CHM1045 GEN CHEM 1 ONLINE 596089 888 ?? Grades Communication?\' Assessments, Too

Question

CHM1045 GEN CHEM 1 ONLINE 596089 888 ?? Grades Communication?' Assessments, Tools? Resourcesv Left:0:30:17 Corey Robinson: Attempt 1 Question 4 (1 point) Pure acetic acid (HC2H302) is a liquid and is known as glacial acetic acid. Calculate the molarity of a solutibn 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. 1.26 x 103 M 21.0 M 0.0210 M 0.350 M 3.50 x 10-4 M F6 ?? O F3 9 4 3

Explanation / Answer

Molarity = number of moles / volume of solution in liters

Weight of acetic acid = 10*1.05 = 10.5 g

Molecular weight of acetic acid = 60.5 gm / mole

Number of moles = 10.5/60.5= 0.1735 moles

Molarity = 0.1735/0.5 = 0.3471M = 0.350 M

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