AP Chemistry
Solution Stoichiometry PRACTICE TEST
Name____________________________________
Date____________________
SECTION I: Multiple Choice (Time: 40 minutes). Select the best answer for each question. You need not show any
work. Calculators are typically not permitted on this section of the test. Today’s test will permit the use of
calculators.
1.
A solid acid HX is mixed with water. Two possible solutions can be obtained. Which of the following is true?
I.
a)
b)
c)
d)
e)
II.
In case I, HX is acting like a weak acid, and in case II, HX is acting like a strong acid.
In case I, HX is acting like a strong acid, and in case II, HX is acting like a weak acid.
In both cases, HX is acting like a strong acid.
In both cases, HX is acting like a weak acid.
HX is not soluble in water.
Ans: b
Algorithm: No
Chapter/Section: 4.2
Difficulty: Moderate
All the acid is ionized in case I, while only 2/3 of the acid is ionized in case II.
2.
A 17.0-g sample of HF is dissolved in water to give 2.0  102 mL of solution. The concentration of the solution is:
a)
0.85 M
b)
0.17 M
c)
0.09 M
d)
4.2 M
e)
8.5 M
Ans: d
Algorithm: Yes
Chapter/Section: 4.3
17.0 g HF x 1 mol HF = 0.84577114 mol HF
20.1g HF
2.0  102 mL x
1L
Difficulty: Easy
= 0.20 L solution
1000 mL
0.8577114 mol HF
0.20 L soln
= 4.2 M soln
NOTE: you are not permitted a calculator on this portion of the actual exam. Begin to calculate values as follows: It’s
easy to see that 17 g of HF is a little less than 1 mole of HF. If that value is divided by 1/5 of a L (0.2), you will get a value
that is around 4-5. D is the only choice in this range.
3.
1.00 mL of a 3.95  10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. Then 2.00 mL
of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. What is the concentration of solution B?
a)
3.95 10–6 M
b)
1.10 10–5 M
c)
7.90 10–5 M
d)
8.78 10–5 M
e)
7.90 10–6 M
Ans: e
Algorithm: Yes
Chapter/Section: 4.3
Difficulty: Moderate
This is a serial dilution, so use M1V1=M2V2 two times!
First dilution: (3.95  10–4 M )(1.00mL)= M2(1.00mL + 9.00mL)
M2 = 3.95  10–5 M
Second dilution: (3.95  10–5 M)(2.00 mL) = M2 (2.00 mL + 8.00 mL)
M2 = 7.90 x 10 -6 M
(OR)THINK: The first dilution creates a solution (A) that is 1/10 of the strength of the original. Solution A is 3.95 x 10-5 M.
The second dilution creates a solution (B) that is 1/5 of the strength of Solution A. 1/5 of 4 is about 0.8. If I divide 3.95 x
10-5 M by 5, I get about 0.8 x 10-5 or 8.0 x 10 -6M. Therefore, e is the best selection.
4.How many grams of NaCl are contained in 350. mL of a 0.334 M solution of sodium chloride?
a)
b)
c)
d)
e)
19.5 g
6.83 g
13.66 g
116.9 g
none of these
Ans: b
Algorithm: Yes
Chapter/Section: 4.3
Difficulty: Moderate
0.350 L x 0.334 mol NaCl x 58.44 g NaCl
=6.83 g NaCl
1 L soln
1 mol NaCl
THINK: the solution is 1/3 M in strength, so 1/3 of the molar mass (58g) means that there are a little less than 20 g in a
liter of solution. You don’t have a liter, however. You have 0.3500 L. 0.350 L is a little more than a third of a L, so you
have a little more than a third of 20 g in that amount of solution. 1/3 of 20 is a little less than 7, so our best answer is b.
5.
Which of the following aqueous solutions contains the greatest number of ions?
a)
400.0 mL of 0.10 M NaCl
b)
300.0 mL of 0.10 M CaCl2
c)
200.0 mL of 0.10 M FeCl3
d)
200.0 mL of 0.10 M KBr
e)
800.0 mL of 0.10 M sucrose
Ans: b
Algorithm: No
Chapter/Section: 4.3
Difficulty: Easy
THINK: Sucrose does NOT form ions, so it is not a correct choice.
Analyze the remaining 4 choices (a-d).
