome. We're also given some multiple choice answers. Let's read them off to see what our final answer might be. And let's also quickly note that all of our units are in nanometers. So A is 0.624 B is 0.315 C is 0.123 and D is 0.973. Awesome. So first off, let us recall that the density is the average mass of the KCL divided by the volume. Let's also recall that the volume equation is V is equal to a cubed and for each of the K and C atoms, it will occupy a cube with a side length of a. Now, we must recall and use the equation to find the density of a compound molecule which states that row is equal to. So the density is equal to M divided by a cut where M is the average of the ionic masses and A is the average spacing. So now at this point, we need to solve for M, so let's find the average of the ionic masses. So M will be equal to. So we need to add the, the mass of the potassium atom and to the mass of the chlorine atom and then divide by two to take the average. So 6.49 multiplied by 10 to the power of negative 26 kg plus 5.89 multiplied by 10 to the power of negative 26 kg all divided by two. So when we perform that calculation in a calculator, we will get that M is equal to 6.19 multiplied by 10 to the power of negative 26 kg. Awesome. So at this stage, we need to rearrange the density equation to isolate and sulfur A which is our end goal. So a cubed will be equal to M divided by row. So the average of the ionic mass is divided by our density. So let's plug in those two variables. So we just found M and that's equal to 6.19 multiplied by 10 to the power of negative 26 kg. And we're also given the density in the prom itself as 1.984 multiplied by 10 to the power of three. And its units are kilograms per meters cubed. So when we plug that into a calculator, we will get 3.12 when we round the two decimal places multiplied by 10 to the power of negative 29 m cubed. But we have to remember that we're trying to get a all by itself. So we have to take the cube route of 3.12 multiplied by 10 of the power of negative 29 m cubed. So when we perform that cube route, we will get when we round the two decimal places again, 3.15 multiplied by 10 to the power of negative 10 m but as you can tell, we have a bit of a problem right now, our units are in meters. But as we can see, all of our multiple choice answers are in units of nanometers. So let's quickly convert our meters to nanometers. You could use dimensional analysis. But I'm just gonna speed things along and just tell you that it's, we just need to multiply it by 10 to the power of nine. And when we do, we will get 0.315 nanometers. And this is our final answer. Hooray, we did it and this answer corresponds to multiple choice answer letter B which is 0.315 nanometers. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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1. Intro to Physics Units

Solving Density Problems

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1. Intro to Physics Units

Solving Density Problems

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Problem 16a

Textbook Question

Potassium bromide (KBr) has a density of $2.75 times 10 cubed$ kg/m³ and the same crystal structure as NaCl. The mass of a potassium atom is $6.49 \times 10^{- 26}$ kg, and the mass of a bromine atom is $1.33 \times 10^{- 25}$ kg. Calculate the average spacing between adjacent atoms in a KBr crystal.

Verified step by step guidance

Step 1: Understand the problem. The goal is to calculate the average spacing between adjacent atoms in a KBr crystal. This involves using the density of the crystal and the masses of the potassium and bromine atoms to determine the volume occupied by each atom pair in the crystal structure.

Step 2: Calculate the mass of one formula unit of KBr. Since KBr consists of one potassium atom and one bromine atom, the mass of one formula unit is the sum of the masses of a potassium atom and a bromine atom: m_{KBr} = m_{K} + m_{Br}. Use MathML:

m_{KBr} = m_{K} + m_{Br}

Step 3: Use the density of KBr to calculate the volume occupied by one formula unit. The density formula is ρ = m/V, where ρ is the density, m is the mass, and V is the volume. Rearrange to find the volume per formula unit: V_{unit} = m_{KBr} / ρ. Use MathML:

V_{unit} = m_{KBr} / ρ

Step 4: Relate the volume of one formula unit to the average spacing between adjacent atoms. In a cubic crystal structure like NaCl (which KBr shares), the volume of one formula unit corresponds to the cube of the average spacing between adjacent atoms: V_{unit} = a^3, where a is the average spacing. Solve for a: a = (V_{unit})^(1/3). Use MathML:

a = {(V_{unit})}^{1 / 3}

Step 5: Substitute the values for m_{K}, m_{Br}, and ρ into the equations to calculate V_{unit} and then find a. Ensure units are consistent throughout the calculation (e.g., kg, m^3). This will yield the average spacing between adjacent atoms in the KBr crystal.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Crystal Structure

Crystal structure refers to the orderly and repeating arrangement of atoms within a crystalline material. In the case of potassium bromide (KBr), it shares the same face-centered cubic structure as sodium chloride (NaCl), where each unit cell contains a specific number of atoms arranged in a three-dimensional lattice. Understanding this structure is crucial for calculating properties like atomic spacing.

Density and Mass Relationship

Density is defined as mass per unit volume and is a critical property in determining how closely packed atoms are in a material. For KBr, the given density allows us to calculate the volume occupied by a certain mass of the substance, which can then be used to find the average spacing between atoms. The relationship between mass, volume, and density is fundamental in solid-state physics.

Atomic Spacing Calculation

Atomic spacing refers to the distance between adjacent atoms in a crystal lattice. To calculate this spacing in KBr, one must consider the total number of atoms in a unit cell and the volume of that cell derived from the density. This calculation is essential for understanding the physical properties of the material, such as its conductivity and reactivity.

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