Question

Assuming that the vinegar solution has a density of 1.05 g/mL, determine the percentage (weight/weight) of acetic acid in the vinegar solution. (It has a 0.275 molarity and there are 1.33 Liters of it, or 22.0 grams of acetic acid in it)

Answer 1

concentration of acetic acid in vinegar = 0.275 M

no. of moles of acetic acid in 1L vinegar solution = 0.275 mol

mass of acetic acid (molar mass = 60.05 g/mol) in 1L vinegar solution = (0.275 mol)(60.05 g/mol) = 16.51 g

density of vinegar solution = 1.05 g/mL

mass of 1 L vinegar solution = density x volume = (1.05 g/mL)(1000 mL) =1050 g

Therefore, 1050 g of vinegar solution has 16.51 g acetic acid.

Hence, percentage of aetic acid(w/w) in vinegar solution = (16.51 g/ 1050g ) x 100 = 1.57 %