Questions: Problem 4.77 Review Constants Periodic Table Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g / mL at 25°C. Part A Calculate the molarity of a solution of acetic acid made by dissolving 28.00 mL of glacial acetic acid at 25°C in enough water to make 290.0 mL of solution. Express your answer using four significant figures.

Problem 4.77
Review
Constants
Periodic Table

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g / mL at 25°C.

Part A
Calculate the molarity of a solution of acetic acid made by dissolving 28.00 mL of glacial acetic acid at 25°C in enough water to make 290.0 mL of solution.
Express your answer using four significant figures.

Transcript text: Problem 4.77 Review Constants Periodic Table Pure acetic acid, known as glacial acetic acid, is a liquid with a density of $1.049 \mathrm{~g} / \mathrm{mL}$ at $25^{\circ} \mathrm{C}$. Part A Calculate the molarity of a solution of acetic acid made by dissolving 28.00 mL of glacial acetic acid at $25^{\circ} \mathrm{C}$ in enough water to make 290.0 mL of solution. Express your answer using four significant figures.

Solution

Solution Steps

Step 1: Calculate the Mass of Glacial Acetic Acid

First, we need to calculate the mass of the glacial acetic acid using its volume and density. The formula for mass is:

\[ \text{mass} = \text{density} \times \text{volume} \]

Given:

Density of acetic acid = $1.049 \, \text{g/mL}$
Volume of acetic acid = $28.00 \, \text{mL}$

\[ \text{mass} = 1.049 \, \text{g/mL} \times 28.00 \, \text{mL} = 29.372 \, \text{g} \]

Step 2: Calculate the Moles of Acetic Acid

Next, we calculate the number of moles of acetic acid using its molar mass. The molar mass of acetic acid ($\text{CH}_3\text{COOH}$) is approximately $60.05 \, \text{g/mol}$.

\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{29.372 \, \text{g}}{60.05 \, \text{g/mol}} = 0.4892 \, \text{mol} \]

Step 3: Calculate the Molarity of the Solution

Finally, we calculate the molarity of the solution. Molarity ($M$) is defined as the number of moles of solute per liter of solution.

Given:

Total volume of the solution = $290.0 \, \text{mL} = 0.2900 \, \text{L}$

\[ M = \frac{\text{moles of acetic acid}}{\text{volume of solution in liters}} = \frac{0.4892 \, \text{mol}}{0.2900 \, \text{L}} = 1.6869 \, \text{M} \]