Chemistry I-Honors
Solution Concentration Problem Set #4
Solution Set
1. If magnesium hydroxide is completely neutralized with 35.0 ml of a 1.25 M solution of hydrochloric acid,
how many equivalents were there of this base?
(0.0350 liter)(1.25 mol/liter) = 0.0438 mol H+ = 0.0438 mol OH-1 = 0.0438 equivalents OH-1
2. How many milliliters of water must be added to a 3.00 M solution to end up with 2.00 liters of a 0.455 M
solution? M1V1 = M2V2 & Vwater = V2 - V1
V1 = (0.455 M)(2000 ml) / 3.00 M = 303 ml , then Vwater = 1697 ml = 1.70 x 103 ml
3. What is the normality of a solution in which 11.2 liters of hydrogen chloride gas (at STP) are dissolved in
enough water to make 400.0 ml of solution?
(11.2 liters / 22.4 liters/mole) = 0.500 mol / 0.400 liter soln = 1.25 M = 1.25 N
4. If a beaker contains 745 ml of a 0.220 M solution of potassium hydroxide, how many grams of the base had
been dissolved?
(0.745 liter)(0.220 mol / liter soln)( 56.1 g/mol) = 9.19 g KOH
5. What is the molality of a solution in which 2.50 moles of sucrose are dissolved in 925ml of ethyl alcohol?
The density of pure ethyl alcohol is 0.870 g/ml.
925 g x 0.870 g/ml = 805 g = 0.805 kg
m = 2.50 mol / 0.805 kg = 3.11 m
6. How many moles of argon are there in a two-gas solution in which the mole fraction of helium (the other
gas) is 0.31? The total number of moles is 12.00.
nAr
if XHe = 0.31, then XAr = 0.69 = ------- then nAr = 8.3
12.00
7. What is the molarity of a 0.436 N solution of aluminum hydroxide?
(0.436 eq / 1 liter soln)( 1 mol / 3 eq ) = 0.145 M
8. A 1.000 M aqueous solution of silver nitrate has a density of 1.113 g/ml. What is the molality of the
solution?
1.000 M solution contains 1 mol AgNO3 = 169.9 grams
1 liter x 1.113 g/ml = 1113 grams solution - 169.9 grams solute = 943 grams water
m = 1.000 mol / 0.943 kg = 1.06 m
9. If 30.0 ml of a 8.05 M solution is diluted to a final volume of 5.00 liters, what is the molarity of the final,
diluted solution?
M2 = (8.05 M)(30.0 ml) / 5000 ml = 0.0483 M
-210. A 4.000-gram mineral sample that contained KCl 16.9% by weight was dissolved in 850.0 ml of water at
room temperature. What was the molality of the solution? The density of the solution was 1.30 g/ml.
[(4.000 g)(16.9/100)(1 mol/74.6 g cpd)] / 0.850 kg =
0.0107 m
11. How many grams of water must be used in order to dissolve 100.0 grams of sodium perchlorate and
produce a 0.335-molal solution?
NaClO4
(100.0 g cpd)( 1 mol cpd/ 122.5 g)( 1000 g H2O / 0.335 mol) = 2440 g H2O
12. How many cubic decimeters of a 0.100 M solution can be prepared by dissolving 200.0 grams of sodium
bicarbonate in water?
1 dm3 = 1 liter
NaHCO3
(200.0 g cpd)(1 mol cpd / 84.0 cpd)( 1 liter / 0.100 mol) = 23.8 dm3
13. What is the normality of a solution in which 45.0 grams of calcium nitrate are dissolved in enough water to
make 2.25 liters of solution?
[(45.0 g cpd)(1 mol cpd / 164.1 g cpd)(2 eq / 1 mol cpd)] / 2.25 liters soln = 0.244 N
14. How many moles of phosphoric acid are in 500.0 ml of a 2.75 M solution? How many equivalents are in
the same solution?
(0.500 liters)(2.75 M) = 1.38 moles acid
=
4.13 eq H+
15. How many grams of water must be used to dissolve 165.0 grams of potassium chloride in order to produce
a 0.750 m solution?
(165.0 g KCl)( 1 mol / 74.6 g KCl)( 1000 g H2O / 0.750 mol) = 2950 gram H2O
16. If 85.0 grams of carbon dioxide gas is dissolved into 1.00 liter of water, what is the mole fraction of the
solute?
(85.0 g / 44.0 g/mol)
1.93
Xcarbon dioxide = ------------------------------------------------------ = ------ = 0.0336
(85.0 g / 44.0 g/mol) + (1000 g / 18.0 g/mol)
57.5
17. A 12.0 M solution of hydrochloric acid is 35.0% acid by weight. What is the density of this solution?
(12.0 mol HCl / 1000 ml soln)(36.5 g HCl / 1 mol HCl)( 100 g soln / 35.0 g HCl) = 1.25 g/ml
18. You are to prepare 750 ml of a 0.50 M nitric acid solution from concentrated nitric acid, which is 16.0 M.
How many milliliters of the concentrated acid are required?
V1 = (0.50 M)(750 ml) / 16.0 M = 23.4 ml