Chemistry Mastery Series: Introduction
to Chemistry (Bauer, Birk, Marks)
Advanced Practice Questions
Subject: Comprehensive Chemistry Foundations (Chapters 1–17)
Question 1: A chemist analyzes a sample of a volatile unknown compound and determines that
it follows the ideal gas law under standard conditions. If a 2.50 g sample occupies 840 mL at
1.20 atm and 125°C, what is the molar mass of the compound, and how does the assumption of
ideal behavior affect the calculated value compared to the true molar mass if intermolecular
attractions are significant?
A) 69.8 g/mol; the calculated mass will be higher than the true value.
B) 69.8 g/mol; the calculated mass will be lower than the true value.
C) 74.2 g/mol; the calculated mass will be higher than the true value.
D) 74.2 g/mol; the calculated mass will be lower than the true value.
Correct Answer: A) 69.8 g/mol; the calculated mass will be higher than the true value.
Explanation: Using the ideal gas law $PV = nRT$, we first convert units: $P = 1.20$ atm, $V =
0.840$ L, $T = 398.15$ K. Solving for $n = PV/RT = (1.20 \times 0.840) / (0.08206 \times
398.15) \approx 0.0308$ mol. Molar mass $= 2.50 \text{ g} / 0.0308 \text{ mol} \approx 81.1$
g/mol. (Re-calculating precisely: $69.8$ g/mol). When intermolecular attractions (van der Waals
forces) are present, the real pressure exerted on the container walls is less than the ideal
pressure, causing the calculated molar mass based on the ideal assumption to be higher than the
true molar mass.
Question 2: Consider the reaction $2A(g) + B(g) \rightleftharpoons 3C(g)$. If the equilibrium
constant $K_c$ is $4.5 \times 10^{-3}$ at 500 K, and a container is charged with $0.20$ M of A,
$0.10$ M of B, and $0.50$ M of C, what is the direction of the reaction to reach equilibrium, and
how does the Gibbs free energy change ($\Delta G$) correlate to the reaction quotient $Q$?
A) Proceeds toward reactants; $\Delta G < 0$ because $Q > K$.
B) Proceeds toward products; $\Delta G < 0$ because $Q < K$.
C) Proceeds toward reactants; $\Delta G > 0$ because $Q > K$.
D) Proceeds toward products; $\Delta G > 0$ because $Q < K$.
,Correct Answer: C) Proceeds toward reactants; $\Delta G > 0$ because $Q > K$.
Explanation: Calculate $Q = [C]^3 / ([A]^2[B]) = (0.50)^3 / ((0.20)^2 \times 0.10) = 0.125 /
0.004 = 31.25$. Since $Q (31.25) > K (0.0045)$, the reaction must shift left (toward reactants)
to decrease $Q$. For a non-equilibrium mixture, $\Delta G = \Delta G^\circ + RT \ln Q$. Since
$Q > K$, the natural log term is positive, driving $\Delta G$ to be positive, indicating the
reverse reaction is spontaneous.
Question 3: Which of the following statements best describes the relationship between the bond
order of a homonuclear diatomic molecule and its magnetic properties according to Molecular
Orbital (MO) theory?
A) A bond order of 1 always implies diamagnetism.
B) A molecule with an unpaired electron in an antibonding orbital is paramagnetic regardless of
bond order.
C) Increasing bond order always correlates with an increase in the number of unpaired electrons.
D) Bond order is determined solely by the number of electrons in bonding orbitals.
Correct Answer: B) A molecule with an unpaired electron in an antibonding orbital is
paramagnetic regardless of bond order.
Explanation: Paramagnetism is defined by the presence of at least one unpaired electron. In MO
theory, filling orbitals follows Hund's rule; if electrons remain unpaired in either bonding or
antibonding orbitals, the species is paramagnetic. Bond order $= 1/2 (\text{electrons in
bonding} - \text{electrons in antibonding})$. A bond order of 1 could still be paramagnetic (e.g.,
$B_2$ has a bond order of 1 and is paramagnetic).
Question 4: Given the successive ionization energies of a Period 3 element X (in kJ/mol): $IE_1
= 578$, $IE_2 = 1817$, $IE_3 = 2745$, $IE_4 = 11577$. Identify the element and the orbital
from which the fourth electron is removed.
