Question

Molarity problem:

Assuming the density of vinegar is 1.0 g/mL, what is the molarity of vinegar? (Use the percent by mass of acetic acid in vinegar from question 1 which is 5.3%). The molar mass of acetic acid is 60.05 g/mol.

Answer 1

Density = 1.0 g/mL

5.3 (w/w)% => 5.3 / 100 => 0.053

molar mass acetic acid = 60.05 g/mol

M = 1000. w . D / molar mass

M = 1000 . 0.053 . 1.0 / 60.05

M = 53 / 60.05

M = 0.882 mol/L

hope this helps!

Answer 2

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution.

5.3 % = 5.3 g acetic acid / 100 g solution

Molarity = 5.3 g acetic acid / 100 g solution ( 1.0 g solution / .001 L solution ) ( 1 mol acetic acid/ 60.05 g acetic acid ) = 0.8826 mol / L solution