Problem 89 - Chemistry: The Central Science in SI Units, Global Edition

Question

A \(30.0-\mathrm{cm}\) -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 \(1.90 \mathrm{~cm}\).

Step 1: Write down the formula for the volume of a cylinder

The volume (V) of a cylinder is given by the formula: V = πr^2h, where r is the radius of the base of the cylinder, and h is its height (or length, in this case).

Step 2: Write down the formula for mass, volume, and density relationship

We know that density (ρ) is related to the mass (m) and volume (V) by the formula: ρ = m/V.

Step 3: Express the volume of the tube in terms of mass and density

Using the mass-volume-density relationship, we can write: V = m/ρ. Given that the mass (m) of acetic acid is 89.24 g, and its density (ρ) is 1.05 g/mL, we can find the volume (V) by plugging the values: V = 89.24 g / 1.05 g/mL = 85 mL

Step 4: Convert the volume to cubic centimeters

Since 1 mL is equal to 1 cm³, the volume in cubic centimeters is: V = 85 cm³

Step 5: Write the equation for the volume of the cylinder using the known length and volume

Now we can write the equation for the volume of the cylinder using the known values: 85 cm³ = πr^2 * 30 cm

Step 6: Solve for the radius of the cylinder

Dividing both sides of the equation by 30π, we can solve for the radius: r^2 = (85 cm³) / (30π) = 85 cm³ / (30 * 3.14) ≈ 0.90 cm² Taking the square root of the result, we find the radius: r = √0.90 cm² ≈ 0.95 cm

Step 7: Calculate the diameter of the cylinder

Knowing the radius, we can easily find the diameter (D) of the cylinder by multiplying the radius by 2: D = 2r = 2 * 0.95 cm ≈ 1.90 cm So, the inner diameter of the tube is approximately 1.90 centimeters.

Key Concepts

Density

Density is an essential concept in understanding how mass and volume relate to each other. It tells you how much mass is present in a given volume and is typically expressed in units like grams per milliliter (g/mL) or kilograms per cubic meter (kg/m³).

The formula for density is simple:

• Density \( (\rho) \) = Mass \( (m) \) / Volume \( (V) \)

Understanding this equation helps in solving problems where you have to determine the missing variable, like mass or volume. For example, if you know the density of a substance, you can easily calculate its volume if you also know its mass.

Density plays a vital role in various fields such as chemistry, physics, and engineering. Having a firm grasp on this concept is critical for calculations involving fluid mechanics and material properties.

Acetic Acid

Acetic acid, commonly known as the main component of vinegar, is a simple organic compound with the chemical formula \( \text{CH}_3\text{COOH} \). It is widely used in chemical reactions and laboratory settings due to its acidic properties and ability to act as a solvent.

Its density of about \( 1.05 \text{ g/mL} \) makes it a bit denser than water, helping it to maintain its properties as a robust liquid acid. This property is beneficial when calculating the volume it occupies inside containers of various sizes.

Whether you're a chemistry student or working on a science project, understanding the properties of acetic acid, such as its density, can be very beneficial. It's not just about memorizing values; it's about comprehending how these values affect real-world applications and experiments.

Geometry Formulas

When solving problems involving three-dimensional objects like cylinders, knowing the correct geometry formulas is crucial. One primary formula is the volume of a cylinder, which combines geometry with concepts from calculus for more complex shapes.

• The volume of a cylinder \( V \) is given by \( V = \pi r^2 h \)
• Where \( r \) is the radius and \( h \) is the height or length

These formulas allow you to compute various attributes of an object, such as volume and surface area, given particular dimensions. Solving for any missing piece, like using volume to find radius, involves re-arranging these formulas.

For instance, if you need to find the radius of the cylinder from the volume and height, you rearrange the formula to solve for \( r \). The use of geometry formulas is important in fields ranging from engineering to everyday household problem-solving, like fitting furniture in a room by calculating space.