Problem 50 - 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

For a sphere with radius \(50\) nm, volume \(5.24 \times 10^{-24}\) m\(^3\) and surface area \(3.14 \times 10^{-14}\) m\(^2\). You need \(1.91 \times 10^{17}\) spheres for \(1\) cm\(^3\) with total surface area \(6.0 \times 10^3\) m\(^2\).

Step 1: Calculate the radius of each sphere

The diameter of each sphere is given as \(1.0 \times 10^2 \text{ nm}\). To find the radius, divide the diameter by 2: \[ r = \frac{1.0 \times 10^2 \text{ nm}}{2} = 50 \text{ nm} = 50 \times 10^{-9} \text{ m} \].

Step 2: Calculate the volume of one sphere

Use the volume formula for a sphere, \(V = \frac{4}{3} \pi r^3\): \[ V = \frac{4}{3} \pi (50 \times 10^{-9})^3 \text{ m}^3\]. Calculate this to find the volume of one sphere.

Step 3: Calculate the surface area of one sphere

Use the surface area formula for a sphere, \(A = 4 \pi r^2\): \[ A = 4 \pi (50 \times 10^{-9})^2 \text{ m}^2\]. Calculate this to find the surface area of one sphere.

Step 4: Calculate the number of spheres for total volume

To find out the number of spheres required to make up a total volume of \(1.0 \text{ cm}^3\), use the formula \( \text{total volume} = n \times \text{volume of one sphere} \), where \(n\) is the number of spheres: \[ n = \frac{1.0 \times 10^{-6} \text{ m}^3}{V} \]. Solve to find \(n\).

Step 5: Calculate the total surface area

Use the total number of spheres \(n\) to find the total surface area: \[ \text{Total Surface Area} = n \times A \]. Substitute the values for \(n\) and \(A\) to find the total surface area in square meters.

Key Concepts

Volume Calculation

Volume calculation is essential in understanding the space an object occupies. When dealing with spheres, the volume can be calculated using the formula \( V = \frac{4}{3} \pi r^3 \). Here, \( r \) is the radius of the sphere. For this exercise, we used a diameter of \(1.0 \times 10^2 \text{ nm}\) to find a radius of \(50 \text{ nm}\). Once the radius is known, it can be substituted back into the volume formula.
The calculation simplifies to \( V = \frac{4}{3} \pi (50 \times 10^{-9})^3 \text{ m}^3\).
This helps us determine the volume of a single sphere. Volume calculations are pivotal, especially when we need to figure out how many such spheres are needed to fill up a larger volume, such as \(1.0 \text{ cm}^3\) in this case.

Surface Area Calculation

The surface area of a sphere gives us information about the total area covered by the surface of the sphere. It is calculated using the formula \( A = 4 \pi r^2 \), where \( r \) is the radius. With the radius determined as \(50 \text{ nm}\), we can directly insert it into the formula. This leads us to \( A = 4 \pi (50 \times 10^{-9})^2 \text{ m}^2\).
Understanding surface area is crucial in fields like physics and chemistry, where reactions occur on surfaces.
In this problem, surface area calculations enable us to determine how much surface is covered when multiple spheres are combined. This total surface area helps in finding out interactions at the surface level in a colloidal dispersion.

Sphere Geometry

Sphere geometry provides a foundation for understanding the symmetric properties and calculations related to spheres. A sphere is a perfectly symmetrical object with all points on its surface equidistant from its center.
This symmetry simplifies computations as everything can be defined in terms of its radius. With just the diameter or radius, both volume and surface area can be calculated efficiently.
The formulas for volume and surface area share a common factor of \( \pi \) and involve raising the radius to a power, highlighting the inherent geometric properties of spheres. In real-world applications, spheres are often used to simplify complex systems into understandable models. Their symmetry makes them essential in physics, chemistry, and numerous other scientific disciplines.

Mathematical Problem Solving

Mathematical problem solving involves following structured processes to arrive at a solution. The exercise exemplifies this with a step-by-step approach: starting with determining the radius from the given diameter, then using this radius for volume and surface area calculations.
Once basic properties—like volume and surface area—are determined, it leads to solving more complex issues, such as figuring out how many spheres fit into a specific volume or the total surface area of multiple spheres.
This methodical approach is key in breaking down seemingly intricate problems into manageable steps, making them solvable with basic mathematical operations and formulas. By laying out each step clearly, we ensure that each part of the problem is addressed and understood, facilitating a comprehensive solution.