Chemistry I-Honors
Solution Concentration Problem Set #1
Solution Set
1. How many grams of NaNO3 must be weighed out to make 50.0 ml of an aqueous solution
containing 70.0 mg Na+1 per ml?
50.0 ml soln
0.0700 g Na+1
x -------------------- x
1 ml soln
85.0 g NaNO3
-------------------23.0 g Na
= 12.9 g NaNO3
2. Exactly 4.00 grams of a solution of sulfuric acid was diluted with water, and an excess of barium
chloride was added. A precipitate of barium sulfate formed. This precipitate was washed, dried,
and weighed. The weight of the barium sulfate was 4.08 grams. What was the percent by weight
of sulfuric acid in the original acid solution. (A toughie!)
(4.08 g BaSO4)(1 mol BaSO4 / 233.4 g BaSO4)(1 mol SO4-2 / 1 mol BaSO4) = 0.0175 mol SO4-2
(0.0175 mol SO4-2 )(1 mol H2SO4 /1 mol SO4-2)( 98.1 g H2SO4/1 mol H2SO4) = 1.71 g
1.71 g / 4.00 x 100 = 42.9% H2SO4 in the sample
3. How many equivalents of nitric acid are needed to neutralize 45.0 ml of a 5.00 M solution of
barium hydroxide?
(0.0450 liter)(5.00 mol/liter)(2 mole OH-1 / 1 mol cpd) = 0.450 mole OH-1
= 0.450 eq OH-1
= 0.450 equivalents of H+1
4. How many grams of calcium chloride should be added to 300.0 ml of water to make up a
2.46-molal solution?
(0.3000 kg H2O)(2.46 mole CaCl2 / 1 kg H2O)(111.1 g CaCl2 / 1 mol cpd) = 81.99 = 82.0 g
5. An aqueous solution labeled 35.0% perchloric acid had a density of 1.251 g/ml.
Calculate the following: A) the molar concentration of the acid solution;
B) the molality of the acid solution.
(1251 g soln / 1 liter soln)(35.0 g HClO4 / 100 g soln)(1 mol acid / 100.5 g acid) = 4.36
35.0 g HClO4 / 100.5 g/mol = 0.348 mol
0.348 mol / 0.0650 kg = 5.36
100 - 35.0 = 65.0 gram H2O = 0.0650 kg
m
6. 6.00 moles of magnesium oxide are added to water to make 2.50 liters of solution.
What is the normality of the solution?
MgO + H2O
Mg(OH)2
6.00 mole cpd / 2.50 liters soln = 2.40 M soln , then [OH-1] = 4.80
N
M
Chemistry I-Honors
Solution Concentration Problem Set #2
Solution Set
1. When 2.24 liters of HCl gas are dissolved enough water to make 2.00 liters of solution at STP
conditions, what is the molarity of the solution?
2.24 liters HCl / 22.4 liter/mol = 0.100 mole HCl
0.100 mole HCl
-------------------------- = 0.0500 M
2.00 liters
2. Assuming the solution to be at room temperature, and the density of the gas to be the same as the density
of air, what is the molality of the solution in #1?
Dair = 1.29 g/liter, then 2.24 liter x 1.29 g/liter = 2.89 g
2.00 liters soln.  2000 g soln, then mass of solvent = 2000 g - 2.89 g = 1.997 kg
m = 0.100 mol / 1.997 kg = 0.0501 m
3. How many grams of ferric hydroxide would be needed to neutralize 28.8 ml of a 3.32 N solution of
sulfurous acid?
(0.0288 liters)(3.32 eq H+/ 1 liter) = 0.0956 eq H+ =
(0.0956 eq OH-1)[1 mole Fe(OH)3][3 moles OH-1](106.8 g / 1 mol cpd) = 3.40 g Fe(OH)3
4. 6.00 grams of a solid sample is dissolved in water and neutralized with 40.00 ml of a 1.50 M solution of
KOH. The solid sample is known to contain formic acid (HCOOH) [MW of the acid is 46.0 g/mole].
The remainder of the sample is inert. What is the percent by weight of acid in the solid sample?
(0.0400 liter)(1.50 M) = 0.0600 mol
0.0600 mol x 46.0 g/mol = 2.76 g acid
(2.76 g / 6.00 g) x 100 = 46.0 %
5. How many equivalents are there in 48.00 grams of strontium hydroxide? Sr(OH)2
(48.00 g / 121.6 g/mol) = 0.3947 moles cpd = 0.7895 moles OH-1 = 0.7895 equivalents
6. If the manganese in KMnO4 is reduced to MnO2, how many moles of electrons (equivalents) are needed
for every mole of KMnO4 being reduced?
+7
+4
KMnO4
MnO2
3 moles of electrons are needed
7. How many moles of ferric hydroxide are needed to neutralize 526 ml of a 1.05 M solution of sulfuric
acid?
