Question

For a solution of acetic acid (CH3COOH) to be called “vinegar,” it must contain 5.00% acetic acid by mass. If a vinegar is made up only of acetic acid and water, what is the molarity of acetic acid in the vinegar? The density of vinegar is 1.006 g/mL

Answer 1

Answer:

• M = 0.838 M

Explanation:

1) Data:

a) Acetic acid CH₃COOH

b) molar mass of CH₃COOH: 60.052 g/mol

c) %: = 5.00%

d) d = 1.006 g/ml

e) M = ?

2) Formulae:

a) M = n of solute / V of solution in liters

b) n = mass in grams / molar mass

c) % = (mass of solute / mass of solution)×100

d) d = mass  / volume

3) Solution:

a) Assume a basis

• V = 1 liter of solution

b) Calculate the mass of that basis (1 liter of solution)

• d = mass / V ⇒ mass = d × V

• mass = 1.006 g/ml × 1 liter × 1,000 ml/liter = 1006 g

c) Calculate the mass of solute

• % = (mass of solute / mass of solution ) × 100 ⇒

        mass of solute = % × mass of solution / 100 = 5.00% × 1,006 g / 100

        mass of solute = 50.3 g

d) Calculate the number of moles of solute

• n = mass in grams / molar mass = 50.3 g / 60.052 g/mol =  0.838 mol

e) Calculate molarity

• M = n / V in liter = 0.838 mol / 1 liter = 0.838 M ← answer