The vinegar solution you can purchase at the store is about 5% by mass acetic acid. it will be easier for you to assess your results if you convert 5% by mass acetic acid from weight percent to molarity since your experimental results will be in molarity. show this conversion in your notebook. you will need to approximate the density of the vinegar solution to complete this conversion.

The answer to your question is: Molarity = 0.83

Explanation:

Data

5% by mass

Molarity = ?

density = 1.05 g/ml

Molecular weight (CH₃COOH) = 24 + 32 + 4 = 60 g

Process

- Percent by mass means 5 g of vinegar in 100 ml of solution

- Calculate the moles of vinegar in the solution

                         60 g of vinegar  1 mol

                           5 g  of vinegar    x

                           x = (5 x 1) / 60

                          x = 0.0833 mol of vinegar

Formula

Molarity = moles of vinegar / volume (liters)

Substitution

Molarity = 0.0833 / 0.1

Molarity = 0.83

The molar concentration of acetic acid in vinegar is 0.877 M.

Let the total weight of the solution be 100 g

Recall that;

%w/w = mass of solute/mass of solution × 100/1

But;

Mass of solution = mass of solute + mass of solvent

Let the mass of acetic acid be x

5  = x/100g  × 100/1

x = 5 g

Mass of water = 100 g - 5 g = 95 g

Since density of water = 1 g/ml, volume of water = 95 ml or 0.095 L

Molar mass of acetic acid = 60 g/mol

Number of moles of acetic acid =   5 g/60 g/mol = 0.0833 moles

But;

number of moles = concentration × volume

concentration = number of moles /volume

concentration = 0.0833 moles /0.095 L

= 0.877 M

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