Question

Silver has a density of \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\). If \(1.00 \mathrm{~mol}\) of silver were shaped into a cube, (a) what would be the volume of the cube? (b) what would be the length of one side of the cube?

Short answer

The volume of the cube is approximately \(10.2733 \, \mathrm{cm^3}\), and the length of one side is approximately \(2.17 \, \mathrm{cm}\).

Step-by-step solution

- Identify given values and relevant formulas

The density (\(d\)) of silver is given as \(10.5 \, \mathrm{g/cm^3}\). The molar mass of silver (\(M\)) is approximately \(107.87 \, \mathrm{g/mol}\). We need to use the formulas for density and volume: \(d = \frac{m}{V}\), where \(m\) is mass and \(V\) is volume.

- Calculate the mass of 1 mole of silver

Using the molar mass of silver: \(m = M\). For 1 mole of silver, \(m = 107.87 \, \mathrm{g}\).

- Calculate the volume of silver

Using the formula for density: \(d = \frac{m}{V}\). Rearrange to solve for volume: \(V = \frac{m}{d}\). Substituting the values: \[ V = \frac{107.87 \, \mathrm{g}}{10.5 \, \mathrm{g/cm^3}} = 10.2733 \, \mathrm{cm^3} \]. Therefore, the volume of the cube is approximately \(10.2733 \, \mathrm{cm^3}\).

- Calculate the length of one side of the cube

The formula for the volume of a cube is \(V = s^3\), where \(s\) is the length of one side. Rearrange to solve for \(s\): \(s = \sqrt[3]{V}\). Using the volume from Step 3: \[ s = \sqrt[3]{10.2733 \, \mathrm{cm^3}} \approx 2.17 \, \mathrm{cm} \]. Therefore, the length of one side of the cube is approximately \(2.17 \, \mathrm{cm}\).

Key concepts

Molar Mass

In chemistry, molar mass is the mass of one mole of a given substance. It is a fundamental concept in stoichiometry, which deals with the quantitative aspects of chemical reactions. The molar mass is typically measured in grams per mole (\text{g/mol}). For example, the molar mass of silver (\text{Ag}) is approximately 107.87 \text{g/mol}. This means that one mole of silver, which contains Avogadro's number of atoms (approximately \(6.022 \times 10^{23}\) atoms), weighs 107.87 grams.
Understanding molar mass is crucial because it helps in converting between the number of moles of a substance and its mass. This is vital for calculations in various chemical processes and reactions. So, by knowing the molar mass of silver, we can determine the mass of any given amount of silver atoms accurately.

Volume of a Cube

The volume of a cube is determined by the formula \( V = s^3 \), where \( s \) represents the length of a side of the cube. This formula is relatively simple but crucial in many real-world applications and scientific calculations.
In the exercise, we calculated the volume of silver when shaped into a cube. Given that the density and mass of silver were known, we first determined the volume before using it to find the side length of the cube.
Keep in mind that the units must be consistent when performing these calculations. If the density is in \( \text{g/cm}^3 \), ensure the volume is in \( \text{cm}^3 \) as well.
Once you have the volume, calculating the side length involves rearranging the volume formula: \( s = \frac{V}{3} \). So, for a cube with a volume of \( 10.2733 \text{cm}^3 \), the side length is \( \text{s} = \frac{10.2733}{3} \text{cm} \), which equals approximately \( 2.17 \text{cm} \), as seen in the original solution.

Density Formula

The density formula is another essential concept in chemistry. It is defined by the equation: \( d = \frac{m}{V} \), where \( d \) is the density, \( m \) is the mass, and \( V \) is the volume. Density indicates how much mass of a substance is contained in a given volume and is often measured in \( \text{g/cm}^3 \) or \( \text{kg/m}^3 \).
To better understand this, consider our exercise where the density of silver is \( 10.5 \text{g/cm}^3 \). Using the formula, we can calculate the volume of a specific mass of silver. Rearranging the formula to solve for volume, \( V = \frac{m}{d} \), we substitute the given values: \( V = \frac{107.87 \text{g}}{10.5 \text{g/cm}^3} \text{cm}^3 \), resulting in a volume of approximately \( 10.2733 \text{cm}^3 \).
Thus, density calculations are fundamental in determining the physical properties of substances and solving various practical problems in the field of chemistry.