400.0 mL NaCl x 1 L
0.10 mol NaCl x
1000mL
1 L soln
6.02 x 10 23 units of NaCl
1 mol of NaCl
x
2 ions
1 unit of NaCl
= 800
300.0 mL CaCl2 x 1 L
1000mL
0.10 mol CaCl2 x
1 L soln
6.02 x 10 23 units of CaCl2
1 mol of CaCl2
x
3 ions
1 unit of CaCl2
= 900
200.0 mL FeCl3 x 1 L
1000mL
0.10 mol FeCl3 x
1 L soln
6.02 x 10 23 units of FeCl3
1 mol of FeCl3
x
4 ions
1 unit of FeCl3
= 800
200.0 mL KBr
0.10 mol KBr
1 L soln
6.02 x 10 23 units of KBr
1 mol of KBr
x
2 ions
1 unit of KBr
= 400
x 1L
1000mL
x
NOTICE that all the conversion factors are identical for the above calculations, EXCEPT the last one (ions/unit of
compound.) So disregard the first 3 conversion factors, BECAUSE THEY ARE NOT AFFECTING YOUR ANALYSIS. The only
parts to take into account are the mL and the ions/unit. So even though the “answer” to the right has no real unit
associated with it, it is a valid comparison of the number of ions in each solution. b is the answer because it has the
greatest number of ions.
6.
Which of the following is not a strong base?
a)
Ca(OH)2
b)
KOH
c)
NH3
d)
LiOH
e)
Sr(OH)2
Ans: c
Algorithm: No
Chapter/Section: 4.2
Difficulty: Easy
THINK: A strong base ionizes fully in water (very soluble). Hydroxides of alkali metals and hydroxides of the heavy
alkaline earth metals are soluble, so a, b, d, and e are strong bases. Ammonia pulls the H+ ion away from water in
solution, but only a very small portion of them. Therefore, it is a weak base.
7.
The man who discovered the essential nature of acids through solution conductivity studies is
a)
Priestly
b)
Boyle
c)
Einstein
d)
Mendeleev
e)
Arrhenius
Ans: e Algorithm: No Chapter/Section 4
8.
Difficulty: Easy
What mass of calcium chloride, CaCl2, is needed to prepare 3.650 L of a 1.75 M solution?
a)
231 g
b)
6.39 g
c)
53.2 g
d)
111 g
e)
709 g
Ans: e
Algorithm: Yes
Chapter/Section: 4.3
Difficulty: Easy
3.650 L x 1.75 mol CaCl2
x
110.98 g CaCl2 = 709 g
1 L sol’n
1 mol CaCl2
THINK: We need to multiply 3.650 x 1.75. This is about equal to the following calculation. 3 ½ x 2, which yields about 7
when we do the math. This is the number of moles of CaCl2 in the given amount of solution. Multiply 7 times the molar
mass, 111 g/mol, and you get in the vicinity of 700 or 800. E is the best answer. (Note how much larger it is than any of
the other answers.)
9.
A 57.17-g sample of Ba(OH)2 is dissolved in enough water to make 1.800 liters of solution. How many mL of this solution must
be diluted with water in order to make 1.000 L of 0.100 M Ba(OH)2?
a)
539 mL
b)
185 mL
c)
18.5 mL
d)
3.34 mL
e)
300 mL
Ans: a
Algorithm: Yes
Chapter/Section: 4.3
Difficulty: Moderate
THINK: This is a dilution problem. Use M1V1=M2V2. We are given M2 and V2, and we can calculate M1 from the mass
of Ba(OH)2 (57g) and the volume of solution made from it (1.800L). We will use these values to solve for V1.
The MM of Ba(OH)2 is about 171 g/mol (calculated from P. Table). 57.17 g is just about 1/3 of a mol of Ba(OH)2. It is
diluted up to 1.800 L. (THINK: almost 2 liters of solution). So you have about 1/3 of a mol of Ba(OH)2 dispersed in 2 liters
of solution, or about 1/6 of a mol of Ba(OH)2 in a liter of solution. OR you could say the molarity of the original solution
is about 1/6 mol/L. Divide the fraction through to find your solution is ~0.17 M.
0.17M (V1) = (0.100M)(1.000L)
V1 = 0.100 mol
0.17M mol/L
V1 ~ 0.100 L
0.200
V1 ~ ½ L
V1 ~ 0.5 L
So the answer is in the vicinity of 0.5 liters. Choice a is 539 mL, which is 0.539 L, which is in our ballpark.) Choice a is our
best answer.
10.
