Pure acetic acid (HC₂H₃O₂) is a liquid and is known as glacial acetic acid. Calculate the molarity of a solution prepared by dissolving 10.00 mL of glacial acetic acid at 25°C in sufficient water to give500.0 mL of solution. The density of glacial acetic acid at 25°C is 1.049 g/mL. A) 21.0 B) 0.350 C) 0.0210 D) 1.26 × 10³ E) 3.50 × 10^-⁴

Question

Quizplus · Accepted Answer

First, we need to convert the volume of glacial acetic acid from mL to L:
10.00 mL = 10.00/1000 L = 0.01000 L

Next, we need to calculate the number of moles of glacial acetic acid in this volume, using its density and molar mass:

mass of glacial acetic acid = volume x density = 0.01000 L x 1.049 g/mL = 0.01049 g

molar mass of glacial acetic acid = 60.05 g/mol

number of moles of glacial acetic acid = mass/molar mass = 0.01049 g/60.05 g/mol = 0.0001748 mol

Finally, we can calculate the molarity of the solution using the volume of the solution in L:

molarity = number of moles/volume of solution = 0.0001748 mol/0.5000 L = 0.3496 M

The answer should be rounded to three significant figures, giving:

Answer: B) 0.350

The other answer choices are either too high (A, D, and E) or too low (C) and do not reflect the correct molarity of the solution.

Pure Acetic Acid (HC2H3O2) Is a Liquid and Is Known

Pure Acetic Acid (HC₂H₃O₂) Is a Liquid and Is Known