Question

The label on a bottle of vinegar indicates that it is 3.5% acetic acid (CH3COOH). If the density of the solution is 1.90 g/mL, what is the molarity of the solution?

Answer 1

The molarity of a 3.5% acetic acid vinegar solution with a density of 1.90 g/mL is approximately 1.107 M. This calculation is based on the conversion of the mass of acetic acid in 1 liter of solution to moles and dividing by the volume in liters.

Step-by-step explanation:

To find the molarity of a vinegar solution that is 3.5% acetic acid (CH3COOH) with a density of 1.90 g/mL, we need to perform several calculations. First, let's determine the mass of acetic acid in 1 liter (or 1000 mL) of the solution, using the percent concentration and the density.

We assume:

• 
  
• 3.5% acetic acid means there are 3.5 grams of CH3COOH per 100 grams of solution.
• 
  
• Since the density is 1.90 g/mL, 1 liter of the solution weighs 1900 grams.

The mass of acetic acid in 1 liter is thus:
(3.5/100) × 1900 g = 66.5 g of CH3COOH

Next, we convert the mass of acetic acid to moles, knowing that the molar mass of CH3COOH is approximately 60.05 g/mol.

Number of moles of acetic acid = 66.5 g / 60.05 g/mol ≈ 1.107 moles

Now, by definition, molarity is the number of moles of solute per liter of solution. Hence, the molarity (M) of the vinegar solution is:

Molarity (M) = Number of moles / Volume (L) = 1.107 moles / 1 L = 1.107 M

Answer 2

[CH₃COOH] = 0.61 M

Step-by-step explanation:

3.5% acetic acid indicates that 3.5mL are contained in 100 mL of solution.

So we need acetic acid, density to solve this.

1.05 g/mL

Acetic acid density = acetic acid mass / acetic acid volume

1.05 g/mL = acetic acid mass / 3.5 mL

Acetic acid mass = 1.05 g/mL . 3.5mL = 3.675 grams

Let's convert this mass in moles ( mass / molar mass)

3.675 g / 60 g/m = 0.061 moles

As this moles, are contained in 100 mL of solution, molarity is (mol /L)

100 mL = 0.1L

Molarity (mol/L) = 0.061 m/ 0.1L = 0.61 M