Question

The density of chloroform, a widely used organic solvent, is \(1.4832 \mathrm{~g} / \mathrm{mL}\) at \(20{ }^{\circ} \mathrm{C}\). How many milliliters would you use if you wanted \(112.5 \mathrm{~g}\) of chloroform?

Short answer

Approximately 75.85 mL of chloroform is needed.

Step-by-step solution

Understand the Relationship Between Mass, Volume, and Density

The problem provides the density of chloroform which is given as \(1.4832\, \mathrm{g/mL}\), and you are asked to find the volume in milliliters for a given mass of \(112.5\, \mathrm{g}\). The key formula that relates these quantities is \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \).

Rearrange the Formula for Volume

To find the volume given the mass and density, rearrange the density formula to solve for volume: \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \). This will allow us to calculate the volume by dividing the mass by the density.

Plug in the Values

Using the rearranged formula, plug in the values provided in the problem: \( \text{Volume} = \frac{112.5\, \mathrm{g}}{1.4832\, \mathrm{g/mL}} \).

Perform the Calculation

Now, perform the division: \( \text{Volume} = \frac{112.5}{1.4832} \approx 75.85 \). Thus, the volume of chloroform needed to obtain \(112.5\, \mathrm{g}\) is approximately \(75.85\, \mathrm{mL}\).

Key concepts

Understanding Chloroform

Chloroform is an organic compound with the chemical formula \(\text{CHCl}_3\). It is a colorless, dense liquid with a sweet smell, often used in laboratories as a solvent or in the production of other chemicals.
Chloroform is part of a group known as halogenated hydrocarbons, characterized by one or more halogens substituting hydrogen in a hydrocarbon.
It has unique properties that make it a valuable solvent, like being able to dissolve a wide range of organic substances easily.
- Chloroform's usefulness in dissolving substances is due to its ability to form hydrogen bonds with other polar molecules.
- It is crucial to handle chloroform with care, as it is volatile and can evaporate easily, posing inhalation risks. These properties make chloroform an essential tool in tasks ranging from extracting organic compounds to serving as a reagent in synthesis.

Mass and Volume Relationship

The relationship between mass, volume, and density is fundamental to understanding how substances are measured and compared. This relationship is defined by the formula:

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

It indicates how much mass of a substance is contained in a specific volume.
This concept is essential in various fields including chemistry, physics, and engineering, as it helps determine how to mix or measure different substances correctly.

Rearranging the Formula

To find the volume when you know the density and mass, you rearrange the formula:

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

This form of the equation allows you to calculate the volume needed, as demonstrated when determining the amount of chloroform required for a given mass. Knowing how to manipulate this fundamental formula is crucial for solving real-world problems.

Organic Solvent Density

Organic solvents like chloroform are essential in various scientific and industrial processes due to their ability to dissolve materials effectively.
The density of a solvent is a significant property, as it influences how substances mix and interact.

The Role of Density

Density provides insights into how compact the molecules of a solvent are, determining its behavior in a mixture:

• A higher density means molecules are closer together, which can affect solubility and reaction rates.
• In the case of chloroform, its density of \(1.4832\, \text{g/mL}\) indicates it is relatively dense compared to other common liquids.

This density impacts how much chloroform is required for specific applications, making accurate density measurement critical in formulation and solvent use. By using density as a benchmark, practitioners can make informed decisions about quantities needed for reactions and solutions, ensuring consistency and effectiveness in their applications.