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

= 20.98 g #tag_title# Step 2: Calculate the moles of acetic acid#tag_content# Now, we can calculate the number of moles of acetic acid using the formula moles = mass / molar mass. Molar mass of acetic acid (CH₃COOH) = 12.01 (C) + 4.03 (H) + 16.00 (O) × 2 = 60.05 g/mol Moles of acetic acid = (20.98 g) / (60.05 g/mol) = 0.349 moles #tag_title# Step 3: Calculate the molarity#tag_content# Finally, we can calculate the molarity using the formula molarity = moles / volume (in L). Volume of the solution = 250.0 mL = 0.250 L Molarity of the acetic acid solution = (0.349 moles) / (0.250 L) = 1.396 mol/L So, the molarity of the acetic acid solution is approximately \(1.396 \mathrm{~mol/L}\).

Step 1: Calculate the mass of acetic acid

We know the volume and density of the acetic acid, so we can calculate the mass using the formula mass = volume × density. Density of glacial acetic acid = 1.049 g/mL Volume of glacial acetic acid = 20.00 mL Mass of acetic acid = (1.049 g/mL) × (20.00 mL)