(a) You are given a cube of silver metal that measures 1.000 \(\mathrm{cm}\) on each edge. The density of silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3} .\) How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that \(74 \%\) of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom. (c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

First, we need to calculate the mass of the silver cube using its density and volume. We are given the density of silver (10.5 g/cm³) and the side length of the cube (1.000 cm). The volume of the cube is given by the formula: Volume = side³ So, the volume of the cube = \(1.000^3 \mathrm{cm}^3 = 1.000 \mathrm{cm}^3\). Now, using the formula: Mass = Volume × Density, We can calculate the mass of silver in the cube: Mass = \(1.000 \mathrm{cm}^3 × 10.5 \frac{\mathrm{g}}{\mathrm{cm}^3} = 10.5 \mathrm{g}\)

Now we have the mass of silver. We can use the molar mass of silver to convert this to the number of silver atoms. Silver's molar mass is approximately \(107.87 \frac{\mathrm{g}}{\mathrm{mol}}\). First, we calculate the number of moles of silver in the cube: Moles of silver = \(\frac{\text{Mass of silver}}{\text{Molar mass of silver}}=\dfrac{10.5\,\mathrm{g}}{107.87\,\mathrm{g}/\mathrm{mol}}=0.0973\,\mathrm{mol}\) Now, using Avogadro's number (\(6.022 × 10^{23} \text{atoms/mol}\)), we can find the number of silver atoms in the cube: Number of silver atoms = Moles of silver × Avogadro's number = \(0.0973\,\mathrm{mol} × 6.022 × 10^{23} \text{atoms/mol} ≈ 5.86 × 10^{22} \text{atoms}\)

We are given that only 74% of the cube's volume is occupied by silver atoms. Knowing the volume of the cube and this percentage, we can find the total volume occupied by silver atoms: Total volume of silver atoms = % of volume filled × Volume of the cube = \(0.74 × 1.000 \mathrm{cm}^3 = 0.740 \mathrm{cm}^3\)

We now have both the total volume occupied by silver atoms and the total number of silver atoms in that volume. We can use this to find the volume of a single silver atom. Volume of a single silver atom = \(\frac{\text{Total volume of silver atoms}}{\text{Number of silver atoms}} = \dfrac{0.740 \,\mathrm{cm}^3}{5.86 × 10^{22} \,\text{atoms}} ≈ 1.26 × 10^{-23} \,\mathrm{cm}^3\)

We will now use the formula for the volume of a sphere to calculate the radius of a silver atom. The formula for the volume of a sphere is: \(V = \frac{4}{3}\pi r^3\) Where V is the volume and r is the radius. We can rearrange the formula to solve for the radius and then insert the volume of a single silver atom: \(r = \sqrt[3]{\dfrac{3V}{4\pi}} = \sqrt[3]{\dfrac{3 × 1.26 × 10^{-23}\, \mathrm{cm}^3}{4\pi}} ≈ 1.44 × 10^{-8}\, \mathrm{cm}\) Finally, we can convert the radius from centimeters to angstroms by using the conversion factor \(1\, \mathrm{cm} = 10^{8}\, \mathrm{angstrom}\). Radius of a silver atom = \(1.44 × 10^{-8}\, \mathrm{cm} × 10^{8}\, \mathrm{angstrom/cm} ≈ 1.44 \mathrm{angstrom}\) So, the radius of a silver atom is approximately 1.44 angstroms.