A solution of vinegar is 5.16% by mass aetic acid, CH3CO2H, in water. Assuming the density of vinegar is 1.00 g/mL, what is the molarity of the acetic acid?
5.16% CH3COOH by mass means 5.16 g CH3COOH in 100 grams of solution. That is 5.16 g CH3COOH/100 g soln.
moles CH3COOH = 5.16/molar mass CH3COOH = ??
g soln = 100 g. Density of 1.00 g/mL means the 100 g has a volume of 100 mL or 0.10 L.
So M = moles from above/0.1 L = ??
To find the molarity of acetic acid in the solution, you need to first determine the mass of acetic acid present in 1 liter of the solution.
Given that the solution is 5.16% acetic acid by mass, we know that for every 100 grams of vinegar, 5.16 grams is acetic acid.
So, in 1 liter (1000 mL) of vinegar, the mass of acetic acid can be calculated as follows:
Mass of acetic acid = (5.16 g/100 g) x 1000 mL = 51.6 g
Now, we can convert the mass of acetic acid to moles using its molar mass, which is 60.05 g/mol.
Moles of acetic acid = (51.6 g) / (60.05 g/mol) = 0.859 moles
Finally, we can find the molarity of acetic acid in the solution by dividing the moles of acetic acid by the volume of the solution in liters. Since the density of vinegar is 1.00 g/mL, the volume of the solution is 1 liter.
Molarity of acetic acid = Moles of acetic acid / Volume of solution
= 0.859 moles / 1 L
= 0.859 M
Therefore, the molarity of acetic acid in the vinegar solution is 0.859 M.
To find the molarity of acetic acid in the solution, we need to first calculate the mass of acetic acid present in a given volume of the solution.
1. We are given that the solution is 5.16% by mass acetic acid. This means that in 100 grams of the solution, there are 5.16 grams of acetic acid.
2. Since the density of vinegar is 1.00 g/mL, we can assume that 100 mL of vinegar weighs 100 grams.
3. Therefore, in 100 mL of the solution, there are 5.16 grams of acetic acid.
4. To find the molarity, we need to know the molecular weight of acetic acid. The molecular weight of CH3CO2H is calculated as follows:
(1 * 12.01) + (3 * 1.01) + 12.01 + 16.00 + (1 * 1.01) + (1 * 16.00) = 60.05 g/mol
5. Now, we can calculate the number of moles of acetic acid in 100 mL of the solution:
(5.16 g / 60.05 g/mol) = 0.08598 mol
6. The molarity is defined as the number of moles of solute divided by the volume of the solution in liters. Since we have the moles and we know that 100 mL is equal to 0.1 L, we can calculate the molarity of acetic acid:
Molarity = (0.08598 mol / 0.1 L) = 0.86 M
Thus, the molarity of acetic acid in the vinegar solution is 0.86 M.