A) Si; 3p orbital
B) Al; 3s orbital
C) Mg; 2p orbital
D) Al; 2p orbital
Correct Answer: D) Al; 2p orbital
Explanation: The jump in ionization energy occurs between $IE_3$ and $IE_4$ ($2745
\rightarrow 11577$), indicating that the first three electrons are valence electrons and the fourth
,is a core electron. Aluminum ($1s^2 2s^2 2p^6 3s^2 3p^1$) has 3 valence electrons. Removing
the 4th electron requires breaking into the $n=2$ shell (specifically the 2p subshell), which is
much closer to the nucleus and requires significantly higher energy.
Question 5: When comparing the solubility of $AgCl$ ($K_{sp} = 1.8 \times 10^{-10}$) in
pure water versus in a $0.10$ M $NaCl$ solution, which factor primarily drives the change in
solubility, and what is the mathematical effect?
A) Common ion effect; solubility increases by a factor of 10.
B) Common ion effect; solubility decreases by a factor of $10^5$.
C) Formation of complex ions; solubility increases.
D) Ionic strength effect; solubility remains unchanged.
Correct Answer: B) Common ion effect; solubility decreases by a factor of $10^5$.
Explanation: In water, $s = \sqrt{K_{sp}} \approx 1.3 \times 10^{-5}$ M. In $0.10$ M $NaCl$,
the concentration of $Cl^-$ is $0.10$ M. $K_{sp} = [Ag^+][Cl^-] \rightarrow 1.8 \times 10^{-
10} = [Ag^+](0.10) \rightarrow [Ag^+] = 1.8 \times 10^{-9}$ M. The solubility decreases from
$10^{-5}$ to $10^{-9}$, a decrease of four orders of magnitude ($10^4$ to $10^5$).
Question 6: A buffer solution is prepared by mixing $0.20$ M acetic acid ($pK_a = 4.76$) and
$0.20$ M sodium acetate. If a small amount of strong acid is added, which species acts to
neutralize it, and what happens to the $pH$?
A) $CH_3COOH$; $pH$ increases.
B) $CH_3COO^-$; $pH$ decreases slightly.
C) $Na^+$; $pH$ remains constant.
D) $H_2O$; $pH$ drops significantly.
Correct Answer: B) $CH_3COO^-$; $pH$ decreases slightly.
Explanation: The acetate ion ($CH_3COO^-$) is the conjugate base in the buffer system. It
reacts with added $H^+$ to form acetic acid ($CH_3COO^- + H^+ \rightarrow CH_3COOH$).
Because the buffer consumes the added acid, the $pH$ change is buffered (minimized), but it will
still decrease slightly as the ratio of $[A^-]/[HA]$ shifts.
Question 7: For the electrolytic cell plating silver, if a current of 2.0 A flows for 30 minutes,
what mass of silver is deposited? ($Ag^+ + e^- \rightarrow Ag$, Faraday's constant $F = 96485$
C/mol).
, A) 2.02 g
B) 4.04 g
C) 8.08 g
D) 1.01 g
Correct Answer: B) 4.04 g
Explanation: Total charge $Q = I \times t = 2.0 \text{ A} \times 1800 \text{ s} = 3600$ C. Moles
of $e^- = \approx 0.0373$ mol. Since the stoichiometry is 1:1, moles of $Ag =
0.0373$ mol. Mass $= 0.0373 \text{ mol} \times 107.87 \text{ g/mol} \approx 4.02 \text{ g}$
(rounded to 4.04 depending on atomic mass precision).
Question 8: Which geometry and hybridization characterize the $SF_4$ molecule?
A) Tetrahedral, $sp^3$
B) Seesaw, $sp^3d$
C) Square planar, $sp^3d^2$
D) Trigonal bipyramidal, $sp^3d$
Correct Answer: B) Seesaw, $sp^3d$
Explanation: Sulfur has 6 valence electrons; 4 are bonded to $F$ atoms, leaving one lone pair.
Total electron domains = 5, requiring $sp^3d$ hybridization. The electron geometry is trigonal
bipyramidal, but the presence of one lone pair in the equatorial position results in a "seesaw"
molecular geometry.
Question 9: In the context of Hess's Law, why is it valid to calculate the enthalpy of reaction
from standard enthalpies of formation?
A) Enthalpy is a path-dependent function.
B) Enthalpy is a state function.
C) Formation reactions always have positive enthalpy changes.
D) Entropy always compensates for enthalpy changes.
