Question

Chemistry question about Molarity?

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25°C. Calculate the molarity of a solution of acetic acid made by dissolving 40.00 mL of glacial acetic acid at 25°C in enough water to make 250.0 mL of solution.

Answers

find mass of acid used:

40.00 ml @ 1.049 g/ml = 41.96 grams of acetic acid

use molar mass to find moles used:

41.96 grams of acetic acid @ 60.05 g/mol = 0.69875 moles of acid

find molarity

0.69875 moles of acid / 0.2500 Litres = 2.7950 Molar

your answer, rounded to 4 sig figs is

2.795 Molar Acetic acid

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