Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
AP Chemistry Molarity and Molality problem?
A solution of acetic acid and water contains 205.0 g L21 of acetic acid and 820.0 g L21 of water.
(a) Compute the density of the solution.
(b) Compute the molarity, molality, mole fraction, and mass percentage of acetic acid in this solution.
(c) Take the acetic acid as the solvent, and do the same for water as the solute.
1 Answer
- Roger the MoleLv 77 years agoFavorite Answer
Supposing "g L21" to mean, bizarrely, "grams per liter":
(a) Supposing there is nothing else in the solution:
(205.0 g/L) + (820.0 g/L) = 1025 g/L
(b)
(205.0 g/L acetic acid) / (60.052 g acetic acid/mol) = 3.414 mol/L acetic acid
(205.0 g acetic acid) / (60.052 g acetic acid/mol) / (820.0 g H2O) x (1000 g H2O) =
4.163 m acetic acid
(820.0 g H2O) / (18.01532 g H2O/mol) = 45.5168 mol H2O
(205.0 g acetic acid) / (60.052 g acetic acid/mol) = 3.41371 mol acetic acid
(3.41371 mol acetic acid) / (3.41371 mol acetic acid + 45.5168 mol H2O) =
0.06977 [the mole fraction of acetic acid]
(205.0 g acetic acid) / (205.0 g acetic acid + 820.0 g H2O) = 0.2000 = 20.00% acetic acid by mass
(c)
(820.0 g/L H2O) / (18.01532 g H2O/mol) = 45.52 mol/L H2O
(820.0 g H2O) / (18.01532 g H2O/mol) / (205.0 g acetic acid) x (1000 g acetic acid) = 222.0 m H2O
Using the value for the mole fraction of acetic acid from above:
1 - 0.06977 = 0.9302 [the mole fraction of H2O]
Using the value for the percent by mass of acetic acid from above:
100% - 20.00% = 80.00% H2O
