Problem 89 - Chemistry The Central Science In Si Units

Question

A 30.0 -cm-long cylindrical plastic tube, sealed at one end, is filled with acetic acid. The mass of acetic acid needed to fill the tube is found to be \(89.24 \mathrm{~g}\). The density of acetic acid is \(1.05 \mathrm{~g} / \mathrm{mL}\). Calculate the inner diameter of the tube in centimeters.

Step-by-Step Solution

Answer

The inner diameter of the tube is approximately 1.90 cm.

Step 1: Understand the Problem

We are asked to find the inner diameter of a cylindrical tube, given the mass and density of the liquid filling the tube. The tube is closed at one end, hence the formula for the volume of a cylinder will be useful.

Step 2: Use the Density Formula

Start by recalling the density formula: \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\). We have \(\text{Density} = 1.05\, \text{g/mL}\) and \(\text{Mass} = 89.24\, \text{g}\). Rearranging for volume gives us: \(\text{Volume} = \frac{\text{Mass}}{\text{Density}}\).

Step 3: Calculate the Volume of Acetic Acid

Substitute the given values into the volume formula: \(\text{Volume} = \frac{89.24 \, \text{g}}{1.05 \, \text{g/mL}}\). Calculate this to find the volume of the acetic acid.

Step 4: Volume of Acetic Acid in mL

Upon calculating: \(\text{Volume} \approx 84.99 \, \text{mL}\). Since 1 mL is equivalent to 1 cm³, the volume of the acetic acid is approximately 84.99 cm³.

Step 5: Use the Volume Formula for a Cylinder

For a cylinder, volume \( V \) is given by \( V = \pi r^2h \) where \( r \) is the radius and \( h = 30.0 \, \text{cm}\) is the height. Substitute the volume you found and the height to find the radius.

Step 6: Rearrange for the Inner Radius

We have \( 84.99 = \pi r^2 \times 30 \). Rearranging for the radius gives \( r^2 = \frac{84.99}{30\pi}\). Calculate \( r \) by taking the square root.

Step 7: Calculate the Inner Radius

Calculate \( r^2 = \frac{84.99}{30\pi} \approx 0.901 \). Thus, \( r \approx \sqrt{0.901} \approx 0.949 \) cm.

Step 8: Find the Inner Diameter

Double the calculated radius to find the diameter: \( d = 2r = 2 \times 0.949 \approx 1.898 \) cm.

Key Concepts

Density Formula

The density formula is a simple yet fundamental concept in understanding how mass and volume relate. At its core, the formula is expressed as

• Density \(\rho = \frac{\text{Mass}}{\text{Volume}}\)

This equation tells us that density is the amount of mass per unit volume. Given this relationship, if you know any two of the three variables (density, mass, volume), you can find the third. In our problem, we were provided with the mass of the acetic acid and its density. By rearranging this basic formula, we determine the volume as:

• \(\text{Volume} = \frac{\text{Mass}}{\text{Density}}\)

Substituting the provided values gives the volume needed to help find the dimensions of the cylinder in the following steps. It demonstrates that having a strong grasp of simple formulas can be incredibly useful in problem-solving.

Cylinder Geometry

Understanding cylinder geometry is essential for solving problems involving cylindrical shapes. A cylinder is defined by its length (or height) and diameter or radius. The basic formula for the volume of a cylinder is:

• \(V = \pi r^2 h\)

In this expression, \(V\) represents the volume, \(r\) is the radius, and \(h\) is the height of the cylinder. In our given problem, the cylinder height is provided, allowing us to express the volume in terms of the radius. Cylinders appear frequently in everyday objects like cans, tubes, and pipes, and understanding their geometry helps in calculating their capacity. For the exercise, identifying radius from volume was key.

Calculation of Volume

The calculation of volume is a crucial step in this exercise. Correctly calculating the cylinder's volume allows us to proceed in determining its other dimensions. We were given that the mass of the acetic acid was 89.24 grams and the density was 1.05 grams per mL. Using the rearranged density formula:

• \(\text{Volume} = \frac{89.24}{1.05} \approx 84.99 \text{ mL} = 84.99 \text{ cm}^3\)

This volume serves as the starting point to plug into the cylinder volume formula \(V = \pi r^2 h\), with \(h\) known (30 cm). It's a straightforward plug-and-solve approach using earlier determined relationships.

Mathematical Problem Solving

Mathematical problem-solving often involves integrating different concepts to arrive at a solution. This exercise required combining knowledge of density, cylinder geometry, and volume calculations. After calculating the volume using the density formula, it was placed into the cylinder volume formula:

• \(84.99 = \pi r^2 \times 30\)
• \(r^2 = \frac{84.99}{30\pi}\)

Solving for \(r\) by isolating it and taking the square root provided the radius. Then calculating the diameter by doubling the radius:

• \(d = 2r \approx 1.898 \text{ cm}\)

This methodical approach of substitution and solving helps in piecing together various mathematical concepts to find the required measurements.