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Chapter 10: Problem 53

Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of \(197 \mathrm{pm},\) calculate the density of solid calcium.

Short Answer

Expert verified
The density of solid calcium, with a cubic closest packed structure and an atomic radius of \(197 \, pm\), can be calculated as approximately \(1.219 \times 10^{7} \ g/m^3\).

Step by step solution

01

Determine the number of atoms in an FCC unit cell

In an FCC lattice, each corner atom is shared by eight adjacent unit cells and each face-centered atom is shared by two adjacent unit cells. There are 8 corner atoms and 6 face-centered atoms within each unit cell. Thus, the total number of atoms in an FCC unit cell can be calculated as: \[\frac{8}{8} (Corner \ atoms) + \frac{6}{2}(Face \ centered \ atoms) = 1 + 3 = 4 \ atoms\]
02

Mass of one unit cell

Now that we know there are 4 calcium atoms in each unit cell, we need to find the mass of these atoms within the unit cell. The atomic mass of calcium is 40.08 atomic mass units (amu) which is equivalent to 40.08 g/mol. Therefore, the mass of one calcium atom can be found using Avogadro's number, \(N_{A} = 6.022 \times 10^{23} \ mol^{-1}\). Mass of one calcium atom: \[\frac{40.08 \ g/mol}{6.022 \times 10^{23} \ atoms/mol} = 6.66 \times 10^{-23} \ g\] To find the mass of one unit cell, we multiply the mass of one atom by the number of atoms in the unit cell (which is 4): Mass of unit cell: \(4 \times 6.66 \times 10^{-23} \ g = 2.664 \times 10^{-22} \ g\)
03

Find the length of the unit cell

We are given the atomic radius of calcium is \(197 \, pm = 1.97 \times 10^{-10} \ m\). In an FCC structure, the diagonal of the face connecting the opposite face-centered atoms is equal to \(4 \times radius\). Using the Pythagorean theorem and considering that the face diagonal forms a right-angled triangle with the two sides and the length of the unit cell, or edge, as 'a', we get: \( (a)^{2} + (a)^{2} = (4 \times 1.97 \times 10^{-10})^{2} \) We can use this relation to find the length of the unit cell.
04

Calculate the edge length and volume of the unit cell

Rewriting the previous equation and solving for 'a' to find the edge length of the unit cell: \( 2a^{2} = (4 \times 1.97 \times 10^{-10})^{2} \) \( a^{2} = \frac{(4 \times 1.97 \times 10^{-10})^{2}}{2} \) \( a = \sqrt{\frac{(4 \times 1.97 \times 10^{-10})^{2}}{2}} = 2.794 \times 10^{-10} \ m \) Now, we can find the volume of the unit cell by cubing the edge length: Volume of the unit cell: \( V = a^3 = (2.794 \times 10^{-10})^3 = 2.186 \times 10^{-29} \ m^3 \)
05

Calculate the density of solid calcium

With the mass and volume of the unit cell, we can now calculate the density of solid calcium: Density: \( \rho = \frac{Mass}{Volume} = \frac{2.664 \times 10^{-22} \ g}{2.186 \times 10^{-29} \ m^3} = 1.219 \times 10^{7} \ g/m^3 \) Finally, we have the density of solid calcium in g/m³: \( \rho \approx 1.219 \times 10^{7} \ g/m^3 \)

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