[2012-Official-AP Practice Exam]__Shreyashkar1301

Question

Is a polar molecule

(A) $\mathrm{BeCl}_2$

(B) $\mathrm{SO}_2$

(D) $\mathrm{O}_2$

(E) $\mathrm{F}_2$

▶️Answer/Explanation

Ans:B

The molecule that is a polar molecule is (B) $\mathrm{SO}_2$.

Explanation:

(A) $\mathrm{BeCl}_2$ is a linear molecule with polar $\mathrm{Be-Cl}$ bonds, but the overall molecule is non-polar due to the symmetry of the molecule.
(B) $\mathrm{SO}_2$ is a bent molecule with polar $\mathrm{S=O}$ bonds, resulting in a polar overall molecule.
(C) $\mathrm{N}_2$ is a linear, non-polar molecule.
(D) $\mathrm{O}_2$ is a diatomic, non-polar molecule.
(E) $\mathrm{F}_2$ is a diatomic, non-polar molecule.

Question2

Is best represented by two or more resonance forms

(A) $\mathrm{BeCl}_2$

(B) $\mathrm{SO}_2$

(D) $\mathrm{O}_2$

(E) $\mathrm{F}_2$

▶️Answer/Explanation

Ans:B

The molecule that is best represented by two or more resonance forms is (B) $\mathrm{SO}_2$.

Explanation:

(A) $\mathrm{BeCl}_2$, (C) $\mathrm{N}_2$, (D) $\mathrm{O}_2$, and (E) $\mathrm{F}_2$ are not resonance structures.
(B) $\mathrm{SO}_2$ can be represented by two resonance structures, with the sulfur atom double-bonded to one oxygen atom and single-bonded to the other.

Question3

Is the molecule in which the intramolecular forces are strongest

(A) $\mathrm{BeCl}_2$

(B) $\mathrm{SO}_2$

(D) $\mathrm{O}_2$

(E) $\mathrm{F}_2$

▶️Answer/Explanation

Ans:C

The molecule in which the intramolecular forces are strongest is (C) $\mathrm{N}_2$.

Explanation:

(A) $\mathrm{BeCl}_2$, (B) $\mathrm{SO}_2$, and (E) $\mathrm{F}_2$ are held together by relatively weak intermolecular forces.
(D) $\mathrm{O}_2$ is held together by strong covalent bonds, but is a diatomic molecule.
(C) $\mathrm{N}_2$ is a diatomic molecule held together by the strongest covalent bonds, resulting in the strongest intramolecular forces among the given molecules.

Question4

Is the major process by which nuclei lose excess energy without a change in atomic number

(A) Alpha-particle emission

(B) Beta-particle emission

(D) Gamma-ray emission

(E) Nuclear fission

▶️Answer/Explanation

Ans:D

The major process by which nuclei lose excess energy without a change in atomic number is (D) Gamma-ray emission.

Explanation:

(A) Alpha-particle emission involves the emission of a helium nucleus, resulting in a decrease in atomic number.
(B) Beta-particle emission involves the emission of an electron or positron, resulting in a change in atomic number.
(C) Electron capture involves the capture of an inner-shell electron, also resulting in a change in atomic number.
(D) Gamma-ray emission involves the emission of high-energy electromagnetic radiation, which allows the nucleus to lose excess energy without changing the atomic number.
(E) Nuclear fission is a process of splitting a heavy nucleus into two or more lighter nuclei, resulting in a change in atomic number.

Question5

Accounts for the transformation of ${ }_{81}^{207} \mathrm{~T} 1$ into ${ }_{82}^{207} \mathrm{~Pb}$

(A) Alpha-particle emission

(B) Beta-particle emission

(D) Gamma-ray emission

(E) Nuclear fission

▶️Answer/Explanation

Ans:B

The process that accounts for the transformation of ${ }_{81}^{207} \mathrm{~T} 1$ into ${ }_{82}^{207} \mathrm{~Pb}$ is (B) Beta-particle emission.

Explanation:

Thallium-207 (${ }_{81}^{207} \mathrm{~T} 1$) is a radioactive isotope that can undergo beta decay to transform into lead-207 (${ }_{82}^{207} \mathrm{~Pb}$).
In beta decay, a neutron in the nucleus is converted into a proton, an electron, and an antineutrino, increasing the atomic number by 1, resulting in the transformation from thallium to lead.
The other options do not account for the transformation from thallium to lead with an increase in atomic number.

