Problem 76 - Chemistry: The Central Science

Question

A \(15.0-\mathrm{cm}\) long cylindrical glass tube, sealed at one end, is filled with ethanol. The mass of ethanol needed to fill the tube is found to be \(11.86 \mathrm{~g}\). The density of ethanol is \(0.789 \mathrm{~g} / \mathrm{mL}\). Calculate the inner diameter of the tube in centimeters.

Step-by-Step Solution

Answer

The inner diameter of the cylindrical glass tube is approximately \(0.572\,cm\).

Step 1: Calculate the volume of ethanol in the tube

We can calculate the volume of ethanol in the tube using the mass and the density of the ethanol. The formula to calculate the volume is: Volume = Mass ÷ Density We are given the mass of the ethanol as 11.86 g and the density as 0.789 g/mL. So, let's plug in the values and calculate the volume. Volume = 11.86 g ÷ 0.789 g/mL = 15.03 mL

Step 2: Convert volume to cubic centimeters (cc)

Since we need the inner diameter of the tube in centimeters, we should convert the volume from milliliters to cubic centimeters. Fortunately, 1 mL is equal to 1 cc, so we don't need to do any conversion. Volume in cc = 15.03 mL = 15.03 cc

Step 3: Use the formula for the volume of a cylinder to find the inner diameter

The formula for the volume of a cylinder is: Volume = π × (radius)^2 × height Our goal is to find the inner diameter of the tube, so we need to rearrange this formula to find the radius. We also know that diameter = 2 × radius. Let's rearrange the formula and solve for the radius first: radius = √(Volume ÷ (π × height)) We are given the height of the tube (15.0 cm) and we just calculated the volume (15.03 cc). Plugging in these values, we get: radius = √(15.03 cc ÷ (π × 15.0 cm)) ≈ 0.286 cm Now let's find the inner diameter: inner diameter = 2 × radius ≈ 2 × 0.286 cm ≈ 0.572 cm The inner diameter of the cylindrical glass tube is approximately 0.572 cm.

Key Concepts

Density Calculation

Density helps us understand how much mass is contained in a given volume of a substance. It is defined as mass per unit volume and is represented by the formula:

• Density = Mass / Volume

In this exercise, we used the density of ethanol, which is 0.789 g/mL. Knowing the density allows us to convert the mass of ethanol to its volume. For example, with a mass of 11.86 g, the volume can be calculated by rearranging the formula to:

• Volume = Mass / Density

This helps us find out how much space the ethanol occupies in the tube. By using this concept, you can easily convert measurements from one property to another, aiding in various calculations.

Unit Conversion

Unit conversion is essential when working with different measurement systems. In this problem, we used the relationship between milliliters (mL) and cubic centimeters (cc). These two units are interchangeable, with 1 mL being exactly equal to 1 cc.

This conversion is seamless, and knowing it helps simplify problems involving liquids and volumes. Understanding unit conversions ensures that calculations are consistent and accurate, especially when results need to be expressed in specific units as seen when determining the inner diameter of the tube in centimeters.

Volume of Cylinder

A cylinder’s volume measures the space it occupies. The formula for calculating the volume of a cylinder is:

• Volume = π × (radius)^2 × height

Here, π (pi) is approximately 3.14159, the radius is half the diameter, and the height is the length of the cylinder.

In this exercise, the volume of ethanol inside the cylinder helps us solve for the radius and subsequently the diameter, using the rearranged formula:

• radius = √(Volume ÷ (π × height))

By understanding this formula, you can solve for any missing dimension if the other two are known. It’s a critical concept in both geometry and practical applications such as this one.

Geometry in Chemistry

Geometry plays a significant role in chemistry, especially when it comes to understanding shapes and spaces of molecules and containers. The cylindrical glass tube in this problem serves as a model for understanding volumetric measurements of substances.

By applying geometric formulas, we can compute how much liquid a tube can hold and deduce dimensions like the diameter. This blend of geometry and chemistry is crucial for experiments and calculations involving reaction vessels, ensuring precise measurements and outcomes. Understanding these principles helps in designing setups and understanding spatial relationships between chemical components.