Question

"Pure acetic acid (HC2H3O2) is a liquid and is known as glacial acetic acid. Calculate the molarity of a solution prepared by dissolving 5.00 mL of glacial acetic acid at 25 °C in sufficient water to give 500.0 mL of solution. The density of glacial acetic acid at 25 °C is 1.05 g/mL. Pure acetic acid (HC2H3O2) is a liquid and is known as glacial acetic acid. Calculate the molarity of a solution prepared by dissolving 5.00 mL of glacial acetic acid at 25 °C in sufficient water to give 500.0 mL of solution. The density of glacial acetic acid at 25 °C is 1.05 g/mL.
A. 3.50 × 10-5 M
B. 126 M
C. 0.0350 M
D. 2.10 M
E. 2.10 × 10-3 M

Answer 1

Answer: The molarity of glacial acetic acid is 0.175 M

Explanation:

To calculate the mass of acetic acid, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of acetic acid = 1.05 g/mL

Volume of acetic acid = 5.00 mL

Putting values in above equation, we get:

$1.05g/mL=\frac{\text{Mass of acetic acid}}{5.00mL}\\\\\text{Mass of acetic acid}=(1.05g/mL\times 5.00mL)=5.25g$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of solute (acetic acid) = 5.25 g

Molar mass of acetic acid = 60.052 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{5.25g\times 1000}{60.052g/mol\times 500.0mL}\\\\\text{Molarity of solution}=0.175M$

Hence, the molarity of glacial acetic acid is 0.175 M