Final answer:
The molarity of 5.0% acetic acid in vinegar is calculated by determining the mass of acetic acid per liter of solution and then converting this mass to moles using the molar mass of acetic acid. The final molarity is determined to be approximately 0.84 M, which is nearest to option b. 0.71 M.
Step-by-step explanation:
The question asks for the molarity of acetic acid in vinegar with a 5.0% acetic acid concentration by mass and a density of 1.01 g/mL. To calculate the molarity, we first find the mass of acetic acid in 1 liter (1000 mL) of vinegar. Since the density is 1.01 g/mL, 1 liter of vinegar weighs 1.01 g/mL × 1000 mL = 1010 g. Thus, 5.0% of this mass is acetic acid, which equals 1010 g × 0.05 = 50.5 g of acetic acid.
Next, we calculate the number of moles of acetic acid using its molar mass, which is 60.05 g/mol. The moles of acetic acid are 50.5 g / (60.05 g/mol) which equals approximately 0.8406 moles. The molarity of the solution is the number of moles of solute (acetic acid) per liter of solution. Therefore, the molarity is 0.8406 moles / 1 liter = 0.84 M, which is closest to the option b. 0.71 M.