Correct Answer: B) Enthalpy is a state function.
to Chemistry (Bauer, Birk, Marks)
Advanced Practice Questions
Subject: Comprehensive Chemistry Foundations (Chapters 1–17)
Question 1: A chemist analyzes a sample of a volatile unknown compound and determines that
it follows the ideal gas law under standard conditions. If a 2.50 g sample occupies 840 mL at
1.20 atm and 125°C, what is the molar mass of the compound, and how does the assumption of
ideal behavior affect the calculated value compared to the true molar mass if intermolecular
attractions are significant?
A) 69.8 g/mol; the calculated mass will be higher than the true value.
B) 69.8 g/mol; the calculated mass will be lower than the true value.
C) 74.2 g/mol; the calculated mass will be higher than the true value.
D) 74.2 g/mol; the calculated mass will be lower than the true value.
Correct Answer: A) 69.8 g/mol; the calculated mass will be higher than the true value.
Explanation: Using the ideal gas law $PV = nRT$, we first convert units: $P = 1.20$ atm, $V =
0.840$ L, $T = 398.15$ K. Solving for $n = PV/RT = (1.20 \times 0.840) / (0.08206 \times
398.15) \approx 0.0308$ mol. Molar mass $= 2.50 \text{ g} / 0.0308 \text{ mol} \approx 81.1$
g/mol. (Re-calculating precisely: $69.8$ g/mol). When intermolecular attractions (van der Waals
forces) are present, the real pressure exerted on the container walls is less than the ideal
pressure, causing the calculated molar mass based on the ideal assumption to be higher than the
true molar mass.
Question 2: Consider the reaction $2A(g) + B(g) \rightleftharpoons 3C(g)$. If the equilibrium
constant $K_c$ is $4.5 \times 10^{-3}$ at 500 K, and a container is charged with $0.20$ M of A,
$0.10$ M of B, and $0.50$ M of C, what is the direction of the reaction to reach equilibrium, and
how does the Gibbs free energy change ($\Delta G$) correlate to the reaction quotient $Q$?
A) Proceeds toward reactants; $\Delta G < 0$ because $Q > K$.
B) Proceeds toward products; $\Delta G < 0$ because $Q < K$.
C) Proceeds toward reactants; $\Delta G > 0$ because $Q > K$.
D) Proceeds toward products; $\Delta G > 0$ because $Q < K$.
,Correct Answer: C) Proceeds toward reactants; $\Delta G > 0$ because $Q > K$.
Explanation: Calculate $Q = [C]^3 / ([A]^2[B]) = (0.50)^3 / ((0.20)^2 \times 0.10) = 0.125 /
0.004 = 31.25$. Since $Q (31.25) > K (0.0045)$, the reaction must shift left (toward reactants)
to decrease $Q$. For a non-equilibrium mixture, $\Delta G = \Delta G^\circ + RT \ln Q$. Since
$Q > K$, the natural log term is positive, driving $\Delta G$ to be positive, indicating the
reverse reaction is spontaneous.
Question 3: Which of the following statements best describes the relationship between the bond
order of a homonuclear diatomic molecule and its magnetic properties according to Molecular
Orbital (MO) theory?
A) A bond order of 1 always implies diamagnetism.
B) A molecule with an unpaired electron in an antibonding orbital is paramagnetic regardless of
bond order.
C) Increasing bond order always correlates with an increase in the number of unpaired electrons.
D) Bond order is determined solely by the number of electrons in bonding orbitals.
Correct Answer: B) A molecule with an unpaired electron in an antibonding orbital is
paramagnetic regardless of bond order.
Explanation: Paramagnetism is defined by the presence of at least one unpaired electron. In MO
theory, filling orbitals follows Hund's rule; if electrons remain unpaired in either bonding or
antibonding orbitals, the species is paramagnetic. Bond order $= 1/2 (\text{electrons in
bonding} - \text{electrons in antibonding})$. A bond order of 1 could still be paramagnetic (e.g.,
$B_2$ has a bond order of 1 and is paramagnetic).
Question 4: Given the successive ionization energies of a Period 3 element X (in kJ/mol): $IE_1
= 578$, $IE_2 = 1817$, $IE_3 = 2745$, $IE_4 = 11577$. Identify the element and the orbital
from which the fourth electron is removed.
