When 80.4 moles of calcium acetate are
dissolved in enough water to make 222.2
milliliters of solution, what is the molarity of acetate ions?
Answer in units of M
I already tried moles/liter and it was wrong
I expect you calculated M Ca(Ac)2. Take your answer for mols Ca(Ac)2 and multiply by 2 to find mols Ac^-.
(80.4/0.2222)*2 = ? and watch the number of significant figures.
To find the molarity of acetate ions, you need to calculate the number of moles of acetate ions and then divide it by the volume in liters.
First, we need to determine the number of moles of calcium acetate. Given that 80.4 moles of calcium acetate are dissolved in 222.2 milliliters of solution, we have the number of moles (80.4) and the volume in liters (0.2222 L) (since there are 1000 milliliters in a liter).
Next, we need to determine the number of moles of acetate ions present in the calcium acetate. Calcium acetate dissociates in water to form one calcium ion (Ca2+) and two acetate ions (C2H3O2-). Since the calcium ion is not relevant to finding the molarity of acetate ions, we can ignore it and focus on the acetate ions.
Since there are two acetate ions for every one calcium acetate molecule, the number of moles of acetate ions is twice the number of moles of calcium acetate.
Therefore, the number of moles of acetate ions is 2 * 80.4 = 160.8 moles.
Finally, to find the molarity of acetate ions, divide the number of moles of acetate ions by the volume in liters:
Molarity = moles of acetate ions / volume in liters
Molarity = 160.8 moles / 0.2222 L
Calculating this gives a molarity of approximately 723.25 M (to four significant figures). Therefore, the molarity of acetate ions in the given solution is 723.25 M.
Sarah, Nick, Joe -- you must be having an identity crisis.
Please use only one name for your posts.