An analytical procedure requires a solution of chloride ions. How many grams of CaCl2 must be dissolved to make 2.15 L of
0.0520 M Cl–?
a)
11.54 g
b)
0.373 g
c)
6.20 g
d)
2.89 g
e)
24.8 g
Ans: c
11.
Algorithm: Yes
Chapter/Section: 4.3
Difficulty: Moderate
You have two solutions of chemical A. To determine which has the highest concentration of A in molarity, what is the minimum
number of the following you must know?
I.
the mass in grams of A in each solution
II.
the molar mass of A
III.
the volume of water added to each solution
IV.
the total volume of the solution
a)
b)
c)
d)
e)
0
1
2
3
You must know all of them.
Ans: c
Algorithm: No
Chapter/Section: 4.3
Difficulty: Easy
THINK: The solute is the same in both solutions. Therefore, the molar mass is not necessary to help us compare these two
solutions, as it is the same for both. If you know the I. the mass in grams of A in each solution and III. The volume of
water added to each solutions, you can determine which has the highest concentration.
12.
What volume of 18 M sulfuric acid must be used to prepare 1.80 L of 0.215 M H2SO4?
a)
22 mL
b)
0.39 mL
c)
2.2  103 mL
d)
4.3 mL
e)
7.0 mL
Ans: a
Algorithm: Yes
Chapter/Section: 4.3
Difficulty: Easy
THINK: This is a dilution problem. Use M1V1 = M2V2.
13.
An analytical procedure requires a solution of chloride ions. How many grams of CaCl 2 must be dissolved to make 2.15 L of
0.0520 M Cl–?
a)
11.54 g
b)
0.373 g
c)
6.20 g
d)
2.89 g
e)
24.8 g
Ans: c
Algorithm: Yes
Chapter/Section: 4.3
Difficulty: Moderate
14.
You have equal masses of different solutes dissolved in equal volumes of solution. Which of the solutes would make the
solution having the highest molar concentration?
a)
NaOH
b)
KCl
c)
KOH
d)
LiOH
e)
all the same
Ans: d
Algorithm: No
Chapter/Section: 4.3
Difficulty: Moderate
THINK: Since you have equal masses of all solutes, the one with the LOWEST molar mass is going to have the highest
molar concentration. Eyeball the periodic table to discover that LiOH has the lowest MM.
15.
Which of the following do you need to know to be able to calculate the molarity of a salt solution?
I.
the mass of salt added
II.
III .
IV.
a)
b)
c)
d)
e)
the molar mass of the salt
the volume of water added
the total volume of the solution
I, III
I, II, III
II, III
I, II, IV
You need all of the information.
Ans: d
Algorithm: No
Chapter/Section: 4.3
Difficulty: Easy
Molarity = moles solute
Liters of solution
You need the mass of the salt (I)and the molar mass of the salt (II) to determine the moles of salt. You need the total
volume of solution as well (IV).
16.
You have exposed electrodes of a light bulb in a solution of H 2SO4 such that the light bulb is on. You add a dilute solution and
the bulb grows dim. Which of the following could be in the solution?
a)
Ba(OH)2
b)
NaNO3
c)
K2SO4
d)
Cu(NO3)2
e)
none of these
Ans: a
Algorithm: No
Chapter/Section: 4.5
Difficulty: Moderate
THINK: Anything that decreases the number of ions in solution will cause the solution to become a poorer electrolyte and
light the bulb less effectively. Solutions made from the choices listed in b, c, and d are all extremely soluble, and will
react with sulfuric acid to produce highly soluble products. Therefore, the number of ions will NOT decrease when any of
these three compounds are added to the sulfuric acid solution. However, barium hydroxide will produce barium sulfate
when placed in sulfuric acid. It is insoluble, so the ions that were previously moving freely in solution are now bound in a
precipitate. Therefore, the solution is a poorer electrolyte than a pure sulfuric acid solution, so a is the best choice.
17.
Aqueous solutions of sodium sulfide and copper(II) chloride are mixed together. Which statement is correct?
a)
Both NaCl and CuS precipitate from solution.
b)
No reaction will occur.
c)
CuS will precipitate from solution.
d)
NaCl will precipitate from solution.
e)
A gas is released.
Ans: c
Algorithm: No
Chapter/Section: 4.5
Difficulty: Easy
THINK: NaCl is very soluble, so it won’t precipitate, so a and d are not a good choices.
18.
