Pure acetic acid (HC₂H₃O₂)is a liquid and is known as glacial acetic acid.Calculate the molarity of a solution prepared by dissolving 1.00 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.

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The correct answer is D.The molarity of a solution is defined as the number of moles of solute per liter of solution. To calculate the molarity of a solution, we need to know the number of moles of solute and the volume of the solution in liters.In this problem, we are given the mass of glacial acetic acid (1.00 g) and the density of glacial acetic acid (1.05 g/mL). We can use these values to calculate the volume of glacial acetic acid in milliliters:Volume of glacial acetic acid = mass of glacial acetic acid / density of glacial acetic acidVolume of glacial acetic acid = 1.00 g / 1.05 g/mLVolume of glacial acetic acid = 0.952 mLWe are also given the volume of the solution (500.0 mL). We can use these values to calculate the molarity of the solution:Molarity of solution = number of moles of solute / volume of solution in litersNumber of moles of solute = mass of solute / molar mass of soluteMolar mass of glacial acetic acid = 60.05 g/molNumber of moles of solute = 1.00 g / 60.05 g/molNumber of moles of solute = 0.0167 molVolume of solution in liters = 500.0 mL / 1000 mL/LVolume of solution in liters = 0.500 LMolarity of solution = 0.0167 mol / 0.500 LMolarity of solution = 0.0350 MTherefore, the molarity of the solution is 0.0350 M.

Pure Acetic Acid (HC2H3O2)is a Liquid and Is Known as Glacial

Pure Acetic Acid (HC₂H₃O₂)is a Liquid and Is Known as Glacial