Problem 56 - 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 is the volume \(\left(V=\frac{4}{3} \pi r^{3}\right)\) and surface area \((A=\) \(\left.4 \pi r^{2}\right)\) 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

Volume: \(5.24 \times 10^{-22} \mathrm{m}^{3}\), Surface Area: \(3.14 \times 10^{-14} \mathrm{m}^{2}\), Need \(1.91 \times 10^{15}\) spheres, giving a total surface area of \(6.00 \mathrm{m}^{2}\).

Step 1: Convert Diameter to Radius

The diameter of the spheres is given as \(1.0 \times 10^{2} \mathrm{nm}\). First, convert this diameter to radius by dividing by 2: \(r = \frac{1.0 \times 10^{2} \mathrm{nm}}{2} = 50 \mathrm{nm}\).

Step 2: Convert Radius to Meters

Since 1 nm = \(1 \times 10^{-9}\) meters, convert the radius to meters: \(50 \mathrm{nm} = 50 \times 10^{-9} \mathrm{m} = 5.0 \times 10^{-8} \mathrm{m}\).

Step 3: Calculate Volume of One Sphere

Use the formula for the volume of a sphere: \(V = \frac{4}{3} \pi r^{3}\). Substitute the radius: \(V = \frac{4}{3} \pi (5.0 \times 10^{-8} \mathrm{m})^{3}\).\[V = \frac{4}{3} \pi (1.25 \times 10^{-22} \mathrm{m}^{3})\] \[V \approx 5.24 \times 10^{-22} \mathrm{m}^{3}\].

Step 4: Calculate Surface Area of One Sphere

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

Step 5: Determine Number of Spheres Required

To find the number of spheres that make up a total volume of \(1.0 \mathrm{cm}^{3}\), which is \(1.0 \times 10^{-6} \mathrm{m}^{3}\), divide the total volume by the volume of one sphere: \[ \text{Number of Spheres} = \frac{1.0 \times 10^{-6} \mathrm{m}^{3}}{5.24 \times 10^{-22} \mathrm{m}^{3}} \] \[ \approx 1.91 \times 10^{15} \].

Step 6: Calculate Total Surface Area for All Spheres

Multiply the surface area of one sphere by the number of spheres calculated: \[ \text{Total Surface Area} = 1.91 \times 10^{15} \times 3.14 \times 10^{-14} \mathrm{m}^{2} \] \[ \approx 6.00 \mathrm{m}^2 \].

Key Concepts

Sphere Volume Calculation

When calculating the volume of a sphere, the formula to use is \( V = \frac{4}{3} \pi r^{3} \). This tells us how much space is inside the sphere. Imagine filling up the sphere with tiny cubes and counting the cubes—that would be its volume.

To apply this formula, first, you need the sphere's radius, denoted by \( r \). In our given problem, the diameter of the sphere is \(1.0 \times 10^{2} \ \text{nm}\). So, the radius is half of the diameter, which gives us \(50 \ \text{nm}\). Converting the radius into meters gives \(5.0 \times 10^{-8} \ \text{m}\).

Now, substitute the radius back into the formula:
\[ V = \frac{4}{3} \pi (5.0 \times 10^{-8})^3 \]
Which calculates to \( V \approx 5.24 \times 10^{-22} \ \text{m}^3 \).

This value represents the volume of one tiny sphere in the colloidal dispersion.

Surface Area Calculation

The surface area of a sphere helps us understand how much of the sphere's outer layer is exposed. The formula used is \( A = 4 \pi r^{2} \).

Similar to the volume calculation, you'll use the radius \(5.0 \times 10^{-8} \ \text{m}\) derived from the diameter. By substituting this radius into the formula, you perform the following steps:
\[ A = 4 \pi (5.0 \times 10^{-8})^2 \]
This results in \( A \approx 3.14 \times 10^{-14} \ \text{m}^2 \).

This is the surface area of one sphere, giving us an idea of how much space the sphere's surface covers in a colloidal system.

• This concept is crucial when considering reactions that occur on the surface.
• More surface area means more space for interaction with surroundings.

Nanometer to Meter Conversion

Scientists often work with very small measurements, such as nanometers (nm). Understanding how to convert these into meters is essential in many calculations.

One nanometer is equivalent to \(1 \times 10^{-9}\) meters. So, when you have a value in nanometers, you multiply it by \(1 \times 10^{-9}\) to convert it to meters.

In our example, the diameter was \(1.0 \times 10^{2} \ \text{nm}\), which converts to \(1.0 \times 10^{2} \times 1 \times 10^{-9} \ \text{m}\), or simply \(1.0 \times 10^{-7} \ \text{m}\).

These conversions are handy in physics and chemistry, allowing scientists to work seamlessly between scales.

• Nano-measurements are used widely in fields like nanotechnology and material science.
• Being comfortable with these conversions is key to understanding and applying scientific concepts accurately.