Problem 77 - Chemistry: The Central Science

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.

Step-by-Step Solution

Answer

The molarity of the acetic acid solution can be calculated using the given information in four steps. First, find the mass of glacial acetic acid: \(Mass = 1.049 \, \frac{g}{mL} \times 20.00 \, mL\). Second, calculate the moles of acetic acid using its molar mass: Moles of acetic acid = \( \frac{Mass}{60.05 \, g/mol} \). Then, convert the total volume of the solution to liters: Total volume = \( 250.0 \, mL \times \frac{1 \, L}{1000 \, mL} \). Finally, calculate the molarity of the acetic acid solution: Molarity = \( \frac{Moles \, of \, acetic \, acid}{Total \, volume \, (L)} \).

Step 1: Find the mass of glacial acetic acid

Using the given density and volume, we can calculate the mass of the glacial acetic acid: Mass = Density × Volume \(Mass = 1.049 \, \frac{g}{mL} \times 20.00 \, mL\)

Step 2: Calculate the moles of acetic acid

Using the molar mass of acetic acid (60.05 g/mol), convert the mass of acetic acid to moles: Moles of acetic acid = \( \frac{Mass}{Molar \, mass} \) Moles of acetic acid = \( \frac{Mass}{60.05 \, g/mol} \)

Step 3: Calculate the total volume of the solution in liters

We know that we have 20.0 mL of glacial acetic acid and enough water is added to make 250.0 mL of solution, so we need to convert the total volume from milliliters to liters: Total volume = \(250.0 \, mL \times \frac{1\, L}{1000 \, mL} \)

Step 4: Calculate the molarity of the acetic acid solution

The molarity is defined as the number of moles of solute (acetic acid) per liter of solution. To find this, we will divide the moles of acetic acid by the total volume of the solution in liters: Molarity = \( \frac{Moles \, of\, acetic\, acid}{Total\, volume\, (L)} \)