Question

Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted with NaOH. a. Suppose that the molarity of acetic acid in vinegar is 0.5 M, how many grams of acetic acid are present in the 5 ml of vinegar solution used in the titration? (molar mass of acetic acid = 60 g/mol). b. If vinegar has a density of 1.01 g/mol. Calculate the % of acetic acid in vinegar by mass.

Answer 1

a) First we need to calculate no. of moles of acetic acid in the 5 ml of vinegar solution.

We have relation, Molarity = No. of moles of solute / volume of solution in L

$\therefore$ No. of moles of acetic acid = [ acetic acid ] $\times$ volume of solution in L

$\therefore$ No. of moles of acetic acid = 0.5 mol / L $\times$ 0.005 L = 0.0025 mol

We know that , no. of moles = Mass / molar mass

$\therefore$ Mass of acetic acid = No. of moles of acetic acid $\times$ Molar mass

$\therefore$ Mass of acetic acid = 0.0025 mol $\times$ 60 g / mol = 0.15 g

ANSWER : 0.15 g acetic acid is present in 5 ml of vinegar solution.

b) We have formula, % mass = [ Mass of solute / mass of solution ] $\times$ 100

$\therefore$ % mass acetic acid = [ Mass of acetic acid / mass of vinegar solution ] $\times$ 100  

We have , density = Mass / volume

$\therefore$ Mass of vinegar solution = density $\times$ volume

$\therefore$ Mass of vinegar solution = 1.01 g /ml $\times$ 5 ml = 5.05 g

$\therefore$ % mass acetic acid = [ 0.15 g / 5.05 g ] $\times$ 100

$\therefore$ % mass acetic acid = 2.97

ANSWER : % mass acetic acid = 2.97