Question

Calculate the molarity of a solution of acetic acid made by dissolving 16.00 mL of glacial acetic acid at 25 ∘C in enough water to make 230.0 mL of solution.

Answer 1

Answer : The molarity of solution is, 1.216 g/mole

Explanation : Given,

Density of acetic acid = 1.049 g/ml  (standard value)

Volume of acetic acid = 16.00 ml

Volume of solution = 230.0 ml

Molar mass of acetic acid $CH_3COOH$ = $2(12)+4(1)+2(16)=60g/mole$

First we have to calculate the mass of acetic acid.

$Mass=Density\times Volume$

$Mass=1.049g/ml\times 16ml=16.784g$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of acetic acid}\times 1000}{\text{Molar mass of acetic acid}\times \text{volume of solution in ml}}$

$Molarity=\frac{16.784g\times 1000}{60g/mole\times 230ml}$

$Molarity=1.216g/mole$

Therefore, the molarity of solution is, 1.216 g/mole

Answer 2

The molarity of solution of acetic acid made by dissolving 16.00mL of glacial acetic acid is 1.22 g/mol.

How to calculate molarity?

The molarity of a solution can be calculated by dividing the number of moles of the substance by its volume.

However, we need to calculate the mass of acetic acid as follows:

• Molar mass of acetic acid = 60g/mol
• Density of acetic acid = 1.049g/mL
• Volume of acetic acid = 46mL

Mass = density × volume

Mass = 1.049 × 16

Mass = 16.784g

Molarity of acetic acid = 16.784 × 1000/60 × 230

Molarity of acetic acid = 1.22M

Therefore, the molarity of solution of acetic acid made by dissolving 16.00mL of glacial acetic acid is 1.22 g/mol.

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