Download Alberta Chemistry 20-30 Sample CAB Questions - McGraw

Alberta Chemistry 20-30 Sample CAB Questions
MULTIPLE CHOICE
1. Use the following information to answer the next question.
Each molecule has a definite geometry because the atoms combine to form a molecule in order to attain a
state of minimum energy. VSEPR (valence shell electron pair repulsion) theory helps to determine the
geometry of molecular structures.
In a molecule, if there are three bonding electron groups around a central atom, what is the maximum
bond angle between the bonds?
A. 180o
B. 120o
C. 109.5o
D. 90o
ANS: B
B
A
120o
B
B
Consider a molecule AB3, in which A is a central atom and B is a surrounding atom. In this molecule, the
central atom is surrounded by three shared pairs of electrons. According to VSEPR theory, these electrons
should be as far apart as possible, so that the electrostatic force of repulsion between them is the
minimum. In such a case, the three electron pairs arrange themselves in trigonal planar geometry, because
it is the best arrangement involving the minimum possible energy. In this case, three atoms bonded to the
central atom, are placed at the vertices of an equilateral triangle. In this arrangement, the maximum angle
between two bonds is 120o. The bond angle can not be less than 120o, because in that case the molecule
will become highly energetic and thus unstable. It cannot be more than 120o, because it will contradict the
properties of geometry.
McGraw Hill Ryerson © 2007
1
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A2.5K
TOP: Diversity of matter
KEY: VSEPR Theory
MSC: Knowledge
2. Use the following information to answer the next question.
The electronegativity values of four elements in the same period of the periodic table are listed below:
P
Q
R
S
3.0
2.0
3.5
4.0
These elements belong to groups 13, 15, 16, and 17, but not in that order.
Which of these elements belongs to group 16?
A. P
B. Q
C. R
D. S
ANS: C
Electronegativity increases along a period from left to right. Arranged in the increasing order, the values
of electronegativity are:
2.0
13
<
3.0
<
15
3.5
<
16
4.0
17
R has an electronegativity of 3.5. Therefore, R belongs to group 16.
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A1.3K
TOP: Diversity of matter
KEY: Electro negativity
MSC: Knowledge
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2
3. An oxygen atom needs two additional electrons to reach the noble gas configuration of neon. Which of
the following Lewis structures represents the oxygen molecule?
A.
B.
C.
D.
ANS: C
+
Each oxygen atom needs two additional electrons to reach the noble gas configuration of neon gas. Two
oxygen atoms share two valence unpaired electrons each and form O = O double bond.
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A2.4K
TOP: Diversity of matter
KEY: Lewis Structure
MSC: Knowledge
4. Use the following information to answer the next question.
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3
Hydrogen fluoride, HF(l), has numerous applications in the petrochemical industry and is an
essential ingredient of super acids. Its boiling point is extremely high. It is a liquid at room
temperature, while other hydrogen halides such as hydrogen chloride, HCl(g), hydrogen
bromide, HBr(g), and hydrogen iodide, HI(g), are all gases at room temperature.
Hydrogen fluoride, HF(l), has a higher boiling point than the other hydrogen halides because the
A. intermolecular hydrogen bonding in HF(l) is stronger than that in other hydrogen halides.
B. intramolecular hydrogen bonding in HF(l) is stronger than that in other hydrogen halides.
C. intermolecular hydrogen bonding in HF(l) is weaker than that in other hydrogen halides.
D. intramolecular hydrogen bonding in HF(l) is weaker than that in other hydrogen halides.
ANS: A
The boiling points of acids decrease as we proceed from HI(g) to HBr(g) to HCl(g). This is because, with
a decrease in molecular mass, London (dispersion) forces decrease. However, HF(l) has an abnormally
high boiling point and is a liquid, whereas HCl(g), HBr(g) and HI(g) are gases. This is because of the
hydrogen bonding in HF(l). Hydrogen bonding does not exist in other hydrogen halides.
H⎯F----- H⎯F----- H⎯F----- H⎯F----- H⎯F----- H⎯F----- H⎯F----- H⎯F----- H⎯F----PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A2.7K
TOP: Diversity of matter
KEY: Intermolecular Bonding
MSC: Knowledge
5. Hydrogen bonds are found in all of the following compounds: glycerol, C3H5(OH)3(l), water, H2O(l),
ethanol, C2H5OH(l), and ethylene glycol, C2H4(OH)2(l). Which of the compounds listed would have the
highest boiling point?
A. water
B. ethylene
C. ethanol
D. glycerol
ANS: D
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4
Glycerol has the highest boiling point due to the presence of three –OH groups.
CH3 – OH
CH – OH
CH3 – OH
H
|
O – CH3
H
|
O – CH
H
|
O – CH3
These –OH groups exhibit very strong hydrogen bonding. This results in a high boiling point.
PTS: 1
DIF: Difficult
OBJ: Section 1.1
LOC: 20–A2.7K
TOP: Diversity of matter
KEY: Intermolecular Bonding
MSC: Knowledge
6. Which of the following compounds is the least ionic in character?
A. MgCl2(s)
B. BeCl2(s)
C. BaCl2(s)
D. CaCl2(s)
ANS: B
Ionic character increases as the difference in electronegativity between the bonded atoms increases.
Electronegativity values decrease as you move down the periodic table.
Be > Mg > Ca > Ba
Therefore, BaCl2(s) is the most ionic in character and BeCl2(s) is the least ionic in character.
PTS: 1
DIF: Average
OBJ: Section 1.1
LOC: 20–A1.4K
TOP: Diversity of matter
KEY: Ionic Bond
MSC: Knowledge
7. Which statement(s) is(are) true?
I. The degree of electronegativity depends predominantly on the size of a cation and an anion.
II. The degree of covalence will be the highest if the cation is smaller when compared to the anion’s size.
III. The larger the size of the cation or anion, the stronger the ionic bond.
A. I and II are true.
B. I and III are true.
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C. II and III are true.
D. All statements are false.
ANS: A
The smaller size of a cation distorts the electron cloud of the anion and pulls the electron density towards
itself. Thus, it introduces some covalent character is the ionic bond. The smaller the size of cation, the
greater is the degree of covalence.
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A1.5K
TOP: Diversity of matter
KEY: Ionic Bond
MSC: Knowledge
8. An ionic compound contains two ions, X and Y. If the charge on X is +3 and that on Y is –2, which of
the following elements could be X and Y respectively?
A. Ga, N
B. Ca, S
C. Na, S
D. Al, O
ANS: D
From the given options, only gallium, Ga, and aluminium, Al, can produce ions with a charge of +3.
Between nitrogen, N, and oxygen, O, only oxygen can produce an ion with a charge of –2. Therefore,
aluminium and oxygen are X and Y and the chemical formula of this compound is Al2O3.
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A1.5K
TOP: Diversity of matter
KEY: Ionic bond
MSC: Knowledge
9. A compound consists of four carbon atoms, ten hydrogen atoms and an oxygen atom. If there is a
C–O–C bond in the compound, what is the molecular formula of this compound?
A. C4H9OH
B. C2H5OC2H5
C. CH3COC2H7
D. C2H5COCH5
ANS: B
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The molecular formula of the given compound will be C2H5OC2H5. Its structural formula will be
H
H
H
H
|
|
|
|
H − C − C − O − C − C − H
|
|
|
|
H
H
H
H
PTS: 1
DIF: Average
OBJ: Section 1.1
LOC: 20–A2.2K
TOP: Diversity of matter
KEY: Molecular formula
MSC: Knowledge
10. Which of the following molecules is non-polar?
A. H2O(g)
B. CO2(g)
C. NH3(g)
D. CH2Cl2(g)
ANS: B
Carbon dioxide is a linear molecule in which both the C = O bonds are oriented at an angle of 180o with
respect to each other. The bonds are polar since the difference in electronegativity between carbon and
oxygen is 0.8, but due to linear geometry of CO2(g), the polarity of the two C = O bonds cancel each other
out. Therefore, CO2(g) is a non-polar.
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A2.9K
TOP: Diversity of matter
KEY: Polarity
MSC: Knowledge
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11. Use the following information to answer the next two questions.
Structural formulae of four compounds are given below:
F
N
I.
II.
H
H
H
F
H
F
Cl
H
III.
IV.
C
C
H
Cl
H
Cl
Cl
H
Which of these compounds is polar?
A. I
B. II
C. III
D. IV
ANS: A
N
H
H
H
NH3(g) is a polar compound because it is pyramidal and asymmetrical so the dipole vectors do not cancel.
PTS: 1
DIF: Average
OBJ: Section 1.1
LOC: 20–A2.9K
TOP: Diversity of matter
KEY: Polarity
MSC: Knowledge
12. Which of the following atoms does not form hydrogen bonds?
A. O
B. N
C. F
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D. Cl
ANS: D
Chlorine, Cl, does not form hydrogen bonds because of its large size. When chlorine, Cl, bonds to a
hydrogen atom, its lower electronegativity, and large atomic size, creates a dipole–dipole interaction that
is not strong enough to produce hydrogen bonding.
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A2.7K
TOP: Diversity of matter
KEY: Hydrogen Bond
MSC: Knowledge
13. Consider the Lewis structure for the following compound, A2.
A
A
If there are two electrons in the inner shell of element A, in which of the following groups of the periodic
table does element A belong?
A. 3
B. 11
C. 13
D. 15
ANS: D
From the above Lewis structure, there are five electrons in the valence shell of element A. Atomic
number of an element = total number of electrons = number of valence electrons + number of electrons in
inner shells = 5 + 2 = 7. Nitrogen has the atomic number of 7 and it corresponds to group 15.
PTS: 1
DIF: Average
OBJ: Section 1.1
LOC: 20–A2.6K
TOP: Diversity of matter
KEY: Lewis Structure
MSC: Knowledge
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14.
Which of the following Lewis structures represents the ethene (ethylene) molecule?
A.
B.
H
H
H
C.
D.
H
H
C
C
H
H
C
C
C
C
H
H
H
H
H
H
H
H
C
C
H
ANS: A
H
H
C
H
C
H
In this structure, all of the carbon and hydrogen atoms have a completed valence shell.
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A2.3K
TOP: Diversity of matter
KEY: Lewis Structure
MSC: Knowledge
15. In a diatomic molecule of an element, each atom has one lone pair and the molecule contains a triple
bond. Which of the following elements does it correspond to?
A. Sulfur
B. Oxygen
C. Nitrogen
D. Fluorine
ANS: C
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A
A
N
≡
N
It is a molecule of nitrogen atom. A nitrogen atom has five valence electrons. Each nitrogen atom shares
three electrons with the other and two electrons are left unshared. Hence, there exists a triple bond in the
molecule.
