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Chapter 10: Problem 51

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 \(197 pm\) is approximately \(1.55 g/cm^{3}\).

Step by step solution

01

1. Calculate the volume of one unit cell

In a cubic closest packed structure, the unit cell has atoms at all corners and the centers of all faces. The edge length (a) of the cubic unit cell can be calculated by knowing the atomic radius (r) from the given information and using the fact that in a ccp structure, the diagonal of the face of the unit cell consists of 4 atomic radii. Thus, \(a = 2\sqrt{2}r\). Given the atomic radius of calcium (r) as \(197 pm\), the edge length of the unit cell (a) can be calculated as follows: \(a = 2\sqrt{2}(197 pm) \approx 556 pm\) Now, to calculate the volume of the unit cell, we can simply cube the edge length (V = a³): \(V = (556 pm)^3 \approx 1.72 * 10^{8} pm^{3}\)
02

2. Configure the mass of one unit cell

Recall that in a ccp structure, there are 4 atoms in the unit cell, so we need to multiply the number of atoms by the molar mass of calcium (40.08 g/mol) to get the total mass per unit cell. Then, convert the result to picograms. Let's first find the mass of one calcium atom by dividing the molar mass by Avogadro's number (\(N_A = 6.022 * 10^{23} \mathrm{atoms/mol}\)): Mass of one atom: \(\frac{40.08\mathrm{g/mol}}{6.022*10^{23}\mathrm{atoms/mol}} \approx 6.66 * 10^{-23} g = 6.66 * 10^{5} pg\) Now, multiply this value by the 4 atoms per unit cell: Mass per unit cell: \(4 * (6.66 * 10^5 pg) = 2.66 * 10^6 pg\)
03

3. Calculate the density

Now that we have the mass and volume of one unit cell, we can calculate the density using the formula for density (\(\rho = \frac{mass}{volume}\)): \(\rho = \frac{2.66 * 10^6 pg}{1.72 * 10^8 pm^{3}} \approx 1.55 g/cm^{3}\) Thus, the density of solid calcium is approximately \(1.55 g/cm^{3}\).

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Most popular questions from this chapter

Some ionic compounds contain a mixture of different charged cations. For example, wüstite is an oxide that contains both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) cations and has a formula of \(\mathrm{Fe}_{0.980} \mathrm{O}_{1.00-}\). Calculate the fraction of iron ions present as \(\mathrm{Fe}^{3+}\). What fraction of the sites normally occupied by \(\mathrm{Fe}^{2+}\) must be vacant in this solid?

What is the formula for the compound that crystallizes with a cubic closest packed array of sulfur ions, and that contains zinc ions in \(\frac{1}{8}\) of the tetrahedral holes and aluminum ions in \(\frac{1}{2}\) of the octahedral holes?

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The molar heat of fusion of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(9.92 \mathrm{~kJ} / \mathrm{mol}\). Its molar heat of vaporization is \(30.7 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to melt \(8.25 \mathrm{~g}\) benzene at its normal melting point. Calculate the heat required to vaporize \(8.25 \mathrm{~g}\) benzene at its normal boiling point. Why is the heat of vaporization more than three times the heat of fusion?

How could you tell experimentally if \(\mathrm{TiO}_{2}\) is an ionic solid or a network solid?
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