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Chapter 10: Problem 60

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 3.41 g/cm³.

Step by step solution

01

Determine the edge length of a unit cell from the atomic radius

In a body-centered cubic structure (BCC), the unit cell contains one atom at each of the eight corners, and one atom in the center. If we draw a straight line connecting the centers of opposite corner atoms, we can form a body diagonal. The length of this diagonal is equal to 4 times the atomic radius. We can use the Pythagorean theorem in 3D to find the edge length 'a' of the unit cell, given by: \(a \sqrt{3} = 4R\) Where 'a' is the edge length, 'R' is the atomic radius, and the \(\sqrt{3}\) factor comes from the geometry of the cubic unit cell.
02

Calculate the edge length of the BCC unit cell

We are given the atomic radius of barium, R = 222 pm. Now we can find the edge length by plugging the known atomic radius into the equation from Step 1: \(a \sqrt{3} = 4 \times 222 pm\) Solving for 'a', we get: \(a = \frac{4 \times 222 pm}{\sqrt{3}} \approx 511.27 pm\)
03

Calculate the volume of the BCC unit cell

We can find the volume of the BCC unit cell by cubing the edge length 'a': Volume = \(a^3\) Volume ≈ \((511.27 pm)^3 = 1.339 \times 10^8 pm^3\)
04

Determine the number of atoms per unit cell in a BCC structure

In a body-centered cubic structure, there are 8 corner atoms (each contributing 1/8 to the unit cell) and 1 center atom (contributing fully to the unit cell), for a total of: Total atoms per unit cell = \(8 \times \frac{1}{8} + 1 = 2\)
05

Calculate the mass of the unit cell

To find the mass of the unit cell, we know there are 2 Barium atoms per unit cell. The molar mass of Barium is 137.327 g/mol. Using the fact that there are \(6.022 \times 10^{23}\) atoms/mol (Avogadro's number), we can find the mass of 2 Barium atoms: Mass of 1 atom of Barium = \(\frac{137.327 g/mol}{6.022 \times 10^{23} atoms/mol} \approx 2.281 \times 10^{-22} g\) Mass of unit cell = \(2 \times 2.281 \times 10^{-22} g \approx 4.562 \times 10^{-22} g\)
06

Calculate the density of solid barium

Finally, we can find the density of solid Barium by dividing the mass of the unit cell by the volume of the unit cell and converting pm^3 to cm^3: Density = \(\frac{mass}{volume} = \frac{4.562 \times 10^{-22} g}{1.339 \times 10^8 pm^3}\) To convert pm^3 to cm^3, use the conversion factor: \(1 cm^3 = 10^{30} pm^3\) Density = \(\frac{4.562 \times 10^{-22} g}{1.339 \times 10^8 pm^3} \times \frac{10^{30} pm^3}{1 cm^3} = 3.41 g/cm^3\) The density of solid barium is approximately 3.41 g/cm³.

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