Problem 86 - Chemistry

Question

What is the density of lead in \(\mathrm{g} / \mathrm{cm}^{3}\) if a rectangular bar measuring \(0.50 \mathrm{~cm}\) in height, \(1.55 \mathrm{~cm}\) in width, and \(25.00 \mathrm{~cm}\) in length has a mass of \(220.9 \mathrm{~g}\) ?

Step-by-Step Solution

Answer

The density of the lead bar is approximately 11.4 g/cm³.

Step 1: Determine the Volume of the Bar

To find the volume of the rectangular bar, use the formula for the volume of a rectangular prism: \( V = ext{height} imes ext{width} imes ext{length} \). Substitute the given values: \( V = 0.50 \, \text{cm} \times 1.55 \, \text{cm} \times 25.00 \, \text{cm} \). Calculate the result.

Step 2: Calculate the Volume of the Bar

Perform the multiplication from Step 1: \( V = 0.50 \, \text{cm} \times 1.55 \, \text{cm} \times 25.00 \, \text{cm} = 19.375 \, \text{cm}^3 \). The volume of the bar is \( 19.375 \, \text{cm}^3 \).

Step 3: Use the Density Formula

The formula for density is \( \text{density} = \frac{\text{mass}}{\text{volume}} \). Insert the known values into the formula: \( \text{density} = \frac{220.9 \, \text{g}}{19.375 \, \text{cm}^3} \).

Step 4: Calculate the Density

Divide the mass by the volume: \( \text{density} = \frac{220.9 \, \text{g}}{19.375 \, \text{cm}^3} \approx 11.4 \, \text{g/cm}^3 \).

Step 5: Conclude the Solution

The density of the lead bar is approximately \( 11.4 \, \text{g/cm}^3 \).

Key Concepts

Volume of Rectangular Prism

Understanding the volume of a rectangular prism can be quite straightforward. A rectangular prism is simply a three-dimensional box shape with six rectangular faces. To find its volume, you multiply its dimensions: height, width, and length. This is expressed by the formula:

• \( V = ext{height} \times ext{width} \times ext{length} \)

To calculate the volume, you use the given measurements of the prism. For instance, in the example problem, the rectangular bar has a height of 0.50 cm, a width of 1.55 cm, and a length of 25.00 cm. By substituting these values into the formula, we get:

• \( V = 0.50 \, \text{cm} \times 1.55 \, \text{cm} \times 25.00 \, \text{cm} = 19.375 \, \text{cm}^3 \)

Let your calculations guide you to the volume, which here is 19.375 cubic centimeters.

Density Formula

Density is a fascinating concept that describes how much mass is contained in a given volume. To calculate density, you'll use the formula:

• \( \text{density} = \frac{\text{mass}}{\text{volume}} \)

This formula shows us that density is equal to the mass of an object divided by its volume.
In the exercise, we're given the mass of the lead bar as 220.9 grams and its volume as 19.375 cm³, calculated earlier. Now, to find the density, you simply divide the mass by the volume:

• \( \text{density} = \frac{220.9 \, \text{g}}{19.375 \, \text{cm}^3} \approx 11.4 \, \text{g/cm}^3 \)

This result means that each cubic centimeter of the lead bar weighs approximately 11.4 grams, helping to illustrate how dense the material is.

Lead Properties

Lead is a dense, heavy metal known for its high atomic number and weight, which gives it unique properties. These properties are crucial in understanding why lead's density is as high as calculated. Some characteristics of lead include:

• High Density: With a density of about 11.4 g/cm³, lead is much denser than many other common metals. This means it has a lot of mass in a small volume.
• Softness and Malleability: Lead is relatively soft, meaning it can be easily shaped and molded without breaking.
• Corrosion Resistance: Lead is resistant to corrosion, which makes it useful in a variety of industrial applications, including batteries and radiation shielding.

Understanding these properties not only highlights the calculated density value but also shows why lead has been historically used in industries where weight and durability are important, yet manageable due to its malleability.