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Find the molarity of the resulting mixture when two solutions of different molarity mixed togather and volume are additive

Submitted by amysmith on Tue, 04/23/2013 - 21:01


Please show me step by step. Thank you.
4. Two sulfuric acid solutions are mixed as follows: 25.0 mL of a 0.50 M sulfuric acid solution are
added to 0.075 L of a 0.25 M sulfuric acid solution. What is the molarity of the resulting mixture?
(Assume volumes are additive) Answer: 0.32 M
6. What is the molarity of a 2.80%(m/v) ammonia solution? (Molar mass NH3 = 17.03 g/mole)
Ans: 1.64 M
8. A solution of ethanol, C2H6O, is prepared by dissolving 14.0 g C 2H6O in 100.0 g water. (Molar mass
ethanol = 46.07 g/mole) a. Find the molality of the solution. Answers: a. 3.04 m
                        b. Find the % (m/m) concentration of the solution . Answers; b. 12.3% (m/m)
10. 76.91 mL of a 0.556 M oxalic acid solution are required to react with 28.43 mL of a sodium
hydroxide solution. What is the molar concentration of the sodium hydroxide solution? (Molar
masses: H2C2O4 = 90.04 NaOH = 40.00 H2O = 18.02 Na2C2O4 = 134.00)
H2C2O4(aq) + 2 NaOH(aq)  2 H2O(l) + Na2C2O4(aq) Answer: 3.01 M
12. 2.489 g of cesium oxalate are dissolved in enough water to give 265.0 mL of a solution whose
density is 1.0528 g/mL. What is the percent by mass concentration of the solution? (molar mass
cesium oxalate = 353.83 g/mole) Answer: 0.8921% (m/m)
14. What is the molar concentration of a 3.296 m aqueous solution of fructose? The density of the
solution is 1.203 g/mL. (molar mass fructose = 180.16 g/mole) Answer: 2.488 M
16. The density of a 60.00 % (m/m) ethanol, C 2H5OH, solution is 0.8937 g/mL. What is the molarity
of the solution? (molar mass = 46.07 g/mole) Answer: 11.64 M
18. An aqueous solution of citric acid, H3C6H5O7, is 0.655 M and has a density of 1.049 g/mL. What
is the molality of the solution? (molar mass = 192.13 g/mole) Answer: 0.710 m
20. Calculate the molality of a 23.0% by mass acetic acid, HC 2H3O2, solution. (molar mass = 60.05
g/mole) Answer: 4.97 M

4.) Solve for the number of moles of sulfuric acid in each solution first:

25.0 mL x (1 L / 1000 mL) x (0.50 mol / L) = 0.0125 mol {Note: 0.013 mol with correct significant figures}

0.075 L x (0.25 mol / L) = 0.01875 mol {Note: 0.019 mol with correct significant figures}

Now, add the moles together to get the total moles of sulfuric acid in the mixture:

0.0125 mol          -OR-          0.013 mol
+0.01875 mol     -OR-          +0.019 mol
= 0.03125 mol    -OR-          =0.032 mol

Finally, divide the total moles by the total volume of the mixture in liters (the sum of the two volumes in L):

0.03125mol / (0.025 L + 0.075 L) = 0.3125 M rounded to 0.31 M with correct significant figures     -OR-
0.032 mol / (0.025 L + 0.075 L) = 0.32 M with correct significant figures.

{Note about significant figures: I always teach my students to NEVER round their answers until the very end of a problem. However, based on the answer you list, your teacher (or textbook) must round after each step in the problem. That is why I showed the work in both ways - in the hopes that you would see where the rounding occurred to give the answer of 0.32 M instead of 0.31 M (which I feel is the more accurate answer).}

Good luck! Sorry I don't have time to answer more of the questions tonight.

~Dr. SmileyTR :)