Pure acetic acid (HC₂H₃O₂)is a liquid and is known as glacial acetic acid. Calculate the molarity of a solution prepared by dissolving 10.00 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)1.26 × 10³ B)21.0 C)0.0210 D)0.350 E)3.50 × 10^-4

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of glacial acetic acid: moles HC₂H₃O₂ = mass ÷ molar mass mass = density × volume = 1.05 g/mL × 10.00 mL = 10.5 g molar mass = 60.05 g/mol moles = 10.5 g ÷ 60.05 g/mol = 0.175 mol Next, we can calculate the molarity: molarity = moles ÷ volume volume = 500.0 mL ÷ 1000 mL/L = 0.500 L molarity = 0.175 mol ÷ 0.500 L = 0.350 M Therefore, the best choice is D because it is the only option that matches our calculation.

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

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