Final answer:
The molarity of 5.55% acetic acid in vinegar with a density of 1.01 g/mL is 0.934 M. This is calculated by converting percentage concentration to grams, then to moles, and finally to molarity.
Step-by-step explanation:
To calculate the molarity of acetic acid in vinegar with a concentration of 5.55%, we need to know the density of vinegar, which is given as 1.01 g/mL. Molarity is defined as moles of solute per liter of solution. Here's how to calculate it:
First, convert the percentage concentration to grams per liter. Since the density is 1.01 g/mL, 100 mL would have a mass of 101 g.
Therefore, 5.55% of 101 g is the mass of acetic acid:
5.55% of 101 g = 5.61 g
Next, convert the mass of acetic acid to moles by dividing by the molar mass of acetic acid, which is approximately 60.05 g/mol:
5.61 g CH3COOH / 60.05 g/mol = 0.0934 moles
Since this is per 100 mL, we need to convert it to liters by dividing by 1000 mL/L to get moles per liter (molarity):
0.0934 moles / 0.1 L = 0.934 M
Therefore, the molarity of acetic acid in vinegar is 0.934 M.