Question

Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80 M acetic acid, what is the % by mass concentration of acetic acid in vinegar?

Answer 1

To find the % by mass concentration of acetic acid in vinegar, calculate the mass of acetic acid in one liter by multiplying its molarity by its molar mass, then divide by the total mass of the solution and multiply by 100%. The result is approximately 4.75%.

Step-by-step explanation:

To calculate the % by mass concentration of acetic acid in vinegar with a molarity of 0.80 M and solution density of 1010 g/L, follow these steps:

First, determine the molar mass of acetic acid (CH3CO2H), which is approximately 60.05 g/mol.

Next, calculate the mass of acetic acid in one liter of solution by multiplying the molarity by the molar mass (0.80 mol/L × 60.05 g/mol), which equals 48.04 g.

Since the solution density is 1010 g/L, the mass of 1 L of the solution is 1010 g. Calculate the % by mass concentration by dividing the mass of acetic acid by the total mass of the solution and then multiply by 100% to obtain the percentage (48.04 g / 1010 g) × 100%.

The result is approximately 4.75% by mass concentration of acetic acid in vinegar.

Answer 2

4.8 %

Step-by-step explanation:

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)

= 1010 g

% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %