Question

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 240.0 mL of solution. Please Explain.

Answer 1

We know that molarity = moles of solute / volume of solution in L

In order to find moles of solute, we need to calculate the mass of the solute first.

We know that density = mass / volume

Hence, mass = volume * density

volume = 10 ml , density = 1.049g/ml

Hence, mass = 10 * 1.049 g = 10.49 g

Now, we obtain the molecular weight of acetic acid = 60 g / mol

Moles = weight of subtance / mol. wt of substance = 10.49 / 60 = 0.175 moles

Thus we have found out the number of moles present.

Now, we calculate molarity using M = moles / volume

volume = 240 mL = 0.24 L

Hence, molarity M = 0.175 / 0.24 = 0.729 M

In case you have any doubts, let me know in the comments.