Question6

Occurs when aqueous solutions of ammonia and vinegar are mixed

(A) Oxidation-reduction reaction

(B) Brønsted-Lowry acid-base reaction

(D) Dehydration

(E) Precipitation

▶️Answer/Explanation

Ans:B

The type of reaction that occurs when aqueous solutions of ammonia and vinegar are mixed is (B) Brønsted-Lowry acid-base reaction.

Explanation:

Vinegar (acetic acid, CH3COOH) is a Brønsted-Lowry acid, and ammonia (NH3) is a Brønsted-Lowry base.
When these two solutions are mixed, a proton transfer occurs, resulting in a Brønsted-Lowry acid-base reaction.

Question7

Occurs when $\mathrm{Al}(s)$ and $\mathrm{CuCl}_2(a q)$ are mixed

(A) Oxidation-reduction reaction

(B) Brønsted-Lowry acid-base reaction

(D) Dehydration

(E) Precipitation

▶️Answer/Explanation

Ans:A

The type of reaction that occurs when $\mathrm{Al}(s)$ and $\mathrm{CuCl}_2(a q)$ are mixed is (A) Oxidation-reduction reaction.

Explanation:

Aluminum (Al) is oxidized, losing electrons and becoming $\mathrm{Al}^{3+}$.
Copper(II) chloride (CuCl2) is reduced, gaining electrons and becoming metallic copper (Cu).
This exchange of electrons is an oxidation-reduction (redox) reaction.

Question8

Occurs when solid sodium acetate, $\mathrm{NaC}_2 \mathrm{H}_3 \mathrm{O}_2(s)$, is added to water

(A) Oxidation-reduction reaction

(B) Brønsted-Lowry acid-base reaction

(D) Dehydration

(E) Precipitation

▶️Answer/Explanation

Ans:B

The type of reaction that occurs when solid sodium acetate, $\mathrm{NaC}_2 \mathrm{H}_3 \mathrm{O}_2(s)$, is added to water is (B) Brønsted-Lowry acid-base reaction.

Explanation:

Sodium acetate is the salt of a weak acid, acetic acid (CH₃COOH).
When sodium acetate dissolves in water, it undergoes hydrolysis, forming a solution with a basic pH.
This is a Brønsted-Lowry acid-base reaction, where the acetate ion (CH₃COO-) acts as a base, accepting a proton from water to form acetic acid and hydroxide ions.

Question9

Behave most like an ideal gas

(A) $\mathrm{H}_2(\mathrm{~g})$ molecules at $10^{-3}$ atm and $200^{\circ} \mathrm{C}$

(B) $\mathrm{O}_2$ (g) molecules at $20 \mathrm{~atm}$ and $200^{\circ} \mathrm{C}$

(D) $\mathrm{NH}_3(\mathrm{~g})$ molecules at $20 \mathrm{~atm}$ and $200^{\circ} \mathrm{C}$

(E) $\mathrm{NH}_3(\mathrm{~g})$ molecules at $20 \mathrm{~atm}$ and $300^{\circ} \mathrm{C}$

▶️Answer/Explanation

Ans:A

The gas molecules that behave most like an ideal gas are (A) $\mathrm{H}_2(\mathrm{~g})$ molecules at $10^{-3} \mathrm{~atm}$ and $200^{\circ} \mathrm{C}$.

Explanation:

Ideal gas behavior is approached under conditions of low pressure and high temperature, where intermolecular interactions are minimized.
At a pressure of $10^{-3} \mathrm{~atm}$, the gas molecules are far apart, and the intermolecular forces have a negligible effect on the gas behavior.
Hydrogen gas ($\mathrm{H}_2$) is also a simple, monatomic gas, which further contributes to its ideal gas behavior.
The other gas molecules (B, C, D, E) are subjected to higher pressures, which can lead to more significant intermolecular interactions and deviations from ideal gas behavior.

