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Chapter 9: Problem 58

Barium has a body-centered cubic structure. If the atomic radius of barium is \(222 \mathrm{pm},\) calculate the density of solid barium.

Short Answer

Expert verified
The density of solid Barium with a body-centered cubic structure and an atomic radius of 222 pm is approximately 3.62 g/cm³.

Step by step solution

01

Determine the unit cell edge length for a body-centered cubic structure

In a body-centered cubic (BCC) structure, the relationship between the atomic radius (r) and the unit cell edge length (a) is given by the formula: \(a = 4r / \sqrt{3}\) We are given that r (the atomic radius of Barium) is 222 pm.
02

Calculate the unit cell edge length

Plug in the value of r into the formula to find the unit cell edge length: \(a = \dfrac{4(222 \,\text{pm})}{\sqrt{3}} \approx 512.66 \,\text{pm}\)
03

Find the number of atoms in a body-centered cubic unit cell

In a BCC structure, there is one atom at each corner of the cube and one atom at the center of the cube: - 8 corner atoms × (1/8) contribution per atom = 1 atom - 1 center atom × 1 contribution per atom = 1 atom Total atoms in the unit cell: 1 (corner atoms) + 1 (center atom) = 2 atoms
04

Find the volume of the unit cell

Now that we have the unit cell edge length, we can find the volume (V) of the unit cell by using the formula for the volume of a cube: \(V = a^3 ≈ (512.66 \,\text{pm})^3 ≈ 1.35 × 10^8 \,\text{pm}^3\) Note that we'll need to convert the volume to \(cm^3\) in the next step, so let's do that now: \(V ≈ 1.35 × 10^8 \,\text{pm}^3 \cdot \left(\dfrac{1 \,\text{cm}}{10^7 \,\text{pm}}\right)^3 ≈ 1.35 × 10^{-23} \,\text{cm}^3\)
05

Calculate the density

To find the density (ρ), we need the atomic weight of Barium and Avogadro's number. The atomic weight of Barium is 137.33 g/mol. Avogadro's number (N_A) is \(6.022 \times 10^{23} \,atoms/mol\). We'll use the following formula: \(ρ = \dfrac{2\,atoms \cdot 137.33 \,\text{g/mol}}{6.022 \times 10^{23} \,atoms/mol \cdot 1.35 \times 10^{-23} \,\text{cm}^3}\)
06

Calculate the density of solid Barium

Plug the values into the formula and calculate the density: \(ρ ≈ \dfrac{2 \cdot 137.33 \,\text{g/mol}}{6.022 \times 10^{23}\,\text{atoms/mol} \cdot 1.35 \times 10^{-23} \,\text{cm}^3} ≈ 3.62 \,\text{g/cm}^3\) Therefore, the density of solid Barium is approximately 3.62 g/cm³.

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