0.526 liter
Fe(OH)3
1.05 moles A
2 mol H+
1 mol OH-1
1 mol Fe(OH)3
x ----------------- x ------------ x -------------- x -------------------
= 0.368 mol
1 liter soln
1 mol A
1 mol H+
-2-
3 mol OH-1
8. In a two-gas system, the mole fraction of argon is 0.233 and the number of grams of carbon monoxide is
67.0. How many moles of argon are there?
if XAr = 0.233, then 0.767 = XCO
(67.0 g / 28.0 g/mol)
XCO = 0.767 = -------------------------(67.0 g / 28.0 g/mol) + nAr
=
2.39 moles
------------- =
2.39 + n
nAr = 0.730 moles
9. How many milliliters of 12.0 M nitric acid are needed to produce 450.0 ml of a 3.40 M solution?
M1V1 = M2V2
V1 = (3.40 M)(450.0 ml) / 12.0 M = 127.5 ml = 128 ml
10. If 750.0 ml of water is added to 250.0 ml of a 2.25 N solution of phosphoric acid, what is the normality
of the final solution?
N1V1 = N2V2
N2 = (2.25 N)(250.0 ml) / 1000.0 ml) = 0.565 N
11. What is the molarity of the final, diluted solution in #10?
0.565 eq
1 mol cpd
----------- x ------------- = 0.188 M
1 liter soln. 3 eq H+
12. How many milliliters of water would have to be added to the initial acid solution in #10 in order to
produce a solution that has a final diluted normality of 0.555N?
(2.26 N)(250.0 ml) / 0.558 N = V2 = 1013 ml, then Vwater = 1013 - 250 = 763
ml H2O
13. What is the molarity, normality, and molality of a solution of sodium phosphate if 75.0 grams are
dissolved in enough water to make 900.0 milliliters of solution? The solution has a density of 1.17
g/ml.
M = (75.0 g / 164.0 g/mol) / 0.900 liter = 0.508
N = 0.508 M x 3 eq / 1 mol = 1.52
M
N
900 ml x 1.17 g/ml = 1053 grams soln - 75 grams solute = 978 grams water
m = (75.0 g / 164.0 g/mol) / 0.978 kg = 0.468 m
Chemistry I-Honors
Solution Concentration Problem Set #3
Solution Set
1. When 25.0 grams of beryllium hydroxide is added to 5.00 liters of a 0.265 M solution of phosphoric
acid, how many equivalents of the acid are used in the reaction?
Be(OH)2
+
H3PO4
(25.0 g / 43.0 g/mol)
(5.00 liters)(0.265 moles/liter)
= 0.581 moles cpd
= 1.33 moles cpd
= 1.16 moles OH-1
= 3.98 moles H+
Base is the L.R, so only 1.16 moles of H+ react.
2. How many equivalents are there in 7.00 grams of chromium (III) oxide if the compound is completely
reduced to oxygen gas and the elemental metal?
+6
0
(ie. Cr2O3 --------> 2 Cr(s) + 3O2
- no need to balance)
(7.00 g)( 1 mol cpd / 152.0 g cpd)(6 moles e-1 / 1 mole cpd)
= 0.276 eq
3. If 3.35 grams of KOH are required to neutralize 55.0 ml of a solution of telluric acid, what was the
molarity of the original acid solution?
H2TeO4
3.35 g KOH / 56.1 g/mol = 0.0597 mol OH-1 = 0.0597 mol H+1
= 0.0299 mol H2TeO4 / 0.0550 liter = 0.543 M
4. When 400.0 ml of a 3.55 N solution phosphoric acid is diluted with 500.0 ml of water, what is the
molarity of the final, diluted solution? N1V1 = N2V2
N2 = ( 3.54 N)(400.0 ml) / 900.0 ml = 1.57 N
1.57 eq / 1 liter x 1 mol / 3 eq = 0.524 M
5. What is the molecular weight of a compound if 6.00 grams of the compound, when dissolved in 45.0
grams of water, produces a 0.615-molal solution?
0.615 m = 6.00 g / M.W. / 0.0450 kg
M.W. = 217 g/mol
-2-
6. What is the mole fraction of the solute if 67.0 grams of lithium fluoride is dissolved in 100.0 grams of
water?
(100.0 g / 18.0 g/mol)
5.56 mol
Xwater = ------------------------------------------------------ = ----------- = 0.682
(100.0 g / 18.0 g/mol) + (67.0 g / 25.9 g/mol)
8.15 mol
Then XLiF = 1 - 0.682 = 0.318
7. What is the density of a sodium chloride solution that is 3.45 N and is 12.0% by weight of solute?
3.45 N = 3.45 M
100 g soln
58.5 g NaCl
------------- x ---------------12.0 g NaCl
1 mol NaCl
3.45 mol NaCl
x ------------------1000 ml soln
= D = 1.68 g / ml
8. What is the molarity of a solution if 333.5 grams of sodium phosphate is dissolved in enough water to
produce 950.0 ml of solution?
