Asked by Aimee
A aqueous solution of acetic acid with a density of 1.25 g/ml is 30.0% by mass HC2h3O2 (MM=60.0) Compute the molarity of this solution.
I figured out that if you divide 30.0 by the molar mass 60.0g which gives you .5 mol. I cannot seem to figure out what i change 1.25 g/ml into for the molarity equation.
I figured out that if you divide 30.0 by the molar mass 60.0g which gives you .5 mol. I cannot seem to figure out what i change 1.25 g/ml into for the molarity equation.
Answers
Answered by
DrBob222
Take a liter of solution. How much does it weigh?
1000 mL x `1.25 g/mL = 1250 grams.
How much of that is HC2H3O2? It is 1250 x 0.30 = ??
How many moles in that?
??/molar mass HC2H3O2.
M = mols/L. You have the moles and you have it in one liter. Voila!
1000 mL x `1.25 g/mL = 1250 grams.
How much of that is HC2H3O2? It is 1250 x 0.30 = ??
How many moles in that?
??/molar mass HC2H3O2.
M = mols/L. You have the moles and you have it in one liter. Voila!
Answered by
Elizabeth
Calculate the molarity of a solution that contains 64.0 g Al2(SO4)3in 340.5 mL water.
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