Question

after titration, 25 ml of a vinegar solution was found to contain 0.03 moles of acetic acid. using a density of 1.00 g/ml, what is the mass percentage of acetic acid in this vinegar sample? report your answer to three decimal places.

Answer 1

The mass percentage of acetic acid in the vinegar sample is 7.204%.

Step-by-step explanation:

To calculate the mass percentage of acetic acid in the vinegar sample, we need to find the mass of acetic acid in the 25 mL of solution. Since the density of the vinegar solution is given as 1.00 g/mL, the mass of the solution is 25 g (25 mL x 1.00 g/mL). From the given information, we know that the solution contains 0.03 moles of acetic acid. The molar mass of acetic acid is 60.052 g/mol. Therefore, the mass of the acetic acid in the solution is 0.03 moles x 60.052 g/mol = 1.801 g.

The mass percentage of acetic acid can be calculated using the equation:

Mass Percentage = (Mass of Acetic Acid / Mass of Solution) x 100

We can substitute the values into the equation:

Mass Percentage = (1.801 g / 25 g) x 100 = 7.204%