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

Step-by-step 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* 5.00mL)=5.25g`

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

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \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}=(5.25g* 1000)/(60.052g/mol* 500.0mL)\\\\\text{Molarity of solution}=0.175M`

Hence, the molarity of glacial acetic acid is 0.175 M