Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25°C. Calculate the molarity of a solution of acetic acid made by dissolving 10.00 mL of glacial acetic acid at 25°C in enough water to make 150.0 mL of solution.
density glacial acetic acid is 1.049 g/mL. How many grams acetic acid do we have in the 10.00 mL glacial acetic acid? That will be 1.049 g/mL x 10.00 mL = 10.49 grams. How many moles is that?
10.49/60 (you look up the molar mass and do it exactly following my estimates) = 0.17 moles.
M = moles/L
M = 0.17/0.150 L = ??
0.350
0.8418
about 1.16 M
To calculate the molarity of a solution, you need to know the moles of the solute (acetic acid) and the volume of the solution in liters.
First, let's calculate the moles of acetic acid:
Given that the density of glacial acetic acid is 1.049 g/mL, we can convert the volume of glacial acetic acid to grams:
10.00 mL x 1.049 g/mL = 10.49 g
Next, we need to convert the grams of acetic acid to moles. To do this, we use the molar mass of acetic acid, which is 60.05 g/mol:
10.49 g / 60.05 g/mol = 0.1748 mol
Now, we calculate the molarity:
Molarity (M) is defined as moles of solute divided by liters of solution. In this case, we have 150.0 mL of solution, which is equal to 0.1500 L.
Molarity (M) = moles of solute / liters of solution
Molarity = 0.1748 mol / 0.1500 L
Molarity = 1.165 M
Therefore, the molarity of the acetic acid solution is 1.165 M.