PTS: 1
DIF: Easy
OBJ: 1.1
LOC: 20–A2.3K
TOP: Diversity of matter
KEY: Lewis Structure
MSC: Knowledge
16. What is the chemical name for CaSO4(s)?
A. Calcium sulfur tetraoxide
B. Calcium sulfate
C. Calcium sulfite
D. Calcium tetraoxysulfur
ANS: B
The chemical name of CaSO4 is calcium sulfate.
CaSO 4
Calcium sulfate
⇌ Ca 2+ + SO 4
Caclium ion
(cation)
2−
Sulfate ion
(anion)
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A2.1K
TOP: Diversity of matter
KEY: Ionic Bond
MSC: Knowledge
17.
What is the chemical name for P4O10(s)?
A. Tetraphosphorus dioxide
B. Tetraphosphorus oxide
C. Phosphorus(IV) decaoxide
D. Tetraphosphorus decaoxide
ANS: D
The name of the compound (P4O10) is tetraphosphorus decaoxide.
McGraw Hill Ryerson © 2007
11
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A2.1K
TOP: Diversity of matter
KEY: Covalent Bond
MSC: Knowledge
18. What of the following is the name of the negative ion, also called the anion, in nitrous acid?
A. nitrite ion
B. nitrate ion
C. pernitrite ion
D. nitride ion
ANS: A
HNO2 dissociates in the following way
HNO2(aq) ⇌ H+ + NO2−
cation
anion
The name of NO2– is ‘nitrite ion’.
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A1.1K
TOP: Diversity of matter
KEY: Ionic Bond
MSC: Knowledge
19. A compound contains an aluminium ion and some chloride ions. If the compound is neutral, how
many chlorine ions will this molecule contain?
A. 1
B. 2
C. 3
D. 4
ANS: C
An aluminium atom releases three electrons to form an Al3+ ion.
Al ⇌ Al3+ + 3e–
A chlorine atom accepts an electron to form Cl– ion
Cl + e– ⇌ Cl–
So when these two atoms combine, three chlorine atoms accept three electrons from an aluminium atom,
to form AlCl3.
Al ⇌ Al3+ + 3e–
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3Cl+ 3e– ⇌ 3Cl–
Al + 3Cl+ 3e– ⇌ Al3+ + 3Cl– + 3e–
Al+ 3Cl⇌ Al3+ + 3Cl–
In the above reaction, the charge has been balanced on both sides. In other words, Al3+ and Cl–1 combine
to form AlCl3. So, there are three chlorine ions in the molecule.
PTS: 1
DIF: Average
OBJ: Section 1.1
LOC: 20–A1.2K
TOP: Diversity of matter
KEY: Ionic Bond
MSC: Knowledge
20. Element X has two valence electrons and element Y has 7 valence electrons. Which of the following
compounds would you expect to form?
A. XY2
B. X2Y
C. X7Y2
D. X2Y7
ANS: A
The number of valence electrons in the element X is 2. It means it is a metal because an element having
less than four valence electrons, generally exhibit metallic character. Moreover, these electrons can also
be easily donated to attain the nearest noble gas configuration.
X –- 2e– ⇌ X2+
The number of valence electrons in the element Y is 7. It can not release 7 electrons to attain
electropositive character. However, it can gain one electron to attain the nearest noble gas configuration.
So, it is a nonmetal.
Y + e-- ⇌ Y–
When one cation and two such anions combine, the compound formed is
X2+ + 2Y– ⇌ XY2
PTS: 1
DIF: Average
OBJ: Section 1.1
LOC: 20–A1.2K
TOP: Diversity of matter
KEY: Ionic Bond
MSC: Knowledge
21. Which of the following sets of bonds are correctly ordered from least to most polar?
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13
A. C–Fe < C–H < C–O < C–F
B. C–H < C–Fe < C–F < C–O
C. C–H < C–Fe < C–O < C–F
D. C–F < C–O < C–Fe < C–H
ANS: C
In the above given bonds, one element is the same (i.e., carbon) and the electronegativity of other
elements increases in the order H < Fe < O < F. So, polar character of the bond increases in the order
C–H < C–Fe < C–O < C–F.
PTS: 1
DIF: Average
OBJ: Section 1.1
LOC: 20–A2.10K
TOP: Diversity of matter
KEY: Electronegativity
MSC: Knowledge
22. The boiling points of water, H2O(l), ammonia, NH3(g), and hydrogen fluoride, HF(l), are higher than
expected due to:
A. strong London (dispersion) forces.
B. strong hydrogen bonding.
C. strong gravitational forces.
D. strong Van der waal’s forces.
ANS: B
Water, H2O(l), ammonia, NH3(g), and hydrogen fluoride, HF(l), contain hydrogen atoms bonded to highly
electronegative elements (N, O, F). In case of such molecules, hydrogen bonding occurs. Because of the
presence of strong hydrogen bonding, these molecules require more energy to move into the vapour phase
from the liquid phase. Therefore, their boiling points are higher than the expected values.
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A2.8K
TOP: Diversity of matter
KEY: Intermolecular forces
MSC: Knowledge
23. Which of the following statements correctly reflects the correlation between boiling point and the
intermolecular forces of attraction?
A. The stronger the intermolecular forces of attraction, the lower the boiling point.
B. The stronger the intermolecular forces of attraction, the higher the boiling point.
C. The weaker the intermolecular forces of attraction, the higher the boiling point.
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D. There is no relationship between intermolecular forces of attraction and boiling point.
ANS: C
Boiling point is strongly related to intermolecular forces of attraction. With an increase in the
intermolecular forces of attraction, more energy is required for the molecules to go into the gas phase and
results in higher boiling point.
PTS: 1
DIF: Average
OBJ: Section 1.1
LOC: 20–A2.8K
TOP: Diversity of matter
KEY: Intermolecular forces
MSC: Knowledge
24. Which of the following compounds is the most polar?
A. H2O(l)
B. H2S(g)
C. HF(l)
D. HCl(g)
ANS: C
All the compounds given above contain one element same that is hydrogen and its electronegativity is
2.20. The electronegativity of other elements, i.e., Cl, F, O, and S is 3.16, 3.98, 3.44, and 2.56,
respectively. The most electronegative element out of these is fluorine. So, the electronegativity
difference in HF is the most, 0.78. Therefore, HF(g) will be the most polar.
PTS: 1
DIF: Average
OBJ: Section 1.1
LOC: 20–A2.10K
TOP: Diversity of matter
KEY: Electronegativity
MSC: Knowledge
25. Use the following information to answer the next question.
The temperature of the Pahoehoe lava can be estimated by observing its color. The results of such
estimation agree significantly with the measured temperatures of lava flows, which are invariably
greater than 1000 degrees Celsius.
Which of the following cannot be a temperature of a burning particle of the Pahoehoe lava?
A. 1010 K
B. 1276 K
C. 1300 K
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D. 1350 K
ANS: A
Lava flows at a temperature of more than 1000 oC, i.e., 1273 K. Hence, lava cannot have a temperature
less than 1273 K. So, 1010 K cannot be a temperature of a burning particle of the lava.
PTS: 1
DIF: Easy
OBJ: Section 1.2
LOC: 20–B1 2K
TOP: Forms of matter–gases
KEY: Celsius and Kelvin scales
MSC: Knowledge
26. In a reaction following the law of combining gas volumes, 1 L of nitrogen gas would react with
i
litres of hydrogen gas, to produce
ii
litres of ammonia gas. The statement given above is
completed by the information in row
Row
i
ii
A
Two
Three
B
Three
Two
C
Three
Four
D
One
One
ANS: B
Reaction between nitrogen and hydrogen takes place as
N2 (g) + 3H2 (g) Æ 2NH3(g)
So, we can say that one litre of nitrogen gas reacts with three litres of hydrogen gas to produce two litres
of ammonia gas.
PTS: 1
DIF: Average
OBJ: Section 1.2
LOC: 20–B1.3K
TOP: Forms of matter–gases
KEY: Law of combining volumes
MSC: Knowledge
27. Use the following information to answer the next question.
Andy took a sample each of milk, silt in water, tomato juice, and wine and tested them for the
presence of suspended particles. He was able to detect the presence of suspended particles in all but
one of the samples.
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16
Which of the above substances will have uniform composition and appearance?
A. Milk
B. Silt in water
C. Tomato juice
D. Wine
ANS: D
Uniform appearance and composition is the property of a homogeneous mixture. Heterogeneous mixtures
consist of visibly different substances and phases. Of the given substances, milk, silt in water, and tomato
juice are examples of suspensions. A suspension is a heterogeneous mixture in which particle sizes are
large; visible and will settle when left undisturbed. These three substances will not have uniform
appearance and composition. Wine is a solution of alcohol and water. All solutions are homogeneous
mixtures and will have uniform appearance and composition. The particles present in homogeneous
mixtures cannot be detected even with microscope.
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C1.1K
TOP: Acids and Bases
KEY: Homogeneous mixtures
MSC: Knowledge
28. Which of the following compounds is not an electrolyte?
A. C2H5OH(aq)
B. H2SO4(aq)
C. HCl(aq)
D. CuSO4(aq)
ANS: A
C2H5OH(aq) is a molecular compound that will dissolve in water, but, as a molecular compound, does not
ionize, and is therefore not an electrolyte.
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C1.4k
TOP: Acid and Bases
KEY: Electrolyte
MSC: Knowledge
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29. Use the following information to answer the next question.
The table shows four experiments, A, B, C, and D, in which a constant mass of sodium chloride,
NaCl(s), is dissolved in the same amount of solvent at different temperatures. The time taken (in
seconds) for the NaCl(s) to dissolve was 10 s, 15 s, 18 s, and 25 s.
Experiment Temperature (oC)
A
37
B
40
C
25
D
58
Which experiment would have taken 18 s to dissolve the NaCl(s) completely?
A. A
B. C
C. B
D. D
ANS: A
Solubility of a compound is directly proportional to the temperature at which salt is dissolved. Therefore,
the higher the temperature, the faster the dissolution. The events, arranged in ascending order of the time
taken to dissolve the salt in water, are C < A < B < D. Therefore, event A took 18 s.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C1.9k
TOP: Acid and Bases
KEY: Solubility
MSC: Knowledge
30. Use the following information to answer the next two questions.
The white washing of walls gives shiny finish to them. Slaked lime is the main ingredient used
for white washing walls. It can be prepared as follows.
CaO(s) + H2O(l) Æ Ca(OH)2(s)
The calcium oxide, CaO(s), must dissolve in water for the reaction to occur.
An increase in pressure will have which of the following effects on the reaction of calcium oxide with
water:
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A. Solubility of calcium oxide will increase.
B. Solubility of calcium oxide will decrease.
C. There will be no effect on the solubility of calcium oxide.
D. The product formed will be different.
ANS: C
The effect of pressure is observed only in case of gases. Since the reaction given above does not include
any gaseous component, the increase in pressure will have no effect on the solubility of calcium oxide.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C1.9K
TOP: Acids and Bases
KEY: Solubility
MSC: Knowledge
31. Use the following information to answer the next question.
NH4ClO4(s) is used as an oxidizer to provide the required thrust for the lift off of the space
shuttle. NH4ClO4(s) reacts with dinitrogen trioxide, to form an acid solution along with
water and nitrogen gas.
What is the name of the acid formed in this process?
A. Chlorous acid
B. Hypochlorous acid
C. Perchloric acid
D. Chloric acid
ANS: C
The following reaction takes place
2 NH4ClO4 + N2O3 ⇌ 2N2 + 2 HClO4 + 3H2O
The acid obtained in the process is HClO4. It contains ClO4– polyatomic ion and its name is perchlorate
ion. According to the nomenclature of acids, the acid obtained in the above process is perchloric acid.
PTS: 1
DIF: Difficult
OBJ: Section 1.3
LOC: 20–C2.1K
TOP: Acids and Bases
KEY: Mono/polyprotic acid
MSC: Knowledge
32. Use the following information to answer the next question.
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A scientist treats dilute sulfuric acid with barium chlorite to produce an acid. The reaction
is shown below:
Ba(ClO2)2(aq)+ H2SO4(aq) Æ BaSO4(s) + 2HOClO(aq)
Name the acid formed in the reaction.
A. chloric acid
B. hypochlorous acid
C. chlorous acid
D. hypobromous acid
ANS: C
The acid obtained in the process is HOClO(aq). It contains ClO2–(aq) polyatomic ion. This ion is named
chlorite. According to the nomenclature of acids, the suffix ‘ite’ is changed to ‘ous’. Therefore the acid
obtained is chlorous acid.
PTS: 1
DIF: Difficult
OBJ: Section 1.3
LOC: 20–C2.1K
TOP: Acids and Bases
KEY: Mono/Polyprotic acid
MSC: Knowledge
33. Use the following information to answer the next question.
A student compares the ionization of monoprotic acids with that of polyprotic acids, and records the
following observations.
Acid
Ka1
Ka2
ethanoic acid
1.76 × 10–5
tartaric acid
6.0 × 10–4
1.5 × 10–5
hydrosulfuric acid
1.3 × 10–7
7.1 × 10–15
phosphoric acid
7.6 × 10–3
6.2 × 10–8
Ka3
2.2 × 10–13
When the above mentioned acids are titrated with NaOH; which of the following acids will have the
highest pH value at the final equivalence point in the titration curve?
A. ethanoic acid
B. tartaric acid
C. hydrosulfuric acid
D. phosphoric acid
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20
ANS: D
pH
Equivalence points
Volume of base
Phosphoric acid, H3PO4(aq), has three ionizable hydrogen atoms, while tartaric acid,
C2H4O2(COOH)2(aq), and hydrosulfuric acid,H2S(aq), have two ionizable hydrogen atoms, and ethanoic
acid, CH3COOH(aq), has one ionization hydrogen atom. In case of phosphoric acid we will have the
maximum number (three) of ionizations and, therefore, the maximum number of equivalence points. So,
out of above acids, phosphoric acid will have equivalence point at the highest pH value, as shown above
in the titration curve. In the case of other acids, the number of equivalence points will be fewer and they
will reach equivalence with the NaOH(aq) at a lower pH value.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C2.11K
TOP: Acids and Bases
KEY: Mono/Polyprotic acids
MSC: Knowledge
34. Strong acids increase the concentration of H3O+(aq) and OH–(aq) ion in aqueous solutions to a large
extent. While weak acids and bases increase the concentration of the respective ions to a small extent. The
relative strength of an acid or a base is generally determined by the dissociation constant Ka or Kb.
Which of the given acids is the strongest acid?
A. HCN(aq)
B. HF(aq)
C. CH3COOH(aq)
D. HNO2(aq)
ANS: B
As per the information, the strength of an acid is directly proportional to the dissociation constant of the
acid (Ka). From the given data
McGraw Hill Ryerson © 2007
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4 × 10–7 < 1.8 × 10–5 < 4.5 × 10–4 < 6.7 × 10–4 < 6.7 × 10–4
HCN < CH3 COOH < HNO2 < HF
∴ HF is the strongest acid.
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C2.10k
TOP: Acid and Bases
KEY: Strong and weak acids
MSC: Knowledge
35. Use the following information to answer the next question.
Acid dissociation constant of some acids
Acids
pKa
hydrofluoric acid
3.3
methanoic acid
3.8
ethanoic acid
4.8
hydrogen sulfide
7.1
Which of the above mentioned acids will give H+ ions most easily?
A. Hydrogen sulfide
B. Ethanoic acid
C. Methanoic acid
D. Hydrofluoric acid
ANS: D
pKa values are related to Ka (dissociation constant) values as follows
pKa = –log10Ka
The stronger the acid, more will be the dissociation constant Ka and lesser will be the pKa values. Now
from the above data we can say that hydrofluoric acid has the smallest pKa values. Hence, it is the
strongest acid out of these and will give H+ ions very easily.
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C2.11K
TOP: Acids and Bases
KEY: Strong acids and bases
MSC: Knowledge
36. Use the following information to answer the next question.
McGraw Hill Ryerson © 2007
22
When HCl(aq), a strong acid, and NaOH(aq), a strong base, are mixed together, a
significant quantity of heat is produced. The products are sodium chloride, NaCl(aq)
and water, H2O(l).The reaction that takes place is:
HCl(aq) + NaOH(aq) Æ NaCl(aq) + H2O(l)
This reaction is a
i
reaction and ii ions are the spectator ions. The statement given above is
completed by the information in row
Row
i
ii
A
hydrolysis
Cl–(aq) and H+(aq)
B
neutralization
H+(aq) and OH–(aq)
C
esterification
Na+(aq) and OH–(aq)
D
neutralization
Na+(aq) and Cl–(aq)
ANS: D
This reaction is called a neutralization reaction as it involves the combination of hydronium ions,
H3O+(aq) and hydroxide ions (OH–) to form water. HCl(aq) and NaOH(aq) ionize and dissociate into ions
in solution according to the following equation:
H3O+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) Æ Na+(aq) + Cl–(aq) + 2H2O(l)
In this reaction Na+(aq) and Cl–(aq) ions remain unchanged during the reaction and they see other ions
react. So, Na+(aq) and Cl–(aq) ions are the spectator ions.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C2.9K
TOP: Acids and Bases
KEY: Neutralization
MSC: Knowledge
37. Use the following information to answer the next question:
A student was asked to identify the three acidic compounds among the six samples listed below.
She applied the Arrhenius concept to classify the acids.
AlCl3(aq), H2SO4(aq) , CuSO4(aq) , HCl(aq) , NH3(aq) , NaOH(aq)
She classified both H2SO4(aq) and HCl(aq) as acids. She could not identify the third acid.
Which of the remaining compounds could she not select as acid?
McGraw Hill Ryerson © 2007
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A. AlCl3
B. CuSO4
C. NH3
D. NaOH
ANS: A
AlCl3 is an acidic compound. However, it does not follow the Arrhenius as it does not contain any proton
(or hydrogen ion). CuSO4 is a salt whereas NH3 and NaOH are basic compounds.
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C2.7k
TOP: Acid and bases
KEY: Arrhenius (modified) theory
MSC: Knowledge
38. Use the following information to answer the next question.
The ionization equations of some acids and bases are given below:
HCl(aq) + H 2O(l ) ⇌ H3O+(aq) + Cl–(aq)
Acid
Base
+
NH 4 (aq) + H 2O (l ) ⇌ H3O+(aq) + NH3(aq)
Acid
Base
2−
H 2O(l ) + CO3 (aq) ⇌ HCO3–(aq) + OH–(aq)
Acid
Base
+
NH 4 (aq) + CH 3COO − (aq) ⇌ CH3COOH(aq) + NH3(aq)
Acid
Base
Which of the above mentioned bases are Bronsted-Lowry bases but not Arrhenius bases?
A. H2O(l) and CO22–(aq)
B. CO32–(aq) and CH3COO–(aq)
C. CO32–(aq) only
D. H2O(l), CO32–(aq), and CH3COO–(aq)
ANS: B
According to the Arrhenius concept, a base is a substance, which can donate OH– ions. Hence, a base
must be a source of OH– ions. But according to the Bronsted-Lowry theory, a base must be a proton
acceptor. H2O(l) can accept a proton. Also it can give OH– ions in the solution. Hence, it is an Arrhenius
as well as a Bronsted base, while CO32–(aq) and CH3COO–(aq) ions are only proton acceptors and not
OH–(aq) donors. Hence, these are Bronsted-Lowry bases but not Arrhenius bases.
McGraw Hill Ryerson © 2007
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PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C2.8K
TOP: Acids and Bases
KEY: Arrhenius (modified) theory
MSC: Knowledge
39. If the pH of a solution, measured with the help of a pH meter, is 5.127, [H3O+] is
A. 5.67 × 10–5 mol/L
B. 7.46 × 10–6 mol/L
C. 5.23 × 10–7 mol/L
D. 7.92 × 10–6 mol/L
ANS: B
Reading on the pH meter = 5.127
pH of the solution = 5.127
We know that pH = –log [H3O+] or log [H3O+] = –pH
log [H3O+] = –5.127 mol/L
[H3O+] = antilog [–5.127] = 7.46 × 10–6 mol/L.
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C2.6K
TOP: Acids and Bases
KEY: Hydronium ion/pH
MSC: Knowledge
40. Use the following information to answer the next two questions.
In an experiment, 30.0 mL of a solution of 0.100 mol/L HCl(aq) is titrated with a standard
0.100 mol/L NaOH(aq) solution. The initial pH of the solution is 1.000.
The pH of the titrant after adding 5.0 mL of 0.100 mol/L NaOH(aq) is
A. 2.334
B. 1.523
C. 1.225
D. 1.143
ANS: D
After the addition of 5 mL of 0.1 M NaOH solution, the volume of the solution becomes
= 30 + 5 = 35 mL
McGraw Hill Ryerson © 2007
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Initially the amount of H+ ions present in the titration flask =
0.1 × 30
1 000
= 3.0 × 10–3 M.
Similarly the amount of OH– ions present in 5 mL of solution =
0.1× 5
.
1 000
= 0.5 × 10–3M
Amount of H+ ions left after neutralization in solution
= 3.0 × 10–3 – 0.5 × 10–3
= 2.5 × 10–3M
Now concentration of H+ ions
[H+] =
2.5 × 10 −3
× 1 000 = 0.072 M
35
pH = –log (.072)
pH = 1.143.
PTS: 1
DIF: Difficult
OBJ: Section 1.3
LOC: 20–C2.5K
TOP: Acids & Bases
KEY: Hydronium ion/pH
MSC: Knowledge
41. In the above experiment, the initial pOH of the HCl(aq) solution is 13.000. The pOH of the solution,
after adding 30.05 mL of NaOH, will be
A. 2.192
B. 3.072
C. 4.079
D. 5.209
ANS: C
After adding 30.0 mL of the NaOH solution the acid is completely neutralized. On adding 30.05 mL of
NaOH, the additional volume of NaOH = 0.05 mL. Now amount of NaOH in 0.05 mL of
NaOH =
0.1
× 0.05 = 5 × 10–6M
1 000
Total volume of solution becomes 30 + 30.05 = 60.05 mL.
Concentration of OH– ions
[OH–] =
=
5 × 10 −6
× 1 000
60.05
5 × 100
× 10–5
60.05
McGraw Hill Ryerson © 2007
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=
500
× 10–5
60.05
= 8.327 × 10–5
pOH = –log [OH–]
= –log [8.327 × 10–5]
= –[log 8.325 – 5]
= –[0.921 – 5]
= 4.079.
PTS: 1
DIF: Difficult
OBJ: Section 1.3
LOC: 20–C2.5K
TOP: Acids and Bases
KEY: Hydroxide ion/pOH
MSC: Knowledge
42. Use the following information to answer the next two questions.
pH value of some of the common substances
Substance
pH value
Battery acid
0.5
Vinegar
2.4 – 3.4
Lemon juice
2.2 – 2.4
Beer
4.0 – 5.0
Milk
6.8
Lime water
10.5
In which of the above solutions will the phenolphthalein indicator be pink?
A. Lemon juice
B. Vinegar
C. Battery acid
D. Lime water
ANS: D
Phenolphthalein is light pink in the pH range 8.3 – 10.0. It is colourless in acidic solutions (low pH); and
in basic solutions it is deeper pink. Only lime water has a pH above 8.3, and will be pink in the presence
of phenolphthalein indicator.
McGraw Hill Ryerson © 2007
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PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C2.2K
TOP: Acids and Bases
KEY: Indicators
MSC: Knowledge
43. In which of the following solutions, will methyl orange be red?
A. Lemon juice
B. Beer
C. Milk
D. Lime water
ANS: A
Methyl orange is red in solutions with a pH less than 3.2, orange in solutions with a pH of 3.2 to 4.4 and
yellow in solutions with a pH greater than 4.4. Methyl orange is red in lemon juice which has a pH of 2.2
to 2.4.
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C2.2K
TOP: Acids and Bases
KEY: Indicators
MSC: Knowledge
44. Use the following information to answer the next question.
A 0.20 mol/L solution of HCN(aq) is prepared. Its pH is 4.92 and [H3O+] is 1.2 × 10–5 mol/L. Students
determined the number of significant digits in the values and recorded them in the table below:
Student Number of significant digits in pH Number of significant digits in concentration
A
2
2
B
3
2
C
1
5
D
0
7
Which of the students reported the pH and concentration of hydronium ion, H3O+(aq), to the correct
number of significant digits?
A. A
B. B
C. C
D. D
McGraw Hill Ryerson © 2007
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ANS: A
The number before the decimal point is not significant in pH, so the value, 4.29 has only 2 significant
digits. In all other values, all numbers except zeroes before the number are significant, and the exponent is
not considered in counting significant digits. Therefore, the value 1.2 x 10–5 has 2 significant digits.
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C2.4K
TOP: Acids and Bases
KEY: Concentration
MSC: Knowledge
45. A student measures 100 mL of a 0.2 mol/L NaCl(aq) solution and dilutes it to 500 mL. The
concentration of NaCl(aq) in the diluted solution will be
A. 0.02 mol/L
B. 0.04 mol/L
C. 0.05 mol/L
D. 0.06 mol/L
ANS: B
Applying the molarity equation,
(concentrated) C1V1 = C2V2 (dilute)
(0.20 mol/L) (100 mL) = C2(500 mL)
C2 =
0.20 × 100
= 0.04 mol/L
500
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C1.11K
TOP: Acids and Bases
KEY: Dilution
MSC: Knowledge
46. Use the following information to answer the next question.
A saturated solution of sodium chloride, NaCl(aq), is prepared with 100 g of water at a
temperature of 90 oC. The solubility of sodium chloride, NaCl(s), at different temperatures
is shown in the graph below:
McGraw Hill Ryerson © 2007
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The Solubility of NaCl in 100 g of
Water at Varying Temperatures
Mass of NaCl (grams)
60
50
40
30
20
10
0
10 20 30 40 50 60 70 80 90 100
Temperature (°C)
How much NaCl(s) would be needed?
A. At least 20 g
B. At least 30 g
C. At least 40 g
D. At least 50 g
ANS: D
Saturation is a point where the system is in equilibrium so that the rate of dissolution is equal to the rate
of crystallization. In the case of NaCl(s) and water this equilibrium is reached on dissolving 50 g of NaCl
in 100 g water at 90 oC. Any further addition of NaCl(s) does not dissolve in water.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C1.10K
TOP: Acids and Bases
KEY: Solubility
47. What amount of NaCl will make the above solution unstable at 90 oC?
A. 20 g
B. 30 g
C. 40 g
D. 50 g
ANS: D
McGraw Hill Ryerson © 2007
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MSC: Knowledge
When 50 g of NaCl is added to 100 g of water, the solution becomes super saturated. Now in a super
saturated solution, the maximum number of solutes is present in the solution. This supersaturated solution
is very unstable and the excess amount of solute will precipitate.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C1.10K
TOP: Acids and Bases
KEY: Solubility
MSC: Knowledge
48. The [H+(aq)] in a 0.0500 mol/L solution of ethanoic acid, CH3COOH(aq) is
A. 4.24 × 10–4 mol/L
B. 9.49 × 10–4 mol/L
C. 9.29 × 10–4 mol/L
D. 5.26 × 10–4 mol/L
ANS: B
The ethanoic acid dissociates as CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO–(aq)
Ionization constant is given as
Ka =
[CH3 COO − ] [H3 O + ]
[CH3 COOH]
The concentration of hydronium ions is equal to the concentration of acetate ions.
Ka =
[H3 O] 2
[CH3 COOH]
or 1.8 × 10–5 =
[H3 O] 2
0.5
[H3O+]2 = 1.8 × 10–5 × 0.05
9.0 × 10–7 mol/L
[H3O+] = (90 × 10–8)1/2 = 9.49 × 10–4 mol L–1.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C1.8K
TOP: Acids and Bases
KEY: Concentration
MSC: Knowledge
49. Which of the following expressions describing the concentration of a solution is not correct?
A. Molality =
Moles of solute
× 1000
Mass of solvent in grams
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B. Molarity =
Moles of solute
Volume of solution in L
C. Normality =
Number of gram equivalents of solute
Volume of solution in mL
D. Formality =
Number of formula masses of solute
Volume of solution in L
ANS: C
The normality of a solution is defined as the number of gram equivalents of a solute dissolved per litre of
a solution.
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C1.5K
TOP: Acids and Bases
KEY: Concentration
MSC: Knowledge
50. Four compounds, carbon tetrachloride, trichlorofluoro methane, phosphorus pentachloride, and
ammonium nitrate, are dissolved in water and the energy changes during the process are noted.
Dissolution of which of the above substances will result in an endothermic process?
A. Carbon tetrachloride
B. Trichlorofluoromethane
C. Ammonium nitrate
D. Phosphorus pentachloride
ANS: C
Dissolving some salts (ionic compound) in water results in an endothermic process. Out of the given
substances, ammonium nitrate is a salt and the rest are covalent compounds. So, when ammonium nitrate
is dissolved in water, it takes in energy. Hence, it is an endothermic process.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C1.3K
TOP: Acids and Bases
KEY: Solubility
MSC: Knowledge
51. Most of the cleaning detergents are in the form of liquids because
A. less quantity is required in the liquid form.
B. in the liquid form, the electrons are in the excited state.
C. in the liquid form, the surface area between the reacting substances is increased.
McGraw Hill Ryerson © 2007
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D. in the liquid form, there is a greater force of attraction between the reactants and the products.
ANS: C
The cleaning power of the detergent increases when the surface area between the reactive chemicals is
increased. Now in liquid form, the substances are broken down into molecules or ions and hence, surface
area between the reacting substances is increased. This results in increasing the cleaning power.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C1.2K
TOP: Acids and Bases
KEY: Dilution
MSC: Knowledge
52. When iron nails are kept in a test tube containing water, they change their color after some time. This
is because a redox reaction takes place. Dissolving reactants in water is often a prerequisite for redox
chemical changes, because
A.
atoms get excited when a substance is dissolved in water
B.
catalyst is more effective when dissolved in water
C.
ionization energy of reactants decreases when they are dissolved in water
D.
the transference of electrons becomes possible when reactants are dissolved in water
ANS: D
Redox reaction involves the transference of electrons between the reactants. These reactions do not occur
by direct contact between the reactants, but by transfer of electron from reactants to water. Ionization of
reactants is possible only in water, such that one reactant is oxidized and the other is reduced. Hence,
chemical reaction takes place faster when reactants are dissolved in water.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C1.2K
TOP: Acids and Bases
KEY: Dilution
MSC: Knowledge
53. Use the following information to answer the next question.
Soda ash, chemically known as sodium carbonate (Na2CO3(s)), is used to soften water for laundering
clothes. When treated with hydrochloric acid, HCl(aq), the reaction produces a gas.
Which of the following gases is released in the above process?
A. Cl2(g)
McGraw Hill Ryerson © 2007
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B. H2(g)
C. CO(g)
D. CO2(g)
ANS: D
Carbonates and bicarbonates react with acids to give CO2 gas
Na2CO3(aq)+ 2HCl(aq) Æ 2NaCl(aq) + H2O(l) + CO2(g)
PTS: 1
DIF: Difficult
OBJ: Section 1.4
LOC: 20–D1.1K
TOP: Chemical Changes
KEY: Chemical equations
MSC: Knowledge
54. Use the following information to answer the next question.
Copper(II) nitrate, Cu(NO3)2(s) is a blue, crystalline solid in its anhydrous form. Nitric acid,
HNO3(aq), is formed when copper(II) nitrate is heated until it decomposes and the resulting gas
fumes are passed directly into water.
What are the products obtained after the decomposition as above?
A. NO(g), O2(g), Cu2O(s)
B. CuO(s), NO2(g), O2(g)
C. NO2(g), O2(g), Cu2O(s)
D. Cu(g), O2(g), NO2(g)
ANS: B
The decomposition reaction of Cu (NO3)2 is
2Cu(NO3)2(s) Æ 2CuO(s) + 4NO2 (g) + O2 (g)
The brown colored gas is NO2(g), nitrogen dioxide.
PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D1.1K
TOP: Chemical Changes
KEY: Chemical Reaction Equations
MSC: Knowledge
55. In a metallurgical process for the generation of tin metal, tin(IV) oxide is reacted with hydrogen gas to
produce tin metal and water vapour. When the chemical equation representing the above reaction is
balanced using lowest whole number coefficients, the coefficient for H2(g) is
McGraw Hill Ryerson © 2007
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A. One
B. Two
C. Three
D. Four
ANS: B
The skeleton equation is
SnO2(s) + H2(g) → Sn(s) + H2O(g)
Balancing the oxygen atom, we get
SnO2(s) + H2(g) → Sn(g) + 2H2O(g)
Now hydrogen atom becomes unbalanced. So, balancing hydrogen atom, we get
SnO2(s) + 2H2(g) → Sn(g) + 2H2O(g)
This is the balanced equation and the coefficient of H2 is two.
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–D1.2K
TOP: Chemical changes
KEY: Chemical reaction equations
MSC: Knowledge
56. A chemist wants to change the oxidation state of iron from +3 to 0. For this purpose, he treats Fe2O3
with carbon. The coefficient of carbon in the balanced chemical equation for the above reaction will be
A. One
B. Two
C. Three
D. Four
ANS: C
(i) The skeleton equation for the above reaction
Fe2O3 + C Æ Fe + CO
(ii) Now skeleton equation along with the oxidation number of each atom is
+3
−2
O
O
+2 −2
Fe2 O2 + C Æ Fe + C O
(iii) The equation may be written as
O.N decreases by 2 per atom
+3
o
Fe2 O3 + C
Æ
o
+2
Fe + CO
O.N decreases by 3 per atom or
3 × 2 per two atom
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(iv) To balance the increase or decrease, multiply C by 3 and Fe2O3 by 1
Fe2O3 + 3C Æ Fe + Co
(v) Balancing all other atoms we get
Fe2O3 + 3C Æ 2Fe + 3CO
Coefficient of carbon is three.
PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D1.2K
TOP: Chemical changes
KEY: Chemical reaction equations
MSC: Knowledge
57. In a metallurgical process, Pb3O4 undergoes a reaction with carbon as shown:
Pb3O4(s) + 2C(s) Æ 3Pb(s) + 2CO2(g)
If 44.5 g of of Pb3O4(s) is, consumed, what mass of lead will be produced?
A. 39.8 g
B. 35.3 g
C. 43.1 g
D. 57.0 g
ANS: B
Molar mass of Pb3O4 = 310 g
According to the balanced chemical equation,
Pb3O4(s) + 2C(s) Æ 3Pb(s) + 2CO2(g)
Amount of Pb produced by 310g of Pb3O4= 246 g
Amount of Pb produced by 1g of Pb3O4=
246
g
310
Amount of Pb produced by 44.5g of Pb3O4 =
246
x 44.5 = 35.3 g
310
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–D1.3K
TOP: Chemical Changes
KEY: Stoichiometry
MSC: Knowledge
58. Use the following information to answer the next question.
Potassium dichromate is used in industries, in calico printing, in tanning of leather. It is used in
photography for hardening of gelatin film.
McGraw Hill Ryerson © 2007
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As an oxidizing agent, acidified potassium dichromate oxidizes tin(II) salts to tin(IV) salts. What is the
balanced ionic equation for this reaction?
A. Cr2O72– + 14H+ + 3Sn2+ Æ 2Cr3+ + 3Sn4+ + 7H2O
B. Cr2O72– + 7H+ + Sn2+ Æ Cr3+ + Sn4+ + 7H2O
C. Cr2O72– + Sn2+ + 7H2 Æ 2Cr3+ + Sn4+ + 7H2O
D. Cr2O72– + 14H+ + 6e– + Sn2+ Æ + 2Cr3+ + 3Sn4+ + 7H2O
ANS: A
Tin(II) salt (Sn2+) is changed to tin(IV) salt (Sn4+) in the following way
Cr2O72– + 14H+ + 3Sn2+ Æ 2Cr3+ + 3Sn4+ + 7H2O
PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D1.4K
TOP: Chemical Changes
KEY: Ionic Equation
MSC: Knowledge
59. Use the following information to answer the next question.
In volumetric analysis of redox reaction, potassium permanganate (KMnO4) is used as an oxidizing
agent. Estimation of ferrous salts, oxalates etc is done in such a reaction between KMnO4 and Mohr’s
salt [FeSO4 (NH4)2 SO4 • 6H2O]. KMnO4 acts as a self indicator.
What is the net ionic equation in this redox reaction?
A. MnO4– + Fe2+ + H2 Æ Mn2+ + Fe3+ + 2H+ + 2O2
B. MnO4– + Fe2+ + 2OH– Æ Mn2+ + Fe3+ + H2O + 3O2
C. MnO4– + Fe2+ + 4H2 Æ Mn2+ + Fe3+ + 4H2O
D. MnO4– + 5Fe2+ + 8H+ Æ Mn2+ + 5Fe3+ + 4H2O
ANS: D
In acidic medium, KMnO4 acts as a strong oxidizing agent.
MnO4– + 5Fe2+ + 8H+ Æ Mn2+ + 5Fe3+ + 4H2O
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–D1.4K
TOP: Chemical Changes
KEY: Ionic Equation
MSC: Knowledge
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60. Use the following information (along with the graphs) to answer the next two questions.
In an experiment of acid - base titration HCl, NaOH, NH3, and CH3COOH are titrated against a strong acid
or a strong base. The data obtained are represented in the following curves.
Graph 1
Graph 2
14
pH Value
pH Value
14
7
7
Volume of acid (in mL)
Volume of base (in mL)
Graph 3
Graph 4
14
pH Value
pH Value
14
7
Volume of acid added (in mL)
7
Volume of base added (in mL)
Match each of the graphs, as numbered above, with the corresponding titration species listed below.
HCl – strong base
_______
NaOH – strong acid
_______
NH3 – strong acid
_______
CH3COOH – strong base
_______
ANS: 2, 1, 3, 4
HCl – strong base
2
NaOH – strong acid
1
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NH3 – strong acid
3
CH3COOH – strong base
4
PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D2.5K
TOP: Chemical Changes
KEY: Titration
MSC: Knowledge
61. Out of the graphs plotted above, which titration curve corresponds to a strong monoprotic base and a
strong monoprotic acid?
A. Graph 1
B. Graph 2
C. Graph 3
D. Graph 4
ANS: A
In the beginning, strong monoprotic base is present, there the pH is high. When a strong monoprotic acid
is run into it, there is a rapid fall of pH in the beginning, but the rate of fall slows down. The equivalence
point in a strong acid/strong base titration is at pH 7.
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–D2.5K
TOP: Chemical Changes
KEY: Titration
MSC: Knowledge
62. A sample mixture of acids, amines and salts is given to students for analysis. Determination of which
of the following characteristics in the above sample will come under the category of qualitative analysis?
A. Number of moles of acids present.
B. Concentration of amines expressed in percentage by mass.
C. Parts per million of salt present.
D. The presence of –COOH group in the sample with the help of spectroscopy.
ANS: D
The purpose of qualitative analysis is to confirm the presence of an element, compound, phase, functional
group, organic compound or liquid in a sample. Qualitative analysis does not involve amounts.
PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D1.3K
TOP: Chemical changes
KEY: Reaction stoichiometry
McGraw Hill Ryerson © 2007
39
MSC: Knowledge
63. Use the following information to answer the next question.
Iron is available in three forms: cast iron, wrought iron and steel. These forms differ from each other
in the carbon content and, of course, traces of certain other elements.
An iron sample of cast iron is 1.5% impure. The following reaction shows the formation of pentacarbonyl
iron:
∆
Fe + 5CO ⎯⎯→
[Fe(CO) 5
Pr essure Pentacarbonyl iron
If 4.2 g of this sample is used in the formation of pentacarbonyl iron, what is the actual yield of
[Fe(CO)5]?
A. 25.2 g
B. 12.9 g
C. 14.5 g
D. 27.1 g
ANS: C
56 g of Fe can give 196 g of [Fe(CO)5].
Since the iron sample is 1.5% impure, mass of impure iron = 4.2 ×
1.5
= 0.063 g
100
Mass of pure iron = 4.2 – 0.063 = 4.137 g
Amount of [Fe(CO)5] produced by 4.136 g of pure iron =
196
× 4.137 = 14.5 g
56
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–D2.3K
TOP: Chemical Changes
KEY: Actual, theoretical and percent yield
MSC: Knowledge
64. Use the following information to answer the next question.
Carbon monoxide gas reacts with solid iron(III) carbonate to produce iron metal and carbon dioxide
gas. The reaction that takes place is
Fe2CO3(s) + CO(g) Æ 2Fe(s) + 2CO2(g)
McGraw Hill Ryerson © 2007
40
Assuming stoichiometric quantities of all reactants and products, what mass of Fe2CO3(s) is consumed
when it is reacted with 5.6 g CO(g), producing 22.4 g of Fe(s) and 17.6 g of CO2(g)?
A. 17.4 g
B. 34.4 g
C. 37.2 g
D. 41.6 g
ANS: B
Fe 2 CO 3 + CO → 2 Fe + 2CO 2
?
5.6 g
22.4 g
17.6 g
According to the law of conservation of mass:
Mass of reactant = mass of products.
Mass of Fe2CO3 + mass of CO = mass of iron + mass of CO2
Mass of Fe2CO3 + 5.6 g = 22.4 g + 17.6 g
Mass of FeCO3 = 40 – 5.6 = 34.4 g
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–D2.1K
TOP: Chemical Changes
KEY: Reaction Stoichiometry
MSC: Knowledge
65. In a fertilizer factory, 410 kg of ammonium hydroxide, NH4OH(aq), reacts with 738 kg of nitric acid,
HNO3(aq), to produce 937.14 kg of the fertilizer ammonium nitrate, NH4NO3(s). Assuming all reactants
and products are in stoichiometric quantities, what is the mass of all other products is formed in this
reaction?
A. 412 kg
B. 383.65 kg
C. 327.1 kg
D. 210.86 kg
ANS: C
The chemical equation is
NH4 OH + HNO 3 → NH 4NO 3 + H 2 O
410 kg
738 kg
937.14 kg
?
According to the law of conservation of mass:
Mass of reactants = mass of products
410 kg+ 738 kg = 937.14 kg + mass of H2O
Mass of H2O = 1148 – 937.14 = 210.86 kg.
McGraw Hill Ryerson © 2007
41
PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D2.1K
TOP: Chemical Changes
KEY: Reaction Stoichiometry
MSC: Knowledge
66. Use the following information to answer the next question.
In a precipitation reaction, 8.5 g of Na2SO4(s) is added to an aqueous solution containing of 15 g of
BaCl2(aq). A white precipitate, insoluble BaSO4(s), is formed.
Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)
Which of the following compounds acts as a limiting reagent in this chemical change?
A. Na2SO4
B. BaCl2
C. BaSO4
D. NaCl
ANS: A
142 g Na2SO4 reacts with BaCl2 = 208 g.
Now 15 g BaCl2 will react with NaSO4 =
142
. × 15 = 10.2 g
208
Mass of Na2SO4 present = 8.5 g.
Mass of Na2SO4 required = 10.2 g.
Hence, Na2SO4 is the limiting reagent.
PTS: 1
DIF: Average-Difficult
OBJ: Section 1.4
LOC: 20–D2.2K
TOP: Chemical Changes
KEY: Limiting and excess species
MSC: Knowledge
67. Iron can be extracted from iron (III) oxide in the reaction Fe2O3(s) + 3C(s) Æ 2Fe(s) + 3CO(g). If
0.50 moles of Fe2O3(s) is present, carbon will be limiting reagent if its number of moles are
A. 0.40 mol or less
B. 1.5 mol or less
C. 1.6 mol or less
D. 2.5 mol or less
ANS: A
McGraw Hill Ryerson © 2007
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We have the following chemical equation
Fe2O3(s) + 3C(s) Æ 2Fe(s) + 3CO(g)
Moles of Fe2O3(s) = 0.5 mol
The mole ratio of reactants in the chemical reaction is given as
Fe2O3: C
or
1:3
C: Fe2O3
1:
1
3
If 0.50 mol Fe2O3(s)is consumed in the reaction, it would require 0.50 mol Fe2O3(s)× 3 mol C(s)/1 mol
Fe2O3(s)= 1.5 moles of C(s). Now if there is only 0.40 mol of C(s) present, this is less than the required
1.5 mol. So if only 0.4 moles of carbon is provided, carbon, C(s), will be the limiting reagent.
PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D2.2K
TOP: Chemical changes
KEY: Limiting and excess species
MSC: Knowledge
68. Which indicator is used for the titration between hydrochloric acid and ammonia?
A. Phenolphthalein
B. Methyl orange
C. Thymol blue
C. Cresol red
ANS: B
In this kind of titration, the equivalence point is below 7 because the salt (NH4Cl) formed at the
neutralization reacts with water to give H+ ions. The equivalence point lies at about pH 5.3. It is,
therefore, necessary to use an indicator with pH range slightly on the acidic side. Therefore, methyl
orange can be used.
PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D2.6K
TOP: Chemical changes
KEY: Titration
MSC: Knowledge
69. While performing an acid–base titration between HCl(aq) and Na2CO3(aq) solutions, a student forgot
to use the indicator. What would be an appropriate indicator to use?
A. Phenolphthalein
B. Methyl red
C. Methyl orange
D. Bromthymol blue
McGraw Hill Ryerson © 2007
43
ANS: C
The titration is between a strong acid (HCl(aq)) and a weak base Na2CO3(aq). The pH at the final
equivalence point will be below 7. The most appropriate indicator will have a Ka halfway between the
hydrogen carbonate ion, HCO3–(aq), and water, which would be methyl orange.
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–D2.6K
TOP: Chemical Changes
KEY: Titration
MSC: Knowledge
70. Solid sulfur, S8(s), and fluorine gas, F2(g), react to produce SF6(s). If 384 grams of S6(s) is reacted
with a limited amount of F2(s) producing 1420 g of SF6(s), what is the percent yield?
A. 44.3%
B. 59.7%
C. 81.1%
D. 93.2%
ANS: C
S 6 + 18F2 → 6SF6
( s)
( g)
( g)
192 g S6 produces SF6 = 876 g.
384 g S6 produces SF6 =
Percent yield =
876
× 384 = 1752 g.
192
Actual yield
1420
× 100 =
× 100 = 81.1%.
1752
Predicted yield
PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D2.4K
TOP: Chemical Changes
KEY: Actual, theoretical and percent yield
MSC: Knowledge
McGraw Hill Ryerson © 2007
44
71. Use the following information to answer the next question.
The pH curve for the titration of HCl(aq) and NaOH(aq).
14
HCl(aq)
4
1
2
7
3
Volume of HCl(aq) added (mL)
The equivalence point in the titration curve between HCl(aq) and NaOH(aq), shown above, is numbered
A. 1
B. 2
C. 3
D. 4
ANS: B
The equivalence point for a strong acid–strong base occurs when pH is equal to 7, labeled #2 on the
graph.
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–D2.7K
TOP: Chemical Changes
KEY: Equivalence point
MSC: Knowledge
McGraw Hill Ryerson © 2007
45
NUMERIC RESPONSE
1. The electron dot diagrams of four elements, A, B, C, and D, which are in the same period of the
periodic table are shown below:
1.
A
2.
B
3.
C
4.
D
The elements in increasing order of electronegativity are _____, _____, _____, and _____, respectively.
ANS: 4, 3, 2, 1
The number of valence electrons of elements increases as you move across a period from left to right. The
electron dot diagram of an atom shows its number of valence electrons. The atom having the fewest
number of valence electrons will be the least electronegative and the element having the highest number
of valence electrons will be the most electronegative. Hence, the elements in the increasing order of
electronegativity are D < C < B < A.
PTS: 1
DIF: Easy
OBJ: Section 1.1
LOC: 20–A2.4K
TOP: Diversity of matter
KEY: Lewis Structure
MSC: Knowledge
McGraw Hill Ryerson © 2007
46
2. Use the following information to answer the next question
In a leading chemical factory, fractional distillation is used to separate ethanol from water. In this
process, ethanol separates from water when the temperature reaches 352 K.
The boiling point of a substance is 22 oC less than the boiling point of ethanol. The boiling point of that
substance is _____K.
(Record your answer in the numerical-response section on the answer sheet.)
ANS: 330 K
The boiling point of the substance is 22 oC less than the boiling point of ethanol on the Celsius scale. So,
the boiling point of the substance on the Celsius scale will be 79 oC – 22 oC = 57 oC, and its boiling point
on the Kelvin scale will be = 57 oC + 273 = 330 K.
PTS: 1
DIF: Easy
OBJ: Section 1.2
LOC: 20–B1.2K
TOP: Forms of matter–gases
KEY: Celsius and Kelvin scales
MSC: Knowledge
3. A 2.00 L flask is filled with 3.20 g H2(g). If the temperature inside of the tube is 27.0 oC, the gas will
exert a pressure of _____.
(Record your answer in the numerical-response section on the answer sheet.)
ANS: 1 × 10–5 atm
The number of hydrogen molecules left in the tube = 4.879 × 1017
The number of moles of hydrogen molecules left =
4.879 × 1017
6.022 × 10
Now we know PV = nRT
P=x
R = 0.082 litre atm K–1 mol–1
V=2L
T = 27 + 273 = 300 K
n = 8.1 × 10–7 mol
Substituting the values
8.1 × 10–7 mol =
x×2
0.082 × 300
x = 1 × 10–5 atm
McGraw Hill Ryerson © 2007
47
23
= 8.1 × 10–7moles
PTS: 1
DIF: Average
OBJ: Section 1.2
LOC: 20–B1.4K
TOP: Forms of matter–gases
KEY: Ideal Gas Law
MSC: Knowledge
4. You want to prepare 1.00 L of a 0.0200 mol/L H2SO4(aq) solution from a 90% (M/M) solution of
H2SO4(aq), with a density 1.80 g/mL. The volume of the 90% H2SO4 solution required to prepare the
1.00 L of 0.0200 mol/L H2SO4(aq) is _____ mL.
(Record your answer in the numerical-response section on the answer sheet.)
ANS: 1.21 mL
Number of moles of H2SO4 =
90
.
98
In 100 mL of solution:
nH2SO4(aq) =
=
m
M
90 g H2SO4
H2SO4
98 g
mol
= 0.918 mol H2SO4(aq)
Volume of solution =
mass of solution
density
Mass of solution = 100 g
Density = 1.80 g mL–1
Volume of solution =
100
= 55.6 mL
1.8 g
Therefore the molar concentration of the original solution is:
CH2SO4(aq) =
=
n
V
0.918 mol H2SO4(aq)
0.0556 L
= 16.5 mol/L H2SO4(aq)
To calculate the quantity of the solution required, use the dilution formula,
C1V1 = C2V2
16.5 mol/L × V1 = 0.0200 mol/L × 1.00 L
V1 =
0.02 × 1 000
= 1.21 mL.
16.5
McGraw Hill Ryerson © 2007
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PTS: 1
DIF: Difficult
OBJ: Section 1.3
LOC: 20–C1.6K
TOP: Acids and Bases
KEY: Dilution
MSC: Knowledge
5. Use the following information to answer the next two questions.
Ammonium chloride, NH4Cl(s), has extensive application in completion of well and prevention of
damage that would result from interaction of clay and water. It is easily soluble in fresh water. The pH
of ammonium chloride brines lies in the range of 5 – 7.
The number of significant digits reported for the concentration of H3O+(aq) ions in a solution of
ammonium chloride with a pH of 5.1 is _____.
(Record your answer on the numerical-response sheet provided).
ANS: 2
pH of the solution is 5.1
pH = –log [H+]
log [H+] = –pH = –5.1
[H+] = antilog [–5.1]
= 7.46 × 10–6
Since pH value has only 1 significant digit, the concentration of [H+] can be represented as
7 × 10–6 mol/L, which has one significant digit.
PTS: 1
DIF: Easy
OBJ: Section 1.3
LOC: 20–C2.4K
TOP: Acids and Bases
KEY: Concentration
MSC: Knowledge
6. Use the following information to answer the next question.
The following steps are used to calculate of volume of dilute solutions from concentrated solutions:
(1) Apply molarity equation.
(2) Determine volume of solution.
(3) Determine the density of acid in solution.
(4) Calculate the molar mass and number of moles of acid.
McGraw Hill Ryerson © 2007
49
The correct order of the above steps for calculating the volume of 70% sulfuric acid, by weight (density =
1.8 g mL–1), required to prepare 500 mL of 0.02 M H2SO4, is _____; _____; _____; and _____.
ANS: 4, 3, 2, 1
70% H2SO4 solution means that 70g of sulfuric acid is present in 100g of solution in water.
Now molecular mass of H2SO4 = 2 + 32 + 64 = 98
Weight of H2SO4 = 70 g.
Moles of H2SO4 present =
70
.
98
Mass of the acidic solution = 100 g.
Density of the acid in solution = 1.8 g mL–1.
Volume of the solution =
Molarity of solution =
Mass
100
=
= 55.5 mL
Density
1.8
70 × 1 000
= 12.9 M
98 × 55.5
Applying molarity equation,
(concentrated) M1V1 = M2V2 (dilute)
12.9 × V1 = .02 × 500
V1 =
0.02 × 500
= 0.77 mL
12.9
So correct order of steps is 4, 3, 2, 1.
PTS: 1
DIF: Difficult
OBJ: Section 1.3
LOC: 20–C1.11K
TOP: Acids and Bases
KEY: Dilution
MSC: Knowledge
7. Use the following information to answer the next question.
The following compounds were provided to students for carrying out various syntheses in a practical
exam. However, when dissolved in water, all the chemicals do not act the same way.
1. Sodium hydroxide
2. Ammonium nitrate
3. Table sugar
The respective order in which the above chemicals will release energy; absorb energy; and neither release
nor absorb energy, is _____, _____, and _____ respectively.
McGraw Hill Ryerson © 2007
50
ANS: 1, 2, 3
Sodium hydroxide, when dissolved in water, releases energy to give an exothermic reaction. Ammonium
nitrate absorbs energy, when it is dissolved in water, to give an endothermic reaction. Table salt neither
absorbs energy nor releases energy, when it is dissolved in water. So the correct order is 1, 2, and 3.
PTS: 1
DIF: Average
LOC: 20–C1.3K
TOP: Acids and Bases
OBJ: Section 1.3
KEY: Solubility
MSC: Knowledge
8. A solution with a concentration of 0.387 mo/L is prepared by dissolving 1.55 g sodium hydroxide,
NaOH(aq) The volume of solution obtained is a.bc × 10–d L. The values of a, b, c, and d are _____,
_____, _____, and _____.
(Record your answer in the numerical-response section on the answer sheet.)
ANS: 1001
The concentation of solution is 0.387 mol/L
mass of NaOH(s) = 1.55 g
nNaOH =
=
m
M
1.55 g NaOH
40.00 g NaOH
mol NaOH
= 0.0388 mol NaOH
CNaOH =
n
V
V=
n
C
0.0388 mol NaOH
0.387 mol NaOH
L
= 0.100 L
When converted to scientific notation, 0.100 L is reported as 1.00 x 10–1 mol/L. The correct answer is
1001.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C1.6K
TOP: Acids and Bases
KEY: Concentration
MSC: Knowledge
McGraw Hill Ryerson © 2007
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9. Use the following information to answer the next question.
In case of a very dilute acidic solution, H+ ion concentrations from acid and water are comparable. In
this case the concentration of H+ ions from water can not be neglected. Therefore,
[H+]total = [H+]acid +[ H+]water
Total concentration of H+ ions in a 10–8 M HCl solution, in scientific notation is abc × 10–d. The values of
a, b, c, and d are _____, _____, _____, and _____, respectively.
ANS: 1, 0, 5, 7
Since HCl is a strong acid and is completely ionized,
[H+] = 1.0 × 10–8
The concentration of H+ ions in water is equal to concentration of OH– ions in water.
[H+ ]H2O = [OH − ]H2O = x
[H + ] Total = 1.0 × 10–8 + x
But [H+] [OH–] = 1.0 × 10–14
(1.0 × 10–8 + x) (x) = 1.0 × 10–14
x2 + 10–8x –10–14 = 0
x=
− 10 −8 ± (10 −8 ) 2 − 4.1.(−10 −14 )
2.1
Solving for x we get
x = 9.5 × 10–8
[H+]total = 1.0 × 10–8 + 9.5 × 10–8
= 10.5 × 10–8
= 1.05 × 10–7
So, the values of a, b, c, and d are 1, 0, 5, and 7, respectively.
PTS: 1
DIF: Difficult
OBJ: Section 1.3
LOC: 20–C2.3K
TOP: Acids and Bases
KEY: Concentration
MSC: Knowledge
10. Use the following information to the answer the next question.
McGraw Hill Ryerson © 2007
52
Sodium acetate (NaCH3COO(aq)) is a strong electrolyte and it dissociates completely when dissolved
in water. Ethanoic acid (CH3COOH(aq)) is a weak electrolyte which ionizes only partially when
dissolved in water.
The concentration of hydronium ion in a mixture of 0.01 M CH3COOH and 0.2 M CH3COONa will be
_____. (Ka for ethanoic acid = 1.8 × 10–5)
(Record your answer in the numerical-response section on the answer sheet.)
ANS: 0.09 × 10–5 mol L–1
From the above information, the ionization of ethanoic acid and sodium acetate may be represented as
CH3COOH + H2O ⇌ CH3COO– + H3O+
CH3COONa Æ CH3COO– + Na+
Now let us assume that the concentration of acetate ions obtained from ethanoic acid is A. Since sodium
acetate is completely ionized, the concentration of acetate ions from sodium acetate is = 0.2 M
Total concentration of acetate ion is = A + 0.2
Concentration of unionized ethanoic acid = 0.01 – A
Now, A is very small as compared to 0.2. So, we can write
[CH3COO–]= 0.2 + A = 0.2 mol L–1
[CH3COOH] = 0.01 – A = 0.01 mol L–1
Now the value of dissociation constant is given as
K=
[H3 O + ] [CH3 COO − ]
[CH3 COOH]
1.8 × 10–5 =
[H3O+] =
[H3 O + ] × 0.2
0.01
1.8 × 10 −5 × 0.01
0.2
[H3O+] = 0.09 × 10–5 mol L–1
PTS: 1
DIF: Difficult
OBJ: Section 1.3
LOC: 20–C1.8K
TOP: Acids and Bases
KEY: Concentration
MSC: Knowledge
11. Use the following information to answer the next question.
A solution is made by mixing two substances A of molecular mass 10 and B of molecular mass 15.
McGraw Hill Ryerson © 2007
53
If the mole fraction of A in a solution with B is 0.30, the mass percentage of A in the solution will be
_____.
(Record your answer in the numerical-response section on the answer sheet.)
ANS: 23.23%
Let the mass percentage of A in solution be x. So 100 g of solution contains x g of A and (100 – x) g of B
Moles of A =
mass of A
x
=
mol. mass of A 10
Moles of B =
100 − x
15
Now mole fraction of A = 0.3
0.3 =
moles of A
moles of A + moles of B
x
10
0.3 =
x 100 − x
+
10
15
x
10
0.3 =
15x + 10(100 − x )
150
0.3 =
150
x
×
15x + 10(100 − x )
10
0.3 =
x × 15
15x + 10(100 − x )
0.3 (15x + 1 000 – 10x) = 15x
0.3 (5x + 1 000) = 15x`
5x + 1 000 =
15x
= 50x
0.3
1 000 = 50x – 5x
45x = 1 000
x=
1 000
= 22.23
45
So, mass percentage of A = 23.23%.
PTS: 1
DIF: Difficult
OBJ: Section 1.3
LOC: 20–C1.5K
TOP: Acids and Bases
KEY: Percent by mass
MSC: Knowledge
McGraw Hill Ryerson © 2007
54
12. Use the following information to answer the next question.
A scientist was working on the ionization of water. He calculated ionic product of water at different
temperatures. At a specific temperature, ionic product of water comes out as 8.25 × 10–14.
The pH of the water at the same temperature will be _____.
(Record your answer in the numerical-response section on the answer sheet.)
ANS: 6.54
Ionic product of water = 8.25 × 10–14 or [H3O+] [OH–] = 8.25 × 10–14
For water, concentration of hydronium ions is equal to concentration of hydroxide ions. So we can write
[H3O+] [H3O+] = 8.25 × 10–14
[H3O+] =
8.25 × 10 −14
= 8.25 × 10–7
= 2.88 × 10–7
Now pH = –log [H3O+]
= –log (2.88 × 10–7)
= –[log 2.88 – 7]
= –[.46 – 7]
= –[–6.54]
pH = 6.54
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C2.4K
TOP: Acids and Bases
KEY: Hydronium ion/pH
MSC: Knowledge
13. The reaction representing the decomposition of nitramide is is shown below.
NH2NO2 (aq) Æ N2O(g) + H2O (l)
If 6.14 g of NH2NO2(aq) decomposes completely, the volume of N2O(g) produced at STP is
____________ L.
(Record your answer in the numerical-response section on the answer sheet.)
ANS: 2.18 L
Molar mass of NH2NO2 = 63 g
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Stoichiometrically,
63 g of NH2NO2 produced N2O at STP is 22.4 L
6.14 g of NH4 NO2 will produce N2O at STP =
22.4
× 6.14 = 2.18 L
63
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–D1.5K
TOP: Chemical Changes
KEY: Stoichiometry
MSC: Knowledge
14. Use the following information to answer the next question.
Eudiometry is a branch of science that deals with the analysis of a gaseous mixture, including the
determination of molecular formula of a gaseous hydrocarbon. The general chemical equation for the
combustion of a hydrocarbon is given as under:
⎛
CxHy + ⎜ x +
⎝
y⎞
y
⎟ O2 Æ xCO2 + H2O
4⎠
2
Butane, a gaseous hydrocarbon is combusted in the presence of oxygen. CO2 gas and 90 g of water are
recovered. How many moles of CO2 gas are recovered?
ANS: 4
The formula of the hydrocarbon = C4H10
Number of moles of H2O =
y
90
= 5 or x = 4, = 5
18
2
x = 4, y = 10
Hence, CxHy = C4H10
Number of moles of CO2 = 4
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–D1.5K
TOP: Chemical Changes
KEY: Stoichiometry
MSC: Knowledge
15. 2.66 g of phosphorus, P4(s), is heated in the presence of 2.43 of oxygen gas, O2(g) The theoretical
yield of phosphorus pentoxide (P2O5(s)) is _____ g.
(Record your answer in the numerical-response section on the answer sheet.)
ANS: 4.31 g
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P4
2.66 g
+ 5O 2 → 2P2 O 5
2.43 g
4.31g
Since Amount of P2O5 produced by 124 g of phosphorus = 2 × 142 g
Amount of P2O5 produced by 2.66 g of phosphorus=
2 × 142
× 2.66 = 6.092 g
124
Theoretical yield of P2O5 = 6.092 g
mass P2O5 (s) produced (if P4 (s) is limiting)
x g P2O5 (s) = 2.66 g P4 (s) ×
1 mol P4 (s)
2 mol P2O5 (s) 141.94 g P2O5 (s)
×
×
123.88 g P4 (s) 1 mol P4 (s)
1 mol P2O5 (s)
= 6.10 g P2O5 (s)
mass P2O5 (s) produced (if O 2 (g) is limiting)
x g P2O5 (s) = 2.43 g O2 (g) ×
1 mol O 2 (g) 2 mol P2O5 (s) 141.94 g P2O5 (s)
×
×
32.00 g O2 (g) 5 mol O2 (g)
1 mol P2O5 (s)
= 4.31 g P2O5 (s)
Since the theoretical mass of P2O5(s) produced with 2.43 g of oxygen gas, O2(g) is less than with
phosphorus, P4(s), oxygen gas is the limiting reagent, and the theoretical mass of P2O5(s) produced is 4.31
g, the theoretical mass produced when all of the limiting reagent is consumed.
PTS: 1
DIF: Difficult
OBJ: Section 1.4
LOC: 20–D2.3K
TOP: Chemical Changes
KEY: Actual, theoretical and percent yield
MSC: Knowledge
16. Use the following information to answer the next question.
In a lab experiment, a solution of an unknown ionic compound AB(s) (molar mass
80 g/mol) is prepared by dissolving 30 g of AB(s) in a 250 mL volumetric flask,
which is then filled with water up to the mark on the neck.
The molar concentration of the solution formed in the above experiment will be
A. 0.015 m
B. 0.025 m
C. 0.030 m
D. 0.035 m
ANS: A
Mass of the compound AB dissolved = 0.30g
Molar mass of AB = 80 g/mol
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Moles of AB dissolved =
.30
80
Volume of solution = 250 mL
So, concentration of the solution will be =
=
0.30 / 80
× 1 000
250
0.30
× 1 000 = 0.015 M.
80 × 250
PTS: 1
DIF: Easy
OBJ: Section 1.4
LOC: 20–C1.5K
TOP: Solutions
KEY: Calculating Concentration
MSC: Knowledge
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WRITTEN RESPONSE
1. Use the following information to answer the next question.
Element Number of Valence Electrons Electronegativity
A
2
2.5
B
7
1.9
C
7
1.5
D
4
1.6
A and B combine to form an ionic compound whereas C and D do not, why? Which will form ionic
bonds, covalent bonds, and polar covalent bonds?
ANS: If difference in electro negativity between two atoms is high, the bonding between them takes place
with complete transfer of electrons, resulting in an ionic bond. In this type of bonding, one of the atoms
loses electrons while the other gains and acquires the nearest noble gas configuration. The bonding
between atoms A and B can be represented by Lewis dot structure as shown below:
A
A2+ , B–
B
[A]2+ 2[B]– or AB2
During the formation of this bond, A will lose its two electrons. Now, B requires only one electron to
complete its octet. So, for the formation of a neutral ionic compound, A will combine with two atoms of
B to form AB2. No such bonding is possible in the case of atoms C and D, which have almost the same
electronegativity. C and D will combine by sharing electrons to form a covalent bond.
PTS: 1
DIF: Difficult
OBJ: Section 1.1
LOC: 20–A1.4K
TOP: Diversity of matter
KEY: Valence electron and Lewis Structure
MSC: Knowledge
2. Use the following information to answer the next question.
Ink and air both are form of mixture. Ink is a liquid, which is composed of various pigments
while air is a mixture of various gases in no definite proportion.
Why ink has visible boundary of separation while air does not?
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59
ANS: Ink is an example of heterogeneous mixture in which one component of mixture is suspended into
the other component while air is an example of non-homogeneous mixture which has same composition
through out. Heterogeneous mixture does contain visible boundary of separation while homogeneous
mixtures do not contain visible boundary of separation. So, ink has visible boundary of separation while
air does not.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C1.1K
TOP: Acids and Bases
KEY: Homogenous mixture
MSC: Knowledge
3. Use the following information to answer the next question.
Most of the common substances used in our home are acidic; slightly acidic; basic; or slightly basic.
The following substances were tested with a pH meter.
Compound
Color
pH value
Soda pop
Red
3.0
Lemon juice
Dark red
2.2
Coffee
Yellow
5.0
Milk
Green
6.5
Sea water
Blue
7.8
(A) Determine the concentration of H+ ions in a sample of milk
(B) Which of the above substances has the highest hydronium ion concentration?
ANS: (A) In a sample of milk, pH paper shows light green color which corresponds to a pH value of 6.5.
Now pH = 6.5 or –log [H3O+] = 6.5
log [H3O+] = –6.5
[H3O+] = antilog (–6.5)
[H3O+] = 3.16 × 10–7
(B) The pH paper turns from blue to red as the acidity of the solution increases. Out of the above
substances lemon juice gives the darkest red color which corresponds to the lowest pH value and the
highest concentration of hydronium ions. So among these substances, lemon juice is the most acidic.
PTS: 1
DIF: Average
OBJ: Section 1.3
LOC: 20–C2.6K
TOP: Acids and Bases
KEY: Concentration
MSC: Knowledge
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60
4. (A) On a graph paper, draw a graph that represents the titration curve of 0.3 M hydrochloric acid and
0.3 M sodium hydroxide solution mark the equivalence point of.
10
8
A.
pH
Equivalence point
6
4
2
0
10
20
30
40
50
Volume of NaOH
(B) In an experiment, titration is being performed between 25 ml of a solution of HCl and a standard
NaOH solution (concentration of both is 0.1M). What will be the value of pH at the equivalence point?
ANS: (A) When we add a solution of NaOH to the solution of HCl, pH progressively increases. This is
because OH– ions from the base will react with the H+ ions from the acid to form water. This decreases
the concentration of H+ ions and pH increases. Near the stoichiometric point there is a sudden jump in pH.
It has been observed that at the equivalence point pH is 7 and the pH increases sharply afterwards.
(B) Initially the pH of HCl is given as
pH = –log [0.1] = 1.0
Now we add 10 ml of 0.1 m NaOH to HCl. The total volume is then = 25 + 10 = 35 ml.
Concentration of H+ ions initially present =
0.1 × 25
= 2.5 × 10–3M
1 000
Concentration of OH– ions in the added NaOH solution =
0.1 × 10
= 1.0 × 10–3M
1 000
After the neutralization of the OH– ions of the added solution, the concentration of H+ ions left
= 2.5 × 10–3 - 1.0 × 10–3
= 1.5 × 10–3 M
Concentration of H+ ions =
1.5 × 10 −3 × 1 000
= 0.043 M
35
pH = –log (0.043) = 1.37
Processing as above, we can calculate the pH of the solution on the addition of 20 ml, 24 ml, 24.9 ml and
24.95 ml of NaOH solution. The pH values come to 1.95, 2.69, 3.70 and 4.00 respectively. Now, on
addition of 25 ml of NaOH, the acid is completely neutralized giving NaCl. The pH of the resulting
solution is 7.
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PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D2.7K
TOP: Chemical changes
KEY: Titration graphs, acids, bases
MSC: Knowledge
5. Use the following information to answer the next question.
In qualitative analysis, group 2 and group 4 cations are precipitated as sulfides by passing H2S(g)
through the aqueous mixture. But in case of group 2, the solution must be acidified with dilute
HCl, before passing the H2S(g), so that the cations of group 4 are not precipitated, if present in
the mixture as sulfide.
Group 2 cations – Hg2+, Pb2+, Bi3+, Cu2+, Cd2+, As3+, Sb3+, Sn2+
Group 4 cations – CO2+, Ni2+, Mn2+, Zn2+
(A) Explain the function of acid in the aqueous mixture solution, along with the relevant chemical
equation(s).
(B) What is the minimum required concentration of S2– ions, to cause precipitation of ZnS from a
1 × 10–3 M Zn (NO3)2 solution? (Ksp for ZnS = 7.0 × 10–16)
ANS: (A) When H2S is passed, the acid is added to suppress the S2– ion (sulfide ion) concentrate in the
solution, due to the common ion effect.
H2s(aq) → 2H+ (aq) + S2–(aq)
HCl(aq) → H+(aq) + Cl–(aq)
common ion
2–
In this way S ion concentration in the solution is controlled and any cation of group 4 is not precipitated
along with the cation of group 2. Also the Ksp (solubility product values) of the sulfides of second group
cations are low and the concentration of S2– ions released by H2S is sufficient to precipitate these sulfides
because the ionic product exceeds the solubility product. As the Ksp values of the sulfides of cations of
fourth group are higher, the ionic products in their case will remain lesser than the solubility products.
Hence, these cations will not be precipitated in the second group. If the medium is not made acidic, the
cations of fourth group may also be precipitated in the second group.
(B) To get precipitates of ZnS, the product [Zn2+] [S2–] should exceed the Ksp for ZnS i.e. 7.0 × 10–16
or [Zn2+] [S2–] > 7.0 × 10–16
As [Zn2+] = 1 × 10–3 mole L–1 in Zn (NO3)2 solution
[Zn2+] [S–2] = 7.0 × 10–16
[1 × 10–3] [S2–] = 7.0 × 10–16
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[S2–] =
7.0 × 10 −16
1 × 10
−3
= 7.0 × 10–3 mole L–1
Thus, minimum concentration of S2– ions required for the precipitation of ZnS is 7.0 × 10–13 mole L–1
PTS: 1
DIF: Average
OBJ: Section 1.4
LOC: 20–D1.3K
TOP: Chemical Changes
KEY: Chemical equations, Stoichiometry
MSC: Knowledge
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