Question

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of \(1.049 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Calculate the molarity of a solution of acetic acid made by dissolving \(20.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) in enough water to make \(250.0 \mathrm{~mL}\) of solution.

Short answer

The molarity of the acetic acid solution is \(1.4 \, M\).

Step-by-step solution

Determine the mass of glacial acetic acid used

To calculate the mass of acetic acid used, we will multiply the density of acetic acid with the volume of glacial acetic acid we are using for the solution: Mass = Density × Volume The density of glacial acetic acid is \(1.049 \, \frac{g}{mL}\), and the volume is \(20.00 \, mL\). Thus, Mass = \(1.049 \, \frac{g}{mL} \times 20.00 \, mL\)

Calculate the mass of acetic acid used

Plug in the given values to compute the mass: Mass = \(1.049 \, \frac{g}{mL} \times 20.00 \, mL = 21.0 \, g\) So, we have \(21.0 \, g\) of glacial acetic acid that will be dissolved in the solution.

Determine the moles of acetic acid in the solution

Now, we will find the number of moles of acetic acid dissolved in the solution. The molar mass of acetic acid is \(60.05 \, \frac{g}{mol}\). To find the moles of acetic acid, we will use the formula: Moles = \(\frac{Mass}{Molar \, Mass}\), where Mass = \(21.0 \, g\) and Molar Mass = \(60.05 \, \frac{g}{mol}\) Moles = \(\frac{21.0 \, g}{60.05 \, \frac{g}{mol}}\)

Calculate the moles of acetic acid in the solution

Plug the given values of acetic acid to get the moles: Moles = \(\frac{21.0 \, g}{60.05 \, \frac{g}{mol}} = 0.35 \, mol\) So there are \(0.35 \, mol\) of acetic acid in the solution.

Calculate the molarity of the solution

Finally, we'll find the molarity of the solution using the formula: Molarity = \(\frac{Moles}{Volume \, of \, solution}\), where Moles = \(0.35 \, mol\) and the Volume of solution = \(250.0 \, mL\) We need to convert mL to L to keep the units consistent: \(250.0 \, mL = 0.250 \, L\) Molarity = \(\frac{0.35 \, mol}{0.250 \, L}\)

Find the molarity of the solution

Divide the given values to compute the molarity of the solution: Molarity = \(\frac{0.35 \, mol}{0.250 \, L} = 1.4 \, M\) The molarity of the acetic acid solution is \(1.4 \, M\).

Key concepts

Glacial Acetic Acid

Glacial acetic acid is an important component in many chemical processes. It is called "glacial" because it solidifies at low room temperatures, forming glacial-like crystals. This type of acetic acid is pure and undiluted, which makes it significantly different from the vinegar you might find in a kitchen.

• Purity: It's nearly 100% acetic acid.
• Physical State: It's a clear, colorless liquid at room temperature, but can freeze at lower temperatures.
• Use: Commonly used in labs for analytical purposes and in industries as a chemical reagent.

Understanding the properties of glacial acetic acid, such as its high purity and freezing behavior, is crucial when calculating molarity in mixed solutions.

Density

Density is a physical property defined as the mass of a substance per unit volume, often represented in units like grams per milliliter (g/mL). For glacial acetic acid, the density is given as 1.049 g/mL at 25 degrees Celsius. Understanding density helps to connect mass and volume – key elements in calculating molarity.

• Formula: Density = \(\frac{Mass}{Volume}\)
• Role in Calculation: By knowing the density, you can find out how much mass a specific volume of glacial acetic acid has.

To calculate the mass of acetic acid, we multiply the volume (20.00 mL) by its density (1.049 g/mL), resulting in a mass of 21.0 g which is useful in further molarity calculations.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It helps in converting mass to moles, which is critical when preparing solutions and finding concentrations.Acetic acid has a molar mass of 60.05 g/mol. This value allows you to convert from grams of acetic acid to moles, using the formula:\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]In this exercise, we use a mass of 21.0 g of acetic acid, and with its molar mass being 60.05 g/mol, we find that there are 0.35 moles of acetic acid in the solution. Converting mass to moles is a foundational step in determining the molarity of a solution.

Solution Preparation

Preparing a solution with a specific molarity involves dissolving a solute, like glacial acetic acid, in a solvent such as water, to reach a desired volume. The final concentration is expressed in terms of molarity, which is moles of solute per liter of solution.To prepare the acetic acid solution:

• Determine Mass: Found using density and volume.
• Convert Mass to Moles: Use the molar mass for conversion.
• Choose Final Volume: Here it is 250.0 mL, which equals 0.250 L.
• Calculate Molarity: \(\frac{\text{Moles}}{\text{Volume of Solution in L}}\)

With 0.35 moles of acetic acid and a total solution volume of 0.250 L, the molarity is calculated to be 1.4 M. This step-by-step approach ensures precise and reliable preparation of solutions, essential in both laboratory and industrial applications.