How many of the following salts are expected to be insoluble in water?
sodium sulfide
barium nitrate
ammonium sulfate
a)
b)
c)
d)
e)
potassium phosphate
none
1
2
3
4
Ans: a
Algorithm: No
Chapter/Section: 4.5
Difficulty: Easy
THINK: Ionic compounds containing alkali metals are always soluble. And ionic compounds containing nitrate ions are
always soluble. Therefore none the compounds are insoluble.
19.
When NH3(aq) is added to Cu2+(aq), a precipitate initially forms. Its formula is:
a)
Cu(NH)3
b)
Cu(NO3)2
c)
Cu(OH)2
d)
Cu(NH3)22+
e)
CuO
Ans: c
Algorithm: No
Chapter/Section: 4.5
Difficulty: Moderate
NH3 is a base. When added to water, it forms NH4OH. When combined with Cu2+ ions, Cu(OH)2 will form, which is
insoluble.
20.
A solution contains the ions Ag+, Pb2+, and Ni2+. Dilute solutions of NaCl, Na2SO4, and Na2S are available to separate the
positive ions from each other. In order to effect separation, the solutions should be added in which order?
a)
Na2SO4, NaCl, Na2S
b)
Na2SO4, Na2S, NaCl
c)
Na2S, NaCl, Na2SO4
d)
NaCl, Na2S, Na2SO4
e)
NaCl, Na2SO4, Na2S
Ans: a
Algorithm: No
Chapter/Section: 4.5
Difficulty: Difficult
THINK: All the available solutions contain Na+, which makes them all soluble in water. However, their anions will be
recombining with other cations when placed in the solution that contains the ions Ag+, Pb2+, and Ni2+. Some of the
recombinations may be insoluble. Sulfide, S 2-, is insoluble with all 3 of the cations. Chloride is insoluble when combined
with Ag+ and Pb2+. Sulfate, SO4 2- , is insoluble only when combined with Pb2+. If you add sulfate first (Na2SO4), only the
Pb2+ ions will be removed from the solution. Next, add the chloride (NaCl). Since only Ag+and Ni2+ remain in solution,
chloride will precipitate the silver out of solution. Finally, add the sulfide (Na2S). It will precipitate out the Ni2+.
Aqueous solutions of barium chloride and silver nitrate are mixed to form solid silver chloride and aqueous barium nitrate.
21.
The balanced molecular equation contains which one of the following terms?
a)
AgCl (s)
b)
2AgCl (s)
c)
2Ba(NO3)2 (aq)
d)
BaNO3 (aq)
e)
3AgCl (aq)
Ans: b
Algorithm: No
Chapter/Section: 4.6
Difficulty: Easy
THINK: Create a molecular equation for this metathesis reaction, then balance.
BaCl2 (aq) + AgNO3 (aq)  Ba(NO3)2 (aq)+ 2AgCl(s)
22.
Consider an aqueous solution of calcium nitrate added to an aqueous solution of sodium phosphate. Write and balance the
equation for this reaction to answer the following question.What is the sum of the coefficients when the molecular equation is
balanced in standard form?
a)
4
b)
5
c)
7
d)
11
e)
12
Ans: e
Algorithm: No
Chapter/Section: 4.6
3Ca(NO3)2 + 2Na3PO4  Ca3(PO4)2 + 6NaNO3
3 +2+1+6 = 12
23.
Difficulty: Moderate
When solutions of carbonic acid and sodium hydroxide react, which of the following are NOT present in the net ionic equation?
I.
hydrogen ion
II.
carbonate ion
III.
sodium ion
IV.
hydroxide ion
a)
I and II
b)
I, II, and III
c)
I and IV
d)
I and III
e)
II and III
Ans: d
Algorithm: Yes
Chapter/Section: 4.8
Difficulty: Moderate
THINK: Create a molecular equation, then a complete ionic equation, then a net ionic equation
Molecular: H2CO3 + 2NaOH  2HOH + Na2CO3
Weak acid strong base
Complete: H2CO3(aq) + 2Na+(aq) + 2OH-(aq)  2H2O + 2Na+ (aq) + CO3 2- (aq)
Net Ionic: H2CO3(aq) + 2OH- (aq)  2H2O + CO3 2- (aq)
Therefore, d is our best choice.
24.
