Problem 79 - General Chemistry

Question

Ethanol has a density of \(0.789 \mathrm{~g} / \mathrm{cm}^{3}\). What volume must be poured into a graduated cylinder to give \(19.8 \mathrm{~g}\) of alcohol?

Step-by-Step Solution

Answer

Pour approximately 25.1 cm³ of ethanol into the cylinder.

Step 1: Understand the Problem

We know the density of ethanol is \(0.789 \text{ g/cm}^3\) and we need to find out what volume is required to have a mass of \(19.8 \text{ g}\). The relationship between mass, density, and volume can be expressed using the formula: \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\).

Step 2: Rearrange the Density Formula

We need to find the volume, so we rearrange the formula for density to solve for volume: \[ \text{Volume} = \frac{\text{Mass}}{\text{Density}}. \]

Step 3: Plug in the Values

Substitute the given mass, \(19.8 \text{ g}\), and density, \(0.789 \text{ g/cm}^3\), into the formula:\[ \text{Volume} = \frac{19.8 \text{ g}}{0.789 \text{ g/cm}^3}. \]

Step 4: Calculate the Volume

Perform the division to find the volume:\[ \text{Volume} = \frac{19.8}{0.789} \approx 25.1 \text{ cm}^3. \]

Step 5: Interpret the Result

The volume of ethanol needed to achieve a mass of \(19.8 \text{ g}\) is approximately \(25.1 \text{ cm}^3\).

Key Concepts

Ethanol

Ethanol, commonly known as alcohol, is a chemical compound often used as a solvent and a fuel. In laboratories, ethanol is used due to its broad solvent properties, allowing it to dissolve a wide range of substances. It is also a major component in alcoholic beverages, and in industrial applications, it is often employed for sanitization and cleaning. One essential property of ethanol is its density, which is the measurement of its mass per unit volume. Density is a critical factor in many scientific calculations, including those related to the preparation and mixing of solutions. Knowing the density of ethanol, which is approximately 0.789 grams per cubic centimeter ( ext{g/cm}^3), provides valuable information that helps in calculating how much volume one needs for a specific mass.

Volume Calculation

Calculating volume involves determining how much three-dimensional space a substance occupies. It is an important measurement for liquids, especially in scientific experiments and industrial applications. In the context of density calculations, volume plays a crucial role.
For ethanol, to find out how much volume is needed for a given mass, we must understand how to rearrange the density formula. The basic relationship used is the mass-density-volume formula:

• The formula for density: \[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]
• To find the volume, rearrange the formula: \[\text{Volume} = \frac{\text{Mass}}{\text{Density}}\]

Thus, when given a mass and knowing the density, you can calculate the volume required. This calculation is crucial in determining the correct measurements to use, ensuring accuracy and consistency in experiments and production.

Mass-Density-Volume Relationship

Understanding the relationship between mass, density, and volume is essential in many scientific fields. This relationship allows for the conversion between these three properties. Here's how each correlates with the others:

• **Density**: defined as mass divided by volume \( \left(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\right) \). This property indicates how tightly the mass is packed within a given volume.
• **Mass**: when rearranging the formula, you can isolate mass to find: \( \text{Mass} = \text{Density} \times \text{Volume} \). This helps in finding out how much a given volume of a substance will weigh.
• **Volume**: to find volume when mass and density are known, use: \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \). This formula lets you determine the space a mass will occupy.

These equations are versatile and can be adapted to myriad situations involving solids, liquids, and gases. They rely on metric units like grams for mass and cubic centimeters for volume, ensuring precise and accurate measurements in scientific work.