Pure acetic acid ( $\mathrm { HC } _ { 2 }$ $\mathrm { H } _ { 3 }$ $\mathrm { O } _ { 2 }$ )is a liquid and is known as glacial acetic acid.Calculate the molarity of a solution prepared by dissolving 50.00 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^-1.

Question

Quizplus · Accepted Answer

To find the molarity, first calculate the mass of acetic acid using its volume and density: $50.00 \, \text{mL} \times 1.05 \, \text{g/mL} = 52.5 \, \text{g}$. Then, convert the mass to moles using the molar mass of acetic acid ($60.05 \, \text{g/mol}$): $52.5 \, \text{g} / 60.05 \, \text{g/mol} = 0.874 \, \text{mol}$. Finally, divide the moles by the volume of the solution in liters to find the molarity: $0.874 \, \text{mol} / 0.500 \, \text{L} = 1.75 \, \text{mol/L}$.

Pure Acetic Acid HC2\mathrm { HC } _ { 2 }HC2​

Pure Acetic Acid $\mathrm { HC } _ { 2 }$