Calculate the molarity of a solution of acetic acid made by dissolving 28.00mL of glacial acetic acid at 25 ∘C in enough water to make 250.0mL of solution?
Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25 ∘C.
Express your answer using four significant figures.
Help please
Calculate the molarity of a solution of acetic acid made by dissolving 28.00mL of glacial acetic acid at 25 ∘C?
2014-03-03 4:49 pm
回答 (1)
2014-03-03 5:14 pm
✔ 最佳答案
28.00 mL HAc X 1.048 g/mL / 60.05 g/mol = 0.4886 mol acetic acid0.04886 mol / 0.2500 L = 1.955 mol/L
收錄日期: 2021-04-20 21:37:06
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