Question10

Have lowest root-mean-square speed

(A) $\mathrm{H}_2(\mathrm{~g})$ molecules at $10^{-3}$ atm and $200^{\circ} \mathrm{C}$

(B) $\mathrm{O}_2$ (g) molecules at $20 \mathrm{~atm}$ and $200^{\circ} \mathrm{C}$

(D) $\mathrm{NH}_3(\mathrm{~g})$ molecules at $20 \mathrm{~atm}$ and $200^{\circ} \mathrm{C}$

(E) $\mathrm{NH}_3(\mathrm{~g})$ molecules at $20 \mathrm{~atm}$ and $300^{\circ} \mathrm{C}$

▶️Answer/Explanation

Ans:C

The gas molecules with the lowest root-mean-square (RMS) speed are:

To determine the gas with the lowest RMS speed, we can use the formula:

RMS speed = $\sqrt{\frac{3RT}{M}}$

Where:

$R$ is the ideal gas constant
$T$ is the absolute temperature
$M$ is the molar mass of the gas

(A) H$_2$(g) at $10^{-3}$ atm and $200^\circ$C:
$T = 200^\circ\mathrm{C} + 273.15 = 473.15\ \mathrm{K}$
$M = 2.016\ \mathrm{g/mol}$
RMS speed = $\sqrt{\frac{3 \times 8.314\ \mathrm{J/mol\cdot K} \times 473.15\ \mathrm{K}}{2.016\ \mathrm{g/mol}}} = 1910\ \mathrm{m/s}$

(B) O$_2$(g) at $20\ \mathrm{atm}$ and $200^\circ$C:
$T = 200^\circ\mathrm{C} + 273.15 = 473.15\ \mathrm{K}$
$M = 32.00\ \mathrm{g/mol}$
RMS speed = $\sqrt{\frac{3 \times 8.314\ \mathrm{J/mol\cdot K} \times 473.15\ \mathrm{K}}{32.00\ \mathrm{g/mol}}} = 574\ \mathrm{m/s}$

(C) SO$_2$(g) at $20\ \mathrm{atm}$ and $200^\circ$C:
$T = 200^\circ\mathrm{C} + 273.15 = 473.15\ \mathrm{K}$
$M = 64.06\ \mathrm{g/mol}$
RMS speed = $\sqrt{\frac{3 \times 8.314\ \mathrm{J/mol\cdot K} \times 473.15\ \mathrm{K}}{64.06\ \mathrm{g/mol}}} = 425\ \mathrm{m/s}$

(D) NH$_3$(g) at $20\ \mathrm{atm}$ and $200^\circ$C:
$T = 200^\circ\mathrm{C} + 273.15 = 473.15\ \mathrm{K}$
$M = 17.03\ \mathrm{g/mol}$
RMS speed = $\sqrt{\frac{3 \times 8.314\ \mathrm{J/mol\cdot K} \times 473.15\ \mathrm{K}}{17.03\ \mathrm{g/mol}}} = 694\ \mathrm{m/s}$

(E) NH$_3$(g) at $20\ \mathrm{atm}$ and $300^\circ$C:
$T = 300^\circ\mathrm{C} + 273.15 = 573.15\ \mathrm{K}$
$M = 17.03\ \mathrm{g/mol}$
RMS speed = $\sqrt{\frac{3 \times 8.314\ \mathrm{J/mol\cdot K} \times 573.15\ \mathrm{K}}{17.03\ \mathrm{g/mol}}} = 851\ \mathrm{m/s}$

The gas molecules with the lowest RMS speed are (C) SO$_2$(g) at $20\ \mathrm{atm}$ and $200^\circ$C, with a speed of $425\ \mathrm{m/s}$.

Question11

Has the highest electronegativity

(A) $\mathrm{Cs}$

(B) $\mathrm{Ag}$

(D) $\mathrm{Br}$

(E) $\mathrm{Se}$

▶️Answer/Explanation

Ans:D

The element with the highest electronegativity is (D) $\mathrm{Br}$.

Explanation:

Electronegativity generally increases from left to right across a period (row) in the periodic table.
Halogens, such as bromine (Br), have the highest electronegativities among the given elements.
The other elements (Cs, Ag, Pb, Se) have lower electronegativities compared to bromine.

Question12

Has the lowest first-ionization energy

(A) $\mathrm{Cs}$

(B) $\mathrm{Ag}$

(D) $\mathrm{Br}$

(E) $\mathrm{Se}$

▶️Answer/Explanation

Ans:A

The element with the lowest first-ionization energy is (A) $\mathrm{Cs}$.

Explanation:

First-ionization energy generally decreases from top to bottom in a group (column) in the periodic table.
Cesium (Cs) is an alkali metal, which have the lowest first-ionization energies among the given elements.
The other elements (Ag, Pb, Br, Se) have higher first-ionization energies compared to cesium.

