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Chapter 9: Problem 124

Aluminum has an atomic radius of \(143 \mathrm{pm}\) and forms a solid with a cubic closest packed structure. Calculate the density of solid aluminum in \(\mathrm{g} / \mathrm{cm}^{3} .\)

Short Answer

Expert verified
The density of solid aluminum in a cubic closest packed structure is approximately 2.71 g/cm³.

Step by step solution

01

Calculate the unit cell volume

Using the cubic closest packed structure, the relationship between the atomic radius (r) and the cubic edge length (a) is given by: \[a = 2\sqrt{2} r\] Given the atomic radius of aluminum, \(r = 143\mathrm{pm}\), let's find the cubic edge length a: \[a = 2\sqrt{2} (143\mathrm{pm}) \approx 404.12\mathrm{pm}\] Now, we will convert pm (picometers) to cm (centimeters) and calculate the unit cell volume (V): \[a = 404.12\mathrm{pm} \times \frac{1\mathrm{cm}}{10^{12}\mathrm{pm}} = 4.0412\times10^{-8}\mathrm{ cm}\] \[V = a^3 = (4.0412\times10^{-8}\mathrm{cm})^3 = 6.587\times10^{-23}\mathrm{cm}^3\]
02

Calculate the number of aluminum atoms in the unit cell

The cubic closest packed structure has 4 atoms per unit cell. Therefore, the number of aluminum atoms in the unit cell is 4.
03

Calculate the mass of the aluminum atoms in the unit cell

First, we will determine the molar mass of aluminum. The molar mass of aluminum is 26.98 g/mol. Next, we will use the number of aluminum atoms in the unit cell and Avogadro's Number (\(6.022\times10^{23}\) atoms/mol) to determine the mass of aluminum atoms in the unit cell: \[\text{Mass of Al atoms} = \frac{4 \text{ atoms} \times 26.98\, \text{g/mol}}{(6.022\times10^{23}\text{ atoms/mol})} \approx 1.791\times10^{-22}\,\text{g}\]
04

Calculate the density of solid aluminum

Finally, we will calculate the density of solid aluminum using the mass and volume of the unit cell: \[\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{1.791\times10^{-22}\,\text{g}}{6.587\times10^{-23}\,\text{cm}^3} \approx 2.71\, \text{g/cm}^3\] So, the density of solid aluminum is approximately 2.71 g/cm³.

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