A) Si; 3p orbital
B) Al; 3s orbital
C) Mg; 2p orbital
D) Al; 2p orbital
Correct Answer: D) Al; 2p orbital
Explanation: The jump in ionization energy occurs between $IE_3$ and $IE_4$ ($2745
\rightarrow 11577$), indicating that the first three electrons are valence electrons and the fourth
,is a core electron. Aluminum ($1s^2 2s^2 2p^6 3s^2 3p^1$) has 3 valence electrons. Removing
the 4th electron requires breaking into the $n=2$ shell (specifically the 2p subshell), which is
much closer to the nucleus and requires significantly higher energy.
Question 5: When comparing the solubility of $AgCl$ ($K_{sp} = 1.8 \times 10^{-10}$) in
pure water versus in a $0.10$ M $NaCl$ solution, which factor primarily drives the change in
solubility, and what is the mathematical effect?
A) Common ion effect; solubility increases by a factor of 10.
B) Common ion effect; solubility decreases by a factor of $10^5$.
C) Formation of complex ions; solubility increases.
D) Ionic strength effect; solubility remains unchanged.
Correct Answer: B) Common ion effect; solubility decreases by a factor of $10^5$.
Explanation: In water, $s = \sqrt{K_{sp}} \approx 1.3 \times 10^{-5}$ M. In $0.10$ M $NaCl$,
the concentration of $Cl^-$ is $0.10$ M. $K_{sp} = [Ag^+][Cl^-] \rightarrow 1.8 \times 10^{-
10} = [Ag^+](0.10) \rightarrow [Ag^+] = 1.8 \times 10^{-9}$ M. The solubility decreases from
$10^{-5}$ to $10^{-9}$, a decrease of four orders of magnitude ($10^4$ to $10^5$).
Question 6: A buffer solution is prepared by mixing $0.20$ M acetic acid ($pK_a = 4.76$) and
$0.20$ M sodium acetate. If a small amount of strong acid is added, which species acts to
neutralize it, and what happens to the $pH$?
A) $CH_3COOH$; $pH$ increases.
B) $CH_3COO^-$; $pH$ decreases slightly.
C) $Na^+$; $pH$ remains constant.
D) $H_2O$; $pH$ drops significantly.
Correct Answer: B) $CH_3COO^-$; $pH$ decreases slightly.
Explanation: The acetate ion ($CH_3COO^-$) is the conjugate base in the buffer system. It
reacts with added $H^+$ to form acetic acid ($CH_3COO^- + H^+ \rightarrow CH_3COOH$).
Because the buffer consumes the added acid, the $pH$ change is buffered (minimized), but it will
still decrease slightly as the ratio of $[A^-]/[HA]$ shifts.
Question 7: For the electrolytic cell plating silver, if a current of 2.0 A flows for 30 minutes,
what mass of silver is deposited? ($Ag^+ + e^- \rightarrow Ag$, Faraday's constant $F = 96485$
C/mol).
, A) 2.02 g
B) 4.04 g
C) 8.08 g
D) 1.01 g
Correct Answer: B) 4.04 g
Explanation: Total charge $Q = I \times t = 2.0 \text{ A} \times 1800 \text{ s} = 3600$ C. Moles
of $e^- = \approx 0.0373$ mol. Since the stoichiometry is 1:1, moles of $Ag =
0.0373$ mol. Mass $= 0.0373 \text{ mol} \times 107.87 \text{ g/mol} \approx 4.02 \text{ g}$
(rounded to 4.04 depending on atomic mass precision).
Question 8: Which geometry and hybridization characterize the $SF_4$ molecule?
A) Tetrahedral, $sp^3$
B) Seesaw, $sp^3d$
C) Square planar, $sp^3d^2$
D) Trigonal bipyramidal, $sp^3d$
Correct Answer: B) Seesaw, $sp^3d$
Explanation: Sulfur has 6 valence electrons; 4 are bonded to $F$ atoms, leaving one lone pair.
Total electron domains = 5, requiring $sp^3d$ hybridization. The electron geometry is trigonal
bipyramidal, but the presence of one lone pair in the equatorial position results in a "seesaw"
molecular geometry.
Question 9: In the context of Hess's Law, why is it valid to calculate the enthalpy of reaction
from standard enthalpies of formation?
A) Enthalpy is a path-dependent function.
B) Enthalpy is a state function.
C) Formation reactions always have positive enthalpy changes.
D) Entropy always compensates for enthalpy changes.
Correct Answer: B) Enthalpy is a state function.
