Problem 52 - Chemistry and Chemical Reactivity

Question

The dispersed phase of a certain colloidal dispersion consists of spheres of diameter \(1.0 \times 10^{2} \mathrm{nm}\) (a) What are the volume \(\left(V=4 / 3 \pi r^{3}\right)\) and surface area \(\left(A=4 \pi r^{2}\right)\) of each sphere? (b) How many spheres are required to give a total volume of \(1.0 \mathrm{cm}^{3} ?\) What is the total surface area of these spheres in square meters?

Step-by-Step Solution

Answer

(a) Volume: \(5.24 \times 10^{-22} \, \mathrm{m}^{3}\), Surface area: \(3.14 \times 10^{-14} \, \mathrm{m}^{2}\). (b) \(1.91 \times 10^{15}\) spheres, total surface area: \(60.0 \, \mathrm{m}^{2}\).

Step 1: Calculate Radius from Diameter

First, find the radius of each sphere. The diameter is given as \(1.0 \times 10^{2} \ \mathrm{nm}\), so the radius \(r\) is half of the diameter. \[ r = \frac{1.0 \times 10^{2}}{2} = 5.0 \times 10^{1} \ \mathrm{nm} = 5.0 \times 10^{-8} \ \mathrm{m} \] (Note: Converted radius to meters for further calculations.)

Step 2: Calculate Volume of Each Sphere

Now, calculate the volume \(V\) using the formula for the volume of a sphere, \( V = \frac{4}{3} \pi r^{3} \):\[ V = \frac{4}{3} \pi (5.0 \times 10^{-8})^{3} \]\[ V = \frac{4}{3} \pi \times 1.25 \times 10^{-22} = 5.24 \times 10^{-22} \ \mathrm{m}^{3} \]

Step 3: Calculate Surface Area of Each Sphere

Use the surface area formula for a sphere, \( A = 4 \pi r^{2} \):\[ A = 4 \pi (5.0 \times 10^{-8})^{2} \]\[ A = 4 \pi \times 2.5 \times 10^{-15} = 3.14 \times 10^{-14} \ \mathrm{m}^{2} \]

Step 4: Determine Number of Spheres for Given Volume

To find out how many spheres are needed for a total volume of \(1.0 \ \mathrm{cm}^{3}\), convert the volume to cubic meters: \(1.0 \ \mathrm{cm}^{3} = 1.0 \times 10^{-6} \ \mathrm{m}^{3}\).Now, divide the total volume by the volume of one sphere:\[ \text{Number of spheres} = \frac{1.0 \times 10^{-6}}{5.24 \times 10^{-22}} \approx 1.91 \times 10^{15} \]

Step 5: Calculate Total Surface Area

To find the total surface area of these spheres, multiply the surface area of one sphere by the number of spheres:\[ \text{Total surface area} = 3.14 \times 10^{-14} \times 1.91 \times 10^{15} \]\[ \text{Total surface area} \approx 6.00 \times 10^{1} \ \mathrm{m}^{2} \]

Key Concepts

Sphere Volume Calculation

The volume of a sphere is an important concept in understanding colloidal dispersions. To calculate the volume, use the formula \( V = \frac{4}{3} \pi r^3 \).

• Here, \( V \) represents the volume of the sphere.
• \( \pi \approx 3.14159 \), a constant in mathematics.
• \( r \) is the radius of the sphere, which you obtain by halving the diameter.

Let's look at an example:Imagine a sphere with a radius of \(5.0 \times 10^{-8} \ \text{m} \) (as derived from a given diameter of \(1.0 \times 10^{2} \ \text{nm}\)). Plug the radius into the volume formula:\[ V = \frac{4}{3} \pi (5.0 \times 10^{-8})^3 \]Calculate the cube of the radius first, which results in \( 1.25 \times 10^{-22} \), and multiply it by \( \frac{4}{3} \pi \). You will get \( 5.24 \times 10^{-22} \ \text{m}^3 \). This is the volume of the sphere.
Understanding this calculation is crucial when dealing with problems that require conversion between amounts of materials represented in spherical shapes.

Surface Area of a Sphere

The surface area of a sphere is another fundamental aspect to study, especially useful in understanding the properties of colloidal particles.To calculate the surface area, use the formula \( A = 4 \pi r^2 \).

• Here, \( A \) represents the surface area of the sphere.
• \( \pi \) remains the same constant.
• \( r \) is the radius as previously defined.

For a sphere with a radius of \(5.0 \times 10^{-8} \ \text{m}\):\[ A = 4 \pi (5.0 \times 10^{-8})^2 \]Calculate the square of the radius, giving \( 2.5 \times 10^{-15} \), and then multiply by \( 4 \pi \). The result is \( 3.14 \times 10^{-14} \ \text{m}^2 \), which is the surface area of the sphere.
Knowing how to find the surface area helps in understanding reactions and interactions at the surface in colloidal systems.

Nanometer to Meter Conversion

When dealing with measurements in nanometers, a common practice is to convert them into meters for ease of calculation. This conversion is particularly handy in scientific calculations involving physics and chemistry.To convert nanometers to meters:

• Remember that \( 1 \ \text{nm} = 1 \times 10^{-9} \ \text{m} \).

Take a given length in nanometers and multiply by \(10^{-9}\) to convert it into meters. For example, if you have a sphere diameter of \(1.0 \times 10^2 \ \text{nm}\), then convert it by:\[ \text{Diameter in meters} = 1.0 \times 10^2 \times 10^{-9} = 1.0 \times 10^{-7} \ \text{m} \]Afterwards, you can find the radius by dividing the converted diameter by 2, giving \(5.0 \times 10^{-8} \ \text{m}\).
This conversion is essential for accurately handling calculations within the metric system, especially when working across different scales of magnitude.