Question13

Has the largest atomic radius

(A) $\mathrm{Cs}$

(B) $\mathrm{Ag}$

(D) $\mathrm{Br}$

(E) $\mathrm{Se}$

▶️Answer/Explanation

Ans:A

The element with the largest atomic radius is (A) $\mathrm{Cs}$.

Explanation:

Atomic radius generally increases from top to bottom in a group (column) in the periodic table.
Cesium (Cs) is an alkali metal, which have the largest atomic radii among the given elements.
The other elements (Ag, Pb, Br, Se) have smaller atomic radii compared to cesium.

Question14

Is isomeric with $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$

(A) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$

(B) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$

(D) $\mathrm{CH}_3 \mathrm{COOH}$

(E) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2$

▶️Answer/Explanation

Ans:C

The compound that is isomeric with $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$ is (C) $\mathrm{CH}_3 \mathrm{COCH}_3$.

Explanation:

Both $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$ and $\mathrm{CH}_3 \mathrm{COCH}_3$ have the molecular formula C₃H₆O, making them isomers.
The difference is in the placement of the functional groups – $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$ has an aldehyde group, while $\mathrm{CH}_3 \mathrm{COCH}_3$ has a ketone group.
The other options (A, B, D, E) are not isomeric with $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$.

Question15

Dissolves in water to form an acidic solution

(A) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$

(B) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$

(D) $\mathrm{CH}_3 \mathrm{COOH}$

(E) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2$

▶️Answer/Explanation

Ans:D

The compound that dissolves in water to form an acidic solution is (D) $\mathrm{CH}_3 \mathrm{COOH}$.

Explanation:

$\mathrm{CH}_3 \mathrm{COOH}$ is acetic acid, a weak acid that dissociates in water to form hydrogen ions (H+), making the solution acidic.
The other compounds (A, B, C, E) are not acids and do not form acidic solutions when dissolved in water.

Question16

Is the LEAST soluble in water

(A) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$

(B) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$

(D) $\mathrm{CH}_3 \mathrm{COOH}$

(E) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2$

▶️Answer/Explanation

Ans:A

The compound that is the LEAST soluble in water is (A) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$.

Explanation:

Alkanes, such as $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$, are non-polar and have low solubility in the polar solvent water.
The other compounds (B, C, D, E) contain polar functional groups (hydroxyl, carbonyl, carboxyl, amino) that increase their solubility in water compared to the non-polar alkane.

Question Part B

Directions: Each of the questions or incomplete statements below is followed by five suggested answers or completions. Select the one that is best in each case and then fill in the corresponding circle on the answer sheet.

Question17

A certain crystalline substance that has a low melting point does not conduct electricity in solution or when melted. This substance is likely to be

(A) a covalent network solid

(B) a metallic solid

(D) an ionic solid

(E) a molecular solid

▶️Answer/Explanation

Ans:E

The crystalline substance that has a low melting point, does not conduct electricity in solution or when melted, is likely to be (E) a molecular solid.

Explanation:

Molecular solids are formed by covalently bonded molecules held together by weak intermolecular forces, such as van der Waals forces or hydrogen bonds.
Molecular solids typically have low melting and boiling points compared to other types of solids.
Molecular solids do not conduct electricity when in solution or in the molten state because they do not have free-moving electrons or ions.
The other options (A, B, C, D) are not consistent with the given information about the substance.

Question18

Solid $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is added to distilled water to produce a solution in which the concentration of nitrate, $\left[\mathrm{NO}_3^{-}\right]$, is $0.10 \mathrm{M}$. What is the concentration of aluminum ion, $\left[\mathrm{Al}^{3+}\right]$, in this solution?

(A) $0.010 \mathrm{M}$

(B) $0.033 \mathrm{M}$

(D) $0.10 \mathrm{M}$

(E) $0.30 \mathrm{M}$

▶️Answer/Explanation

Ans:B

The concentration of aluminum ion, (Al³⁺) , in the solution of solid Al(NO₃)₃in distilled water is (B) 0.033 M.

Explanation:

The dissociation of solid Al(NO₃)₃ in water produces aluminum ions (Al³⁺) and nitrate ions (NO₃^–).
The concentration of nitrate ions is given as 0.10 M.
Since the aluminum nitrate dissociates completely, the concentration of aluminum ions (Al³⁺) is one-third of the nitrate ion concentration, which is 0.033 M.

Question19

Which of the following is a weak acid in aqueous solution?

