Problem 78 - Chemistry The Central Science In Si Units

Question

Glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) is a substance used extensively in the manufacture of cosmetic s, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid with a density of \(1.2656 \mathrm{~g} / \mathrm{mL}\) at \(15^{\circ} \mathrm{C}\). Calculate the molarity of a solution of glycerol made by dissolving \(50.000 \mathrm{~mL}\) glycerol at \(15^{\circ} \mathrm{C}\) in enough water to make \(250.00 \mathrm{~mL}\) of solution.

Step-by-Step Solution

Answer

The molarity of the glycerol solution is 2.748 M.

Step 1: Calculate Mass of Glycerol

First, calculate the mass of glycerol using its volume and density. The volume of glycerol is given as 50.000 mL, and the density is 1.2656 g/mL. Use the formula: \( \text{Mass} = \text{Density} \times \text{Volume} \), so \( \text{Mass of glycerol} = 1.2656 \, \text{g/mL} \times 50.000 \, \text{mL} = 63.280 \, \text{g} \).

Step 2: Calculate Moles of Glycerol

Next, convert the mass of glycerol to moles. The molar mass of glycerol (\( \text{C}_3\text{H}_8\text{O}_3 \)) is calculated as follows: \( 3 \times 12.01 \, \text{g/mol (Carbon)} + 8 \times 1.008 \, \text{g/mol (Hydrogen)} + 3 \times 16.00 \, \text{g/mol (Oxygen)} = 92.094 \, \text{g/mol} \). The moles of glycerol are: \( \text{Moles of glycerol} = \frac{63.280 \, \text{g}}{92.094 \, \text{g/mol}} = 0.6869 \, \text{mol} \).

Step 3: Calculate Molarity of the Solution

Finally, calculate the molarity of the solution using the formula: \( \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \). The total volume of the solution is 250.00 mL or 0.25000 L. Thus, \( \text{Molarity} = \frac{0.6869 \, \text{mol}}{0.25000 \, \text{L}} = 2.748 \, \text{M} \).

Key Concepts

Glycerol Properties

Glycerol, known by its chemical formula \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\), is a versatile compound utilized in a variety of industries. It is often a key ingredient in cosmetics, due to its hydrating properties.
It attracts moisture from the air and onto the skin, making it an effective moisturizer. In food, glycerol acts as a sweetener and preservative.
It helps to retain moisture in food products.

In addition to its uses in cosmetics and food, glycerol plays a role in producing antifreeze due to its freezing point depression capabilities. Glycerol is also characterized by being colorless, odorless, and non-toxic. An important physical property of glycerol is its relatively high density, which is \(1.2656 \, \mathrm{g/mL}\) at \(15^{\circ} \mathrm{C}\). This density is significant because it influences how glycerol behaves in solution and impacts the accuracy of volume-to-mass conversions.

Density and Volume Relationship

Understanding the relationship between density and volume is essential in chemistry, particularly when needing to convert between mass and volume. Density is defined as mass per unit volume, represented mathematically as \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \).
In the case of glycerol with a density of \(1.2656 \, \mathrm{g/mL}\), it means every milliliter of glycerol weighs \(1.2656\) grams.

To determine the mass of a liquid, you can rearrange the density formula to \( \text{Mass} = \text{Density} \times \text{Volume} \).
This relationship is pivotal in calculating how much of a liquid you have when given its volume or mass. For example, knowing that a 50 mL volume of glycerol will give you a mass of \(63.280 \, \mathrm{g}\), shows how effectively density allows us to switch between these two measurements depending on experimental needs.
This is an integral step before proceeding to molarity calculations, helping to ensure accuracy and precision in the laboratory.

Mole Conversion

Conversion between mass and moles is fundamental in chemistry for preparing solutions and understanding chemical reactions. The key to this conversion lies in the molar mass, which is the mass of one mole of a substance. For glycerol, the molar mass is \(92.094 \, \mathrm{g/mol}\).
Knowing this allows scientists to convert a specific mass of glycerol into moles by using the formula \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \).

In this exercise, for instance, the mass of glycerol is \(63.280 \, \mathrm{g}\), which can be converted by dividing by the molar mass, resulting in \(0.6869 \, \text{mol}\) of glycerol.
This calculation is critical because moles serve as a bridge to determine the number of particles, atoms, or molecules in a sample, vital for understanding reaction stoichiometry and preparing solutions with particular concentrations in the lab.

Solution Preparation

Preparing a solution with a specific molarity involves careful calculation and accuracy. Molarity is defined as moles of solute per liter of solution, represented mathematically as \( \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \).
In applications such as the one at hand, creating a glycerol solution at a desirable concentration requires dissolving a calculated amount of glycerol into a precise volume of water.

First, the volume of the solution must be converted into liters. In our exercise, the final solution volume is \(250.00 \, \mathrm{mL}\) or \(0.25000 \, \mathrm{L}\).
With \(0.6869 \, \text{mol}\) of glycerol, you then calculate molarity using the formula, yielding a molarity of \(2.748 \, \text{M}\).
Understanding how to accurately prepare solutions is crucial in laboratory settings, as it directly affects experiment outcomes, ensuring the correct proportions of solute to solvent.