When solutions of acetic acid and copper(II) hydroxide react, which of the following are spectator ions?
a)
hydrogen ion
b)
acetate ion
c)
copper(II) ion
d)
hydroxide ion
e)
none of these
Ans: e
Algorithm: Yes
Chapter/Section: 4.8
Difficulty: Moderate
Molecular: 2HC2H3O2(aq)+ Cu(OH)2(s) Cu(C2H3O2)2 (aq) + 2H2O (l)
Complete Ionic: 2HC2H3O2 (aq)+ Cu(OH)2(s) Cu2+ + 2C2H3O2- (aq) + H2O (l)
Net Ionic: same as Complete Ionic! (No spectator ions.)
25. What is the net ionic equation for the reaction that occurs when aqueous copper (II) sulfate is added to excess 6molar ammonia?
(A) Cu2+ + SO42− + 2 NH4+ + 2 OH−  (NH4)2SO4 + Cu(OH)2
(B) Cu2+ + 4 NH3 + 4 H2O  Cu(OH)42− + 4 NH4+
(C) Cu2+ + 2 NH3 + 2 H2O  Cu(OH)2 + 2 NH4+
(D) Cu2+ + 4 NH3  Cu(NH3)42+
(E) Cu2+ + 2 NH3 + H2O  CuO + 2 NH4+
Ans: C
Molecular: CuSO4 (aq) + 2 NH3 + 2 H2O  (NH4)2SO4 (aq)+ Cu(OH)2 (s) (remember, NH3 is weak, so it is still mostly NH3
in aq soln.)
Complete Ionic: Cu2+ + SO42− + 2 NH3 + 2 H2O  2NH4+ + SO42- + Cu(OH)2(s)
Net Ionic: Cu2+ + 2 NH3 + 2 H2O  2NH4+ + Cu(OH)2(s)
SECTION II: Free Response
(Total time—45 minutes) Part A -YOU MAY USE YOUR CALCULATOR FOR PART A. CLEARLY SHOW THE METHOD USED
AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is to your advantage to do this, since you may obtain
partial credit if you do and you will receive little or no credit if you do not. Attention should be paid to significant figures.
Be sure to write all your answers to the questions on the lined pages following each question in this booklet. Answer
Questions 1, 2, and 3. The Section II score weighting for each question is 20 percent.
1. A metal cylinder with a volume of 5.25 L contains 34.1 g of N2 (g) at 15.0° C.
a) Calculate the total pressure in the cylinder.
THINK: Use PV =nRTP = nRT
V
V=5.25 L
T = 15.0 + 273 = 288K
R = any acceptable R value b/c pressure is what you are solving for. I select R = 0.0821 L atm / mol·K
n= is not given. But I know grams of N2 gas, so I can calculate my n value.
Step 1: n =34.1 g N2 x 1 mol N2 = 1.22 mol N2
28.02 g N2
Step 2: P = (1.22mol)(0.0821 L·atm/mol·K)(288K)
(5.25 L)
P = 5.48 atm
b) A 1.50 L container holds a 9.62 g sample of an unknown gaseous saturated hydrocarbon at 30 C and 3.62 atm. Its
molar mass is 44.1 g/mol.
i)
Calculate the density of the gas.
9.62 g = 6.41 g/L
1.50 L
ii)
Write the formula of the hydrocarbon.
Saturated hydrocarbons have the generic formula CnH2n+2.
Therefore, let 44.1 g = n(12) + (2n+2)(1) =14n +2
42.1 =14n
3=n
The formula is therefore 3 (CnH2n+2) = C3H8
2. Aqueous solutions of zinc hydroxide and hydrochloric acid are combined.
a) Write the balanced chemical equation for this reaction.
Zn(OH)2 (aq) + 2HCl(aq)ZnCl2 (aq) + 2H2O (l)
b) Write the net ionic equation for this reaction.
(Complete: Zn(OH)2 (aq) + 2H+ (aq) + 2Cl−(aq)  Zn2+ (aq) + 2Cl- (aq)+ 2H2O(l))
Net: Zn(OH)2 (aq) + 2H+ (aq)  Zn2+ (aq) + 2H2O(l)
c) Explain why a reaction occurs when these two compounds are combined.
A change has to take place in order for a reaction to occur. The main driving force, or change, in this reaction is the
formation of water.
d) How many grams of water are produced if 0.350 L of a 2.00-M zinc hydroxide solution is combined with excess
HCl?
0.350 L Zn(OH)2 sol’n x 2.00 mol Zn(OH)2 x 2 mol H2O x
16.02 g H2O = g H2O
1 L sol’n
1 mol Zn(OH)2
1 mol H2O