(A) $\mathrm{HCl}$

(B) $\mathrm{HClO}_4$

(D) $\mathrm{H}_2 \mathrm{~S}$

(E) $\mathrm{H}_2 \mathrm{SO}_4$

▶️Answer/Explanation

Ans:D

The weak acid in aqueous solution is (D) $\mathrm{H}_2 \mathrm{~S}$.

Explanation:

Hydrogen chloride (HCl), perchloric acid $\mathrm{HClO}_4$, and nitric acid $\mathrm{HNO}_3$ are all strong acids that completely dissociate in water.
Sulfuric acid $\mathrm{H}_2 \mathrm{SO}_4$ is also a strong acid.
Hydrogen sulfide $\mathrm{H}_2 \mathrm{~S}$ is a weak acid that partially dissociates in water, forming a solution with a pH greater than 0.

Question20

In $1.00 \mathrm{~mol}$ of potassium zirconium sulfate trihydrate, $\mathrm{K}_4 \mathrm{Zr}\left(\mathrm{SO}_4\right)_4 \cdot 3 \mathrm{H}_2 \mathrm{O}$, there are

(A) $3 \times 6.02 \times 10^{23}$ hydrogen atoms

(B) $6.02 \times 10^{23}$ sulfur atoms

(D) 4 moles of oxygen atoms

(E) 4 moles of zirconium atoms

▶️Answer/Explanation

Ans:C

To determine the number of atoms of each element present in $1.00 \, \mathrm{mol}$ of potassium zirconium sulfate trihydrate, we need to consider the chemical formula and stoichiometry of the compound.

The chemical formula is $\mathrm{K}_4 \mathrm{Zr}\left(\mathrm{SO}_4\right)_4 \cdot 3 \mathrm{H}_2 \mathrm{O}$.

This means:

There are $4$ potassium atoms ($\mathrm{K}$) per formula unit.
There is $1$ zirconium atom ($\mathrm{Zr}$) per formula unit.
There are $4$ sulfate groups ($\mathrm{SO}_4$) per formula unit, which means $4 \times 4 = 16$ oxygen atoms ($\mathrm{O}$).
Each sulfate group contains $1$ sulfur atom ($\mathrm{S}$).
Each formula unit is associated with $3$ water molecules, which contain a total of $3 \times 2 = 6$ hydrogen atoms ($\mathrm{H}$).

Now, let’s check each option:

(A) $3 \times 6.02 \times 10^{23}$ hydrogen atoms – This is incorrect. The number of hydrogen atoms is $6$ times Avogadro’s number per formula unit, not $3$ times.

(B) $6.02 \times 10^{23}$ sulfur atoms – This is incorrect. There is $1$ sulfur atom per formula unit, not $6.02 \times 10^{23}$.

(C) $4 \times 6.02 \times 10^{23}$ potassium atoms – This is correct. There are $4$ potassium atoms per formula unit.

(D) $4$ moles of oxygen atoms – This is incorrect. We calculated that there are $16$ oxygen atoms per formula unit. To convert this to moles, we need to divide by Avogadro’s number.

(E) $4$ moles of zirconium atoms – This is incorrect. There is $1$ zirconium atom per formula unit, not $4$ moles.

Therefore, the correct answer is (C) $4 \times 6.02 \times 10^{23}$ potassium atoms.

Question21

$
\mathrm{X}+2 \mathrm{Y} \rightarrow \mathrm{Z}+3 \mathrm{Q}
$

For the reaction represented above, the initial rate of decrease in [X] was $2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. What was the initial rate of decrease in $[\mathrm{Y}]$ ?

(A) $7.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

(B) $1.4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

(D) $5.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

(E) $1.1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

▶️Answer/Explanation

Ans:D

To determine the initial rate of decrease in $[\mathrm{Y}]$, we can use the stoichiometry of the reaction.

From the balanced equation:

\[ \mathrm{X} + 2 \mathrm{Y} \rightarrow \mathrm{Z} + 3 \mathrm{Q} \]

We see that for every one mole of $\mathrm{X}$ consumed, two moles of $\mathrm{Y}$ are consumed. Thus, the rate of decrease of $\mathrm{Y}$ is related to the rate of decrease of $\mathrm{X}$ by a factor of 2.

Given that the initial rate of decrease in $\mathrm{X}$ is $2.8 \times 10^{-3} \, \mathrm{mol \cdot L^{-1} \cdