8General chemistry discretes test1 w solutions

General Chemistry
Discretes Test
Time: 30 Minutes
Number of Questions: 30

This test consists of 30 discrete questions—questions
that are NOT based on a descriptive passage. These
discretes comprise 15 of the 77 questions on the
Physical Sciences and Biological Sciences sections of
the MCAT.


MCAT

GENERAL CHEMISTRY
DISCRETES TEST
DIRECTIONS: The following questions are not based
on a descriptive passage; you must select the best
answer to these questions. If you are unsure of the best
answer, eliminate the choices that you know are
incorrect, then select an answer from the choices that
remain. Indicate your selection by blackening the
corresponding circle on your answer sheet. A periodic
table is provided below for your use with the questions.

PERIODIC TABLE OF THE ELEMENTS
1
H
1.0

2

He
4.0

3
Li
6.9

4
Be
9.0

5
B
10.8

6
C
12.0

7
N
14.0

8
O
16.0

9
F
19.0

10
Ne
20.2

11
Na
23.0

12
Mg
24.3

13
Al
27.0

14
Si
28.1

15
P
31.0

16
S
32.1

17

Cl
35.5

18
Ar
39.9

19
K
39.1

20
Ca
40.1

21
Sc
45.0

22
Ti
47.9

23
V
50.9

24
Cr
52.0

25
Mn
54.9

26
Fe
55.8

27
Co
58.9

28
Ni
58.7

29
Cu
63.5

30
Zn
65.4

31
Ga
69.7

32

Ge
72.6

33
As
74.9

34
Se
79.0

35
Br
79.9

36
Kr
83.8

37
Rb
85.5

38
Sr
87.6

39
Y
88.9

40
Zr
91.2

41
Nb
92.9

42
Mo
95.9

43
Tc
(98)

44
Ru
101.1

45
Rh
102.9

46
Pd
106.4

47

Ag
107.9

48
Cd
112.4

49
In
114.8

50
Sn
118.7

51
Sb
121.8

52
Te
127.6

53
I
126.9

54
Xe
131.3

55
Cs
132.9

56
Ba
137.3

57
La *
138.9

72
Hf
178.5

73
Ta
180.9

74
W
183.9

75
Re
186.2

76

Os
190.2

77
Ir
192.2

78
Pt
195.1

79
Au
197.0

80
Hg
200.6

81
Tl
204.4

82
Pb
207.2

83
Bi
209.0

84
Po
(209)

85
At
(210)

86
Rn
(222)

87
Fr
(223)

88
Ra
226.0

89
Ac †
227.0

104
Unq
(261)

105

Unp
(262)

106
Unh
(263)

107
Uns
(262)

108
Uno
(265)

109
Une
(267)

*

58
Ce
140.1

59
Pr
140.9

60

Nd
144.2

61
Pm
(145)

62
Sm
150.4

63
Eu
152.0

64
Gd
157.3

65
Tb
158.9

66
Dy
162.5

67
Ho
164.9

68
Er
167.3

69
Tm
168.9

70
Yb
173.0

71
Lu
175.0

†

90
Th
232.0

91
Pa
(231)

92
U
238.0

93
Np
(237)

94
Pu
(244)

95
Am
(243)

96
Cm
(247)

97
Bk
(247)

98
Cf
(251)

99
Es
(252)

100

Fm
(257)

101
Md
(258)

102
No
(259)

103
Lr
(260)

GO ON TO THE NEXT PAGE.

2

as developed by


General Chemistry Discretes Test
1 . Which of the following statements about molecular
and empirical formulas is (are) FALSE?
I. A given compound can have the same
molecular and empirical formula.
II. The empirical formula is a whole number
multiple of the molecular formula.
III. H2O2 represents the empirical formula of

hydrogen peroxide.
A.
B.
C.
D.

III only
I and II only
II and III only
I, II, and III

2 . All of the following statements are consistent with
Bohr’s model of the atom EXCEPT:
A . an electron may assume an infinite number of
velocities.
B . an electron is most stable in its ground state.
C . the electron shell numbers represent the principal
energy levels.
D . electrons in orbitals closest to the nucleus have
the lowest energy.

4 . What volume of water would be needed to dilute
50mL of 3M H2SO4 to 0.75M?
A.
B.
C.
D.

50 mL
100 mL

150 mL
200 mL

5 . Among the elements with partially filled p subshells,
atomic radii decrease significantly from left to right
across a period while the atomic radii of transition
elements change only slightly from left to right
across a period. This difference is most likely due to:
A . the presence of greater numbers of electrons in
the outer shells of transition elements.
B . the presence of greater numbers of electrons in
the outer shells of elements with partially filled p
subshells.
C . the greater nuclear charges of transition elements.
D . the presence of d electrons in the second
outermost shell in transition elements.

6 . H2O has a higher boiling point than HF because:
3 . Which of the following statements is NOT true of a
gas at pressures greater than 400 atm and temperatures
close to 0 K?
A . The pressure of the gas is always lower than that
predicted by the ideal gas law.
B . The extent to which the gas deviates from the
ideal gas law depends on the molecules’ size and
polarity.
C . Intermolecular forces become significant.
D . The ideal gas equation is less accurate under these
conditions than at higher temperatures and lower
pressures.

A.
B.
C.
D.

H2O is more polar than HF.
H2O can form more hydrogen bonds.
H2O has a higher molecular weight.
H2O has more atoms.

7 . Which of the following compounds contains the
greatest percentage of oxygen by weight?
A.
B.
C.
D.

C3H6O5Cl
C3H6O2
C5H10O5
C4H8O3

GO ON TO THE NEXT PAGE.

KAPLAN

3

MCAT
8 . Which of the following conditions guarantees a
spontaneous reaction?
A.
B.
C.
D.

Positive ∆H, positive ∆S
Positive ∆H, negative ∆S
Negative ∆H, negative ∆S
Negative ∆H, positive ∆S

9 . Which of the following statements concerning
equilibrium constants is true?
A . An equilibrium constant that is close to 1 means
that the reaction does not favor either reactants or
products heavily.
B . Equilibrium constants have units of L/mol.
C . The stoichiometric coefficients in a reaction are
not part of the equilibrium expression.
D . The equilibrium constant for a reaction is the
same for any temperature.

1 3 . The Ksp of Mg(OH)2 in water is 1.2 × 10–11
mol3/L3. If the Mg 2+ concentration in an acid
solution is 1.2 × 10–5, what is the pH at which
Mg(OH)2 just begins to precipitate?
A.
B.

C.
D.

3
4
5
11

1 4 . The rate law expression for the following reaction:
N2 + 3 H2 → 2 NH3
A.
B.
C.
D.

can be represented by rate = [NH3]2 / [N2][H2]3.
can be represented by rate = k[NH3]2.
can be represented by rate = k[NH3]2 / [N2][H2]3.
cannot be determined from the information given.

1 5 . K+ and Cl– have the same:
1 0 . If the pressure of a gas sample is doubled at constant
temperature, the volume will be:

A.
B.
C.
D.

atomic weight.

electronic configuration.
ionization constant.
number of protons and neutrons.

A . 4 times the original.
B . 2 times the original.

1 6 . Which of the following
characterizes a galvanic cell?

A.
B.
C.
D.

relates resistance and voltage.
relates voltage and current.
relates resistance and concentration.
relates concentration and voltage.

1 2 . The melting points of ionic solids are:
A.
B.
C.
D.

4

high because of their electrostatic attractions.
high because of their high densities.

low because of their intermolecular attractions.
low because of their high dissociation constants.

correctly

I. Oxidation occurs at the anode, which is
negative.
II. Oxidation occurs at the anode, which is
positive.
III. Reduction occurs at the cathode, which is
positive.

1
of the original.
2
1
D.
of the original.
4
C.

1 1 . The Nernst equation:

statements

A.
B.
C.
D.

II only
III only
I and III only
I, II, and III

1 7 . In order to make a buffer solution, a weak monoprotic
acid could be added to:
A.
B.
C.
D.

another acid.
another base.
the salt of its conjugate base.
the salt of its conjugate acid.

GO ON TO THE NEXT PAGE.

as developed by


General Chemistry Discretes Test
1 8 . When chromium metal is used to form K2Cr2O7 the
oxidation state of chromium changes from:
A.
B.
C.
D.

0
3
2
0

to
to
to
to

4.
6.
6.
6.

1 9 . Which of the following molecules contains both
ionic and covalent bonds?
A.
B.
C.
D.

C6H14
MgCl2
(NH4)2SO4
H2O

2 3 . What is the range of possible values for the
[OH–] / [H+] ratio in an aqueous acid solution?
A.

B.
C.
D.

0–1
0 – 14
1 – 14
1–∞

2 4 . A flask contains three times as many moles of H2 gas
as it does O2 gas. If hydrogen and oxygen are the only
gases present, what is the total pressure in the flask if
the partial pressure due to oxygen is Pl?
A . 4P 1
B . 3P 1
4
P
3 1
3
D.
P
4 1
C.

2 0 . Which of the following will increase the rate at which
ice melts in a closed container if all other parameters
are kept constant?
A.
B.
C.

D.

Adding water with a temperature of 0°C
Lowering the temperature below 0° C
Lowering the pressure
Raising the pressure

2 5 . Which of the following molecules is polarized?
A.
B.
C.
D.

BH3
NF3
C2H6
SF 6

2 1 . BaCl2 dissociates in water to give one Ba2+ ion and
two Cl – ions. If concentrated HCl is added to this
solution:
A.
B.
C.
D.

[Ba2+] increases.
[Ba2+] remains constant.
[OH–] increases.
the number of moles of undissociated BaCl2

increases.

2 2 . Avogadro’s number is NOT equal to:
A.
B.
C.
D.

2 6 . Which of the following is NOT a true statement
about the entropy of a system?
A . Entropy is a measure of the randomness in a
system.
B . The entropy of an amorphous solid is greater
than that of a crystalline solid.
C . The entropy of a spontaneous reaction cannot
decrease.
D . At constant temperature and pressure, the entropy
of a system will spontaneously increase.

the number of atoms in 11.2 L of O2 at STP.
the number of atoms in 1 mole of He at STP.
the number of electrons in 96,500 coulombs.
the
number
of
SO
GO ON TO THE NEXT PAGE.
ions in 1 L of 1 N sulfuric acid.

KAPLAN

5


MCAT
2 7 . From the following reaction and thermodynamic data,
what is the energy released when a C—H bond forms?
(Note: ∆Hf° CO2(g) = –393 kJ/mol, and ∆Hf° H2O(g)
= –242 kJ/mol.)
CH4 + 2 O2(g) → CO2(g) + 2 H2O(g) ∆H°rxn = – 802
kJ/mol
A.
B.
C.
D.

–75
–19
75
167

kJ/mol
kJ/mol
kJ/mol
kJ/mol

2 9 . What would be the stoichiometric coefficient of
hydrochloric acid in the following equation?
Cl2 + H2O → HCl + HClO3
A.

B.
C.
D.

1
3
5
10

3 0 . Which of the following compounds can act as a
Lewis base?
2 8 . Which of the following generalizations CANNOT be
made about the phase change of a pure substance from
solid to liquid?
A.
B.
C.
D.

6

It involves a change in potential energy.
It involves no change in temperature.
It involves a change in kinetic energy.
It involves a change in entropy.

A . HClO2
B . NH2NH2
C . NH4+
D . BF3

END OF TEST

as developed by


General Chemistry Discretes Test

THE ANSWER KEY IS ON THE NEXT PAGE

KAPLAN

7


MCAT
ANSWER KEY:
1. C
11. D
2. A
12. A
3. A
13. D
4. C
14. D
5. D
15. B
6.
7.
8.
9.

10.

8

B
C
D
A
C

16.
17.
18.
19.
20.

C
C
D
C
D

21.
22.
23.
24.
25.

D
D

A
A
B

26.
27.
28.
29.
30.

C
B
C
C
B

as developed by


General Chemistry Discretes Test
GENERAL CHEMISTRY DISCRETES TEST EXPLANATIONS
1.
The answer to question 1 is choice C . To answer this, you must understand the differences between
empirical and molecular formulas. The empirical formula of a compound is the simplest whole number ratio of the atoms in
it, whereas the molecular formula represents the actual whole number ratio of the atoms. The molecular formula is usually a
whole number multiple of the empirical formula, although sometimes the empirical and molecular formulas are the same.
Given this information, let's go through the statements to determine which of them is false. Statement number I is true. For
instance, water, H2O, has the same empirical and molecular formula. Statement II is false because it is the molecular formula
which is usually a whole number multiple of the empirical formula, not the other way around. Finally, statement III is false.
H2O2, hydrogen peroxide, has an empirical formula of HO, so the empirical formula is different from the molecular formula

given. Another is glucose, C 6H 12O 6, which has an empirical formula of CH2O. Since statements II and III are both
incorrect, choice C is the answer we're looking for.
2.
In question 2, the right answer is choice A. Let's go through each of the choices and see how they apply
to Bohr's model of the atom. Choice A, which says that an electron may assume an infinite number of different velocities, is
true in classical mechanics but not in Bohr's model. Bohr used quantum theory in developing his atomic model and placed
specific conditions on the possible values of the electron velocity. Since we're looking for the incorrect statement, A is the
correct answer. Choice B, which says that an electron is most stable in its ground state, is true. The ground state of an
electron is its lowest possible energy state. From thermodynamics, we know that systems tend to be the most stable at low
energy; thus, an electron is most stable in the ground state. Choice C says the electron shell numbers represent the principal
energy levels. This is true; the principal quantum number of a given electron is the same as the number of its electron shell
in the Bohr model. Choice D says that electrons closest to the nucleus have the lowest energy. As an electron absorbs
energy, it jumps to a higher orbital and increases its distance from the nucleus. The electron drops to a lower orbital releasing
energy or it escapes the pull of the nucleus, making the atom a cation. Again, choice A is the correct answer.
3.
For question 3, the correct choice is A . One is asked in this question to determine which of the choices
contains a false description of the gas. Under conditions of high pressure and low temperature, the gas is not behaving ideally
and corrections must be made for the volume of the gas molecules and the intermolecular forces between them. Therefore,
choice D, which states that the ideal gas equation would not describe the gas accurately is quite correct. Choice C is also a
characteristic of the gas since the high pressure reduces the volume and thus the distance between the molecules.
Intermolecular forces, therefore, have a greater effect on each gas molecule. Now we have narrowed our choices down to A
and B. The van der Waals equation is useful in describing and predicting the behavior of non-ideal gases, and one of its
features is the introduction of two parameters a and b, which account for intermolecular forces and excluded volumes
respectively. The values of these parameters depend on the gas in question; more specifically, the magnitude of a is
determined by how strongly the gas molecules attract one another, and b is influenced by the size of the molecules. The
higher the values of a nd b, the more the behavior of the gas is non-ideal. (Recall that if a =b=0, one gets back the ideal gas
law.) Therefore choice B is also correct. By elimination, one can hence conclude that the correct answer is choice A because it
contains a false statement. In rationalizing the van der Waals equation, one observes that the presence of intermolecular
attraction leads to a lower pressure than would be expected from using the ideal gas law, but when coupled with the effect of
excluded volume, no generalization can be made as to whether the actual or ideal pressure would be greater. (This holds for gas

volume as well, unless if condensation has taken place, in which case the volume occupied by the liquid will be much smaller
than that predicted by any gas law.)
4.
For question 4, which is a simple dilution problem, the correct answer is C . The number of
moles of solute in the solution will be the same after dilution as before, and the number of moles in each case is equal to the
molar concentration multiplied by the volume of solution. This means that the initial concentration times the initial volume
will be equal to the final concentration times the final volume. So, the final volume will equal the initial concentration
times the initial volume divided by the final concentration. If we work out the math, we find that the final volume of the
diluted solution will be 0.2 liters, or 200 milliliters. Since we started out with 50 milliliters of the solution, we would have
to add 150 milliliters of water to get a final volume of 200 milliliters. Therefore, the correct answer is choice C.
5.
Question 5 is correctly answered by choice D. To answer this question, we have to think about the thing
that determines atomic radius: how strongly the outermost electron shell is attracted to the nucleus. Electrons are attracted to
the nucleus by its positive charge. There are two factors that affect the strength of that attraction: more protons for a
stronger pull by the nucleus, which reduces the atomic radius, and the shielding of the outermost electron shell by the
intervening filled electron shells, which increases the atomic radius. So trends in atomic radii result from interaction between
these two competing effects. The key thing here is that the electrons in the outer shell are not very effective at shielding each
other from the nuclear pull, while electron shells further in toward the nucleus do have a strong shielding effect on electrons
in the outer shell.
Now, because of the order in which electron orbitals fill, as you go from left to right across the table, each
subsequent transition element generally has one more d electron but doesn't change the number of electrons in the outermost s
subshell. However, when you look at the elements with the partially filled p subshells, each subsequent element has one
more p electron. The d electrons are in the next-to-last shell, so they shield the valence s shell pretty effectively. Thus, even
though the number of protons in the nucleus increases from left to right across the table, the fact that the number of d
electrons is increasing at the same time means that the atomic radii don't decrease significantly. Therefore, choice D is the
correct answer. By contrast, the elements with the partially filled p subshells are adding electrons to their p subshell, which
is their outermost subshell. Thus, choice B is wrong because it does not explain the strange trends in transition metal radii.
KAPLAN

9

MCAT
Choices A and C are wrong because within a given period, the transition elements have fewer electrons and smaller nuclear
charges that the elements with partially filled p orbitals. Once again, choice D is the right answer.
6.
The correct answer to question 6 is choice B. This question asks you to determine why water has a
higher boiling point than hydrogen fluoride. Choice A says that water has a higher boiling point because it is more polar.
Before you decide whether polarity has an effect on the boiling point of a substance, you should be able to eliminate this
choice by realizing that water is not more polar that HF. Hydrogen fluoride has a greater charge separation between its
constituent atoms and thus has a more polar bond. Choice B says that water has a higher boiling point because it can form
more hydrogen bonds. Hydrogen bonding does affect boiling points by increasing the attraction between the molecules of a
compound. Water is capable of forming as many as four hydrogen bonds per molecule while hydrogen fluoride can only form
2. This intermolecular attraction leads to a complexation of water molecules and contributes to the high boiling point of
water. Thus choice B is correct. Although we have found the right answer, let's look at the rest of the choices. Choice C
says that water will have a higher boiling point because it has a higher molecular weight. Well, this statement is also false
in and of itself because hydrogen fluoride, at 20 grams per mole, has a greater molecular weight than water, which is only 18
grams per mole. Finally, choice D says that water has more atoms than hydrogen fluoride. Well, this is true, but it has
nothing to do with boiling point, so is not the answer we're looking for. So, the correct answer is choice B.
7.
For question 7, the correct answer is C . You need to first calculate the weight of oxygen in each
compound and then divide that value by the compound's molecular weight. The weight of oxygen in choice A is 80 grams
per mole and the molecular weight is 157 grams per mole. Dividing this out gives 51% oxygen. For B, the weight of
oxygen is 32 grams per mole and the molecular weight is 74 grams per mole. This gives you 43% oxygen. Choice C
contains 80 grams of oxygen per mole and its molecular weight is 150 grams per mole. Here, you get 53% oxygen.
Finally, choice D contains 48 grams of oxygen per mole and its molecular weight is 104 grams per mole, which yields 46%
oxygen. Obviously, choice C contains the most oxygen by weight and is the correct answer.
8.
The correct answer for question 8 is D . This question asks you to predict which combination of ∆H and
T∆S values will always give a spontaneous reaction. Recall that spontaneous reactions have negative values of Gibbs free

energy, ∆G, and that ∆G equals ∆H – T∆S, where ∆H is the change in enthalpy, ∆ S is the change in entropy, and T is the
absolute temperature. From this equation, it's clear that the best way to guarantee a negative ∆G is to have a negative ∆H
value and a positive ∆S value, since T, always expressed in Kelvin, is always positive. This corresponds to choice D, so that
is the right answer.
9.
For question number 9, choice A is correct. To answer this question, you need to understand the concept
of equilibrium and know the expression for the equilibrium constant. Let's go through all of the choices. Choice A says that
if an equilibrium constant is close to 1, then neither reactants nor products are favored heavily. This is correct because the
equilibrium constant expresses a ratio of product concentrations to reactant concentrations. If the reaction is heavily in favor
of products, the value of the equilibium constant will be much greater than 1.. Conversely, if the reaction favors reactants,
then the equilibrium constant would have a value much less than one. Choice B says that equilibrium constants have units of
L/mol. This is incorrect; the equilibrium constant is actually a dimensionless number. Choice C states that stoichiometric
coefficients from the reaction have nothing to do with the equilibrium expression. The equilibrium expression is derived by
multiplying the concentrations of the products raised to their stoichiometric coefficients together and then dividing by the
product of the reactant concentrations raised to their stoichiometric coefficients. So we know choice C is wrong. Finally,
choice D says that the equilibrium constant of a reaction is the same for any given temperature. This is wrong because an
equilibrium constant is a characteristic of a system at a particular temperature. Changing temperatures changes the system
identity, so it also changes the equilibrium constant. Again, the correct answer is choice A.
10.
For question 10, the right answer is choice C . The question is an application of Boyle's Law. This
states that at constant temperature, the pressure and volume of a gas are inversely proportional to each other. Therefore, since
the pressure of the gas in the question increased, the volume must decrease. Since the pressure and volume are inversely
related, when the pressure is doubled, the volume is halved. This is choice C.
11.
The answer to question 11 is choice D . The Nernst equation expresses the relationship between the
observed potential of a redox system and the concentrations of the reactants and the products involved. It is expressed as
follows: the potential of the cell, E, is equal to the standard potential of the cell minus the quantity RT over nF times the
natural log of the reaction quotient, Q. R is the universal gas constant expressed in joules per mole Kelvin, T is the absolute
temperature, n is the number of moles of electrons exchanged in the redox reaction, and F is Faraday's constant. Q is just a
representation of the extent to which a reaction has proceeded. Since potential is expressed in volts and Q is expressed in

concentration units, the equation relates voltage and concentration and the correct answer is D.
12.
Moving on to question 12, the correct answer is A. Because an electron is transferred from one atom to
another in an ionic bond rather than being shared as in a covalent bond, the atoms in an ionic solid exist as ions and are held
together by very strong electrostatic forces. Because these forces hold them in tight crystalline patterns, ionic solids generally
have high melting points. This makes choice A correct. Density, as in choice B, doesn't explain the high melting points of
ionic solids since substances can be very dense and still melt easily. The concept of intermolecular forces, choice C, do not
really apply in the case of ionic solids because there are no discrete molecules within such a solid: it is a giant lattice of
positive and negative ions. Further, choice D is also wrong because even though ionic solids do tend to dissociate easily, this

10

as developed by


General Chemistry Discretes Test
has nothing to do with melting points. For instance, table salt, NaCl, will readily dissociate and dissolve in water, but if you
were to simply place the solid on a hot plate, it would take a long time to melt. Therefore, choice A is the best answer.
13.
Question 13 is appropriately answered by choice D . This question deals with solubility constants and
the common ion effect. The first step to solving this problem is to express the solubility product constant, which is the ion
product of the saturated solution, as the product of the concentration of Mg2+ ion and the concentration of the hydroxide ion
squared. Second, you must determine the minimum concentration of hydroxide necessary to precipitate the Mg(OH)2. From
the equation just given, you can solve for the concentration of hydroxide, which is the square root of the K sp divided by the
concentration of magnesium. Substituting in the values provided in the question, you should find that the concentration is
equal to ten to the minus 3 moles per liter. That means that the pOH of the solution, the negative log of the hydroxide ion
concentration, will be 3. Since the pOH plus the pH of any solution must equal 14, the pH must be 11. This corresponds to
choice D.
14.
The correct answer to 14 is D . This question asks you to determine the rate law of the reaction for the

formation of ammonia, NH3. A rate law is an equation which gives the relationship between the rate of the reaction and the
concentration of the reactants, each raised to an appropriate power which will depend on the exact reaction. For example, the
rate of this reaction is equal to k times [N2] raised to the x power times [H2] raised to the y power, where k is the rate
constant and x and y are real numbers. x is called the partial order of the reaction with respect to N 2 and y is the partial order
of reaction with respect to H2. x plus y gives the overall order of reaction. k, x, and y can only be determined experimentally
by systematically varying the initial concentration of one reactant. In particular, there is no relationship between the
coefficients of the balanced equation and the order of reaction. In this question, choices A, B, and C show rate laws which
contain the concentrations of the product, ammonia, and the actual values for the partial orders of reaction. However, since
there is no experimental data, there is no way for us to know what these values are, so these three choices are wrong. The
only correct answer is D, which says that the rate law cannot be determined from the information given.
15.
The right answer for question 15 is B. The atomic weight of an element depends on the number of protons
and neutrons in its nucleus; those numbers are always unique to a particular element, regardless of the number of electrons in
the species. Therefore, choices A and D are incorrect. The ionization constant, choice C, depends chiefly on the radius of the
parent atom and the effective charge of the nucleus. Since K+ and Cl – have different numbers of protons by definition, they
will have different effective charges, different atomic radii, and therefore different ionization constants, so choice C is wrong.
Choice B, however, says that the two ions have the same electronic configuration. Well, chlorine belongs to the third period
and potassium belongs to the fourth, so in their unionized forms, chlorine's third shell contains seven electrons and
potassium's fourth shell contains one electron. If chlorine gains one electron and potassium loses one electron, both will
have eight electrons in the third shell, which becomes the valence shell. So the potassium and chlorine ions described in the
question both contain the same number of electrons and the same number of occupied orbitals and thus share the same
electronic configuration. Again, choice B is the correct answer.
16.
For question 16, the correct answer is choice C . This question asks which of the given statements
describes a galvanic cell. This is a roman numeral question, so remember that one or more of the answers may be correct.
The first thing to remember is that galvanic cells are capable of spontaneous reaction. Let's go through each of the choices to
see which of them are correct. Statement I says that oxidation occurs at the anode, which is negative. In all electrochemical
cells, oxidation occurs at the anode and reduction occurs at the cathode. In addition, the anode in a galvanic cell is negative,
meaning that it is a source of electrons. Since a species loses electrons when it is oxidized, this should make sense. There is
a trick to remembering these facts. In a galvanic cell, oxidation occurs at the anode, which is negative. Alphabetically, anode

comes before cathode, oxidation before reduction, and negative before positive. However, this little trick only works for
galvanic cells, but that's what we've got here. Thus, statement I is correct and statement II is wrong. That eliminates choices
A, B, and D, so the answer must be C. Statement II says that the anode is positive in a galvanic cell. We just talked about
that being wrong, but you may have gotten confused since electrolytic cells have positive anodes. Statement III is the
compliment of statement I, so it must also be true. That makes choice C the right answer.
17.
The correct answer to number 17 is choice C. Buffers are solutions made from a mixture of a weak acid
and a salt containing its anion. They resist pH change due to the addition of acid or base. Thus, to make a buffer solution
with a weak monoprotic acid, the addition of the corresponding salt is required. This salt must contain the anion, or
conjugate base, of the acid. Therefore, the way to make a buffer solution when you start with a weak monoprotic acid is to
add the salt of the conjugate base to the solution. This is choice C, which is thus the right answer. If you add a strong acid
to a weak acid, you just get an acidic solution, so choice A is wrong. If you add a weak acid to a strong base, the base will
neutralize the acid, so choice B is wrong. Finally, the only way that an acidic species can have a conjugate acid is if it is
polyprotic. For example, the species HSO4– is acidic, but is the conjugate base of the acid H2SO4. We know that our acid is
monoprotic, so choice D is way off the mark. Again, the correct answer is choice C.
18.
Question 18 is choice D . This question asks you to determine the change in the oxidation number of
chromium when it goes from its elemental state into K 2Cr2O 7. By definition, the oxidation number of any element in its
elemental form is zero. Choices B and C do not have zero in them, so they can easily be eliminated. Next, you should
calculate the oxidation number of chromium in K 2Cr2O 7. Since K 2Cr2O 7 is a neutral molecule, the sum of the oxidation
numbers of all of the atoms in it must be zero. K is in group one A, so it has an oxidation number of +1. In most
compounds, oxygen has an oxidation number of –2. Generally, any time you have to figure out the oxidation state of oxygen
in a compound, it is safe to assume that it is –2 unless you have reason to suppose otherwise. Okay, if you set up the
KAPLAN

11


MCAT
calculation, you get (2 × +1) + (7 × –2) for a total of –12. That means that between them, the two chromium atoms have to

add up to +12 to balance out the molecular charge. That means that in K2Cr2O7, each chromium atom has an oxidation state
of +6, so chromium has gone from 0 to +6, which is choice D.
19.
The correct answer for question 19 is choice C. This question asks you which of the molecules given as
answer choices contains both ionic and covalent bonds. Choice A is hexane, which is a hydrocarbon. This is wrong because
all the bonds in hexane are covalent carbon-carbon or carbon-hydrogen bonds. Choice B is magnesium chloride, which is a
salt. This only has ionic bonds, so choice B is wrong. Choice C is ammonium sulfate. The bonds between nitrogen and
hydrogen in the ammonium ion and between sulfur and oxygen in the sulfate ion are all covalent. However, the bond
between the ammonium ion and the sulfate ion is an ionic bond. Thus, ammonium sulfate has both ionic and covalent
bonds, so choice C is correct. Finally, choice D is water. The oxygen-hydrogen bonds in water are covalent, so choice D is
incorrect. Again, choice C is the right answer.
20.
For question 20, answer choice D is correct. This question asks you which of the choices will increase
the rate at which ice melts in a closed container. Okay, the first two choices both deal with temperature -- and of course it's
intuitively obvious that temperature changes can alter the melting rate of ice. Now, choice A says that the melting rate will
be increased if water at zero degrees Celsius is added to the container. But zero degrees Celsius is the temperature at which the
solid and liquid forms of water are in equilibrium, that is, the melting or freezing point of water, so adding water at that
temperature won't change the melting rate at all. Choice B says that the rate will increase if the temperature is dropped below
zero degrees Celsius. This is even worse. As I just said, zero degrees Celsius is the point at which the solid and liquid are in
equilibrium, so if you lower the temperature, you are tipping the equilibrium toward the ice. The last two choices are less
intuitive. They deal with pressure, and to evaluate them, we have to apply Le Chatelier’s Principle, which says that a system
in equilibrium that is subject to stress will shift its equilibrium so as to relieve the stress. If the pressure is lowered, as in
choice C, the system will counteract the change in pressure by shifting its equilibrium toward the phase that is less dense. In
the case of water, that is the ice. Remember that water has a strange property in that the solid form, ice, at zero degrees
Celsius is less dense than the liquid phase, water, at that temperature. Since the reduction of pressure drives the system to
produce ice, this is the wrong answer. However, choice D, an increase in pressure, will have the opposite effect and the water
will be produced preferentially. That means that the ice is melting faster and D is the correct answer.
21.
Choice D is the right answer for question 2 1 . To answer this question, you should know that when
hydrochloric acid, a very strong acid, is added to a solution, it will dissociate completely into hydrogen ions and chloride ions.

This increases the concentration of chloride ions already in the solution from the barium chloride. According to Le Chatelier's
Principle, if the concentration of one reaction species is increased, the reaction will be driven in the opposite direction. For
this example and the dissociation reaction of barium chloride, if you increase the concentration of chloride ion, you will drive
the reaction in the direction of reassociation of barium chloride. Therefore, barium chloride will precipitate out of the
solution. That means that choice A, which says that the concentration of barium ion increases, and choice B, which says that
the concentration of barium ion stays constant, are wrong. Choice C says that the concentration of hydroxide ion will
increase. This is wrong because it will actually decrease so as to neutralize the hydrogen ions added by the hydrochloric acid.
Choice D says that the number of moles of undissociated barium chloride increases. This is the same as saying that barium
chloride precipitates out of solution, so choice D is the correct answer.
22.
The correct answer for question 22 is choice D. Avogadro's number, 6.02 times ten to the twenty-third,
is the number of molecules in one gram molecular weight of a substance, or one mole of the substance. So the correct
answer choice is the choice that does not describe one mole of a species. You probably remember that a mole of gas at STP
occupies a volume of 22.4 liters. Choice A gives the volume of a half a mole of oxygen molecules, but since each molecule
of oxygen gas has two oxygen atoms, choice A represents Avogadro's number of atoms. Choice B, one mole of helium, is
obviously a mole. Choice C is the definition of a Faraday, which is one mole of electrons. Choice D, the number of sulfate
ions in a 1 normal sulfuric acid solution is not a mole. The reason is that there are two equivalents of hydrogen in each mole
of sulfuric acid. A one normal solution contains one mole of hydrogen ions, but only half a mole of sulfate ions. One mole
of sulfate ions from sulfuric acid would have to be 2 normal. So the right answer is choice D.
23.
The correct answer to question 23 is choice A . The question asks you to give the range of possible
values for the ratio of hydroxide ion concentration to hydrogen ion concentration in an aqueous acid solution. Well, for a
solution to be acidic, it must have a pH between about 1 and 6.999. That means that the pOH of these solutions will be
between 13 and 7.001. We know that pH is equal to the negative log of the hydrogen ion concentration in solution and the
pOH is equal to the negative log of the hydroxide ion concentration in solution. From this we can determine that the range of
possible hydrogen ion concentrations in an acidic solution will be about 10 –1 to 10 –6.999 . The corresponding range of
concentrations of hydroxide ions in the same solutions will be 10–13 to 10–7.001. Now let's calculate the range of the ratio of
hydroxide ions to hydrogen ions. For the most acidic solution, the ratio will be 10 –13 divided by 10 –1 , which is equal to
10–12, or approximately 0. For a slightly acidic solution, one very close to neutrality, the ratio will be 10 –7.001 divided by
10–6.999, which is nearly equal to one. Thus, the range of the hydroxide ions to the hydrogen ions in an acidic solution is 0

to 1, which corresponds to choice A. From a purely mathematic point of view, it's even easier to see that in an acidic
solution, the concentration of H+ will always be greater than the OH– concentration. If the denominator of a fraction is
always greater than the numerator, but both are positive numbers, then the limits will be 0 and 1. Again, choice A is the
correct answer.

12

as developed by


General Chemistry Discretes Test
24.
The answer for question 24 is choice A. You are asked to determine the total pressure in the flask in terms
of the partial pressure of the oxygen gas. To do this, you need to use Dalton's Law of partial pressures. This law says that
the sum of the partial pressures of all the gases in a given vessel is equal to the total pressure. Since the partial pressure of
oxygen is P 1 and you know that there are three times as many moles of hydrogen as oxygen in the flask, then the partial
pressure of hydrogen must be 3P1. Thus, the total pressure in the flask is P1 + 3P1, or 4P1, which corresponds to choice A.
25.
The answer to question 25 is choice B. To answer questions about a molecule's polarity, you must think
about the geometry of the molecule as well as the polarity of the individual bonds. Of the four molecules given in the
choices, three are symmetrical and so, even if their bonds are polar, the molecules themselves will not have a dipole moment.
According to the Valence Shell Electron Pair Repulsion theory, the three N-F bonds in NF 3 point toward the vertices of a
tetrahedron, and the nitrogen's lone pair of electrons points toward the fourth vertex, so the molecule has a trigonal pyramidal
conformation. The NF3 molecule is thus asymmetrical and will have a net dipole moment. The electronegativity difference
between nitrogen and fluorine shows that fluorine exerts a greater pull on the electrons in the N-F bonds, so the dipole
moment will put a partial negative charge on the fluorine end of the molecule and a partial positive charge on the nitrogen,
despite its lone pair of electrons. Again, the correct answer is choice B.
26.
For question 26, the correct answer is choice C . This question gives four statements concerning the
entropy of a system and asks you to determine which one is incorrect. Let's examine each one of the choices. Choice A says

that entropy is a measure of the randomness of a system. This is the definition of entropy and is thus correct. Choice B says
that the entropy of an amorphous solid is greater than that of a crystalline solid. This is true because an amorphous solid is
less ordered than a crystalline solid and thus has greater entropy. Choice C says that a spontaneous reaction will never have a
decrease in entropy. This is a false statement. Depending on the temperature and the enthalpy change of a reaction, the
entropy change can be negative. For instance, in phase changes from gas to liquid or liquid to solid, a decreasing temperature
spontaneously drives the system into the more ordered state. That would be a decrease in entropy, so choice C is the answer
to the question. Although we found the answer, let's look at choice D to see how it applies to entropy. Choice D says that
at constant temperature and pressure, a system will spontaneously lose its order. This is a basic premise of entropy because
the second law of thermodynamics states that the entropy of the universe will increase. Hence, choice D is a correct statement
concerning entropy. Again, the answer to this question is C.
27.
Looking at question 27, we get B as the right answer. This question presents you with the combustion
reaction for methane and asks you to calculate the energy of a carbon-hydrogen bond. To begin this problem, you must use
Hess' Law, which says that the heat of reaction is equal to the sum of the heats of formation of the products minus the sum
of the heats of formation for the reactants. The heat of reaction therefore equals the heat of formation of CO 2 plus twice the
heat of formation of H2O minus the heat of formation of CH4 minus twice the heat of formation of O2. The heat of reaction
is –802 kJ/mol, the heats of formation of CO 2, H 2O, and O 2 are –393 kJ/mol, –242 kJ/mol, and 0 kJ/mol respectively.
Now you want to solve for the heat of formation for CH4. Since the heat of formation of a compound in its elemental state
is always 0 kJ/mol, the factor in our equation that allows for the heat of formation of O2 drops out. Given this information,
the equation to be solved is –802 kJ/mol = –393 kJ/mol plus 2 times –242 kJ/mol minus x, where x is the heat of formation
of methane. Solving for x, we get –75 kJ/mol for the methane heat of formation. Since there are four equal carbon-hydrogen
bonds in methane, the energy of one bond is –75 kJ/mol divided by 4, or about –19 kJ/mol. This is choice B.
28.
The correct answer to question 28 is choice C . This question asks you which generalization CANNOT
be made about the phase change of a pure substance from solid to liquid. Choice A says that the phase change involves a
change in potential energy. When a phase change occurs, the internal energy of the system, that is, the total energy contained
in the system, will change. The potential energy of the system during the phase change is the same as the internal energy.
Therefore, when a solid melts into a liquid, the potential energy of the substance will change. Choice B says that the phase
change will occur at a constant temperature. Solids have a defined temperature, known as a melting point, at which they
change to liquid. At this temperature, any energy added to the solid will go toward changing the phase, not changing the

temperature, until all the solid has changed to liquid. Thus choice B is true. Choice C says that the phase change involves a
change in kinetic energy. We have already shown that the phase change occurs at a constant temperature, and because a
change in kinetic energy is associated with a temperature change, the kinetic energy the solid to liquid transition will remain
the same. Hence, choice C is the answer to this question. Choice D says that the solid to liquid phase change will involve a
change in entropy. This is true because molecules in the liquid state have more vibrational freedom and will therefore exhibit
a greater degree of disorder. Once again, the correct choice is C.
29.
In question 29, the answer is C. To answer this question, you must balance the equation that's given to find
the stoichiometric coefficient for hydrochloric acid. Let's start with oxygen since it's the only element that is present in only
one compound on each side of the equation. Since there are three oxygens on the right side of the equation, you have to place
a 3 before the water molecule on the left. There are now six hydrogen atoms on the left side of the equation, so a 5 must be
placed before the HCl on the right side for a total of six hydrogens on that side also. Before jumping right to the answer, we
should balance the entire equation, since the answer might be five or some multiple of five, depending on the other
compounds. There are six chlorines on the right, so you must place a three before the Cl 2 on the left. The equation is now
fully balanced and it's clear that the answer is indeed 5, which is choice C.
30.
The correct answer to question 30 is choice B. A Lewis base is a compound that can donate an electron
pair. Choice A, chlorous acid, will not act as a Lewis base. Rather, it is an Arrhenius acid, capable of donating a hydrogen
ion in solution. Choice B, hydrazine, consists of two nitrogens bonded to each other and to two hydrogen atoms each. A
KAPLAN

13


MCAT
nitrogen atom has a valence of five, and when it's bonded to another nitrogen and two hydrogens, it's only using three of its
five valence electrons. Therefore, each nitrogen will have a pair of unbonded electrons. This makes hydrazine able to act as
an electron-pair donor, or Lewis base, so choice B is the correct answer. Looking at the other choices, choice C, an
ammonium ion, is not a Lewis base. In fact, it's the conjugate acid of ammonia. Choice D is boron trifluoride. The boron
atom has three valence electrons, each of which is involved in a covalent bond with a fluorine atom, and thus has no electron

pairs to donate. Once again, choice B is the correct answer.

14

as developed by