(333.5 g / 164.0 g/mol)
-------------------------= 2.141 M
0.9500 liter
9. How many grams of calcium sulfate must be weighed out in order to end up with 67.0 grams of an
aqueous solution that is 32.8% by weight?
(67.0 g soln)( 32.8 g CaSO4 / 100 g soln) = 21.976 g = 22.0 g CaSO4
10. How much water is needed if dissolving 50.0 grams of potassium chloride produces a 2.28-molal
solution?
(50.0 g KCl )( 1 mol KCl / 74.6 g KCl )( 1 kg H2O / 2.28 mol KCl ) = 0.294 kg H2O
= 294 grams H2O
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
-2-
10. 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
Chemistry I-Honors
Colligative Problem Set #1
Solution Set
1. What is the boiling point of an aqueous solution if 3.00 moles of a nonelectrolyte is dissolved
in 400.0 g of water?
m = 3.00 mole / 0.400 kg = 7.50 m
t = kb. m = (0.512oC/ m)(7.50 m) = 3.84oC, then B. Pt = 103.84oC
2. What is the freezing point for the solution in problem #1?
t = kf. m = (1.86oC/ m)(7.50 m) = 13.95oC, then F. Pt = -14.0oC
3. If 30.0 grams of magnesium chloride (a strong electrolyte) is dissolved in 500.0 ml of water, what is the
freezing point of this solution? (Careful!)
MgCl2 M.Wt = 95.3 g/mole
m = 30.0 grams / 95.3 g/mole / 0.500 kg = 0.630 m

t = ikf. m = 3(1.86oC/ m)(0.630 m) = 3.51oC,
then F. Pt = -3.51oC
4. What is the boiling point constant for benzene (an organic solvent) if 4.98 grams of glucose, whose
molecular weight is 180.0 g/mole, is dissolved in 50.0 grams of benzene? The normal boiling point for
pure benzene is 80.1oC and the boiling point of the resulting solution is 81.5?
g x kb
MW = ---------- then
t x kg
MW x t x kg
kb = --------------------g
=
(180.0 g/mol)(1.4oC)(0.0500 kg)
----------------------------------------4.98 g
kb = 2.53 oC/ m
5. What is the molecular weight of a nonelectrolyte if 4.000 grams of this compound, when dissolved in
20.00 grams of water, will lower the freezing point of the solution by 3.80oC ?
g x kf
MW = ---------t x kg
then
(4.0000 g)(1.86oC/ m )
MW = ----------------------------(3.80oC)(0.0200 kg)
= 97.9 g/mol
6. An aqueous solution of ethanol (a nonelectrolyte) freezes at a temperature of -5.50oC. How many grams
of water were used as solvent if 16.0 grams of ethanol were added to make the solution? The formula for
ethyl alcohol (ethanol) is C2H5OH.
g x kb
MW = ---------- then
g x kf
(16.0 g)(1.86oC / m )
kg = -------------- = ------------------------
= 0.118 kg = 118 g H2O
t x kg
t x MW
(5.50oC)(46.0 g/mol)
7. What is the vapor pressure over an aqueous solution of glucose ( MW = 180 amu) if 90.0 grams of the
monosaccharide is added to 0.100 liter of water at 20.0oC? The vapor pressure of pure distilled water at
that temperature is 17.5 mm Hg.
VPsoln.
(100 g H2O / 18.0 g/mol)
= 17.5 mm ---------------------------------------------------------(100 g H2O / 18.0 g/mol) + (90.0 g/ 180 g/mol)
VPsoln.
= 16.1 mm Hg
= 17.5 mm (5.56 / 6.056)
8. What is the osmotic pressure (in mm Hg) of an aqueous solution of NaCl if 0.230 grams of the salt is
dissolved in enough water to make 350.0 ml of solution at 25oC? (Remember that NaCl is a strong
electrolyte!)
 = i MRT = 2[(0.230 g/ 58.5 g/mol) / 0.3500liter](0.0821(298 K)
= 0.5496 atm = 418 mm Hg
9. A student adds 1.000 gram of a compound to 75.00 ml of water. The osmotic pressure of the aqueous
solution of the unknown molecular nonelectrolyte is 23.3 mm Hg at standard temperature. What is the
molecular weight of the unknown compound.
(1.000g)(0.0821)(273 K)
MW = gRT / V = --------------------------------------- = 9.75 x 103 g/mol
(23.3 mm / 760 mm)(0.0750 liter)
10. When 19.00 grams of sucrose ( MW = 342 g/mole) is added to 10.00 g of water at a given temperature,
the vapor pressure of the solution is measured to be 20.00 mm Hg. What would be the vapor pressure of
pure distilled water at that same temperature?
(10.0 g H2O / 18.0 g/mol)
19.00 mm = VPpure solvent x ---------------------------------------------------= VPpure (0.556 / 0.611)
(10.00 g H2O / 18.0 g/mol) + (19.00 g/ 342 g/mol)
VPpure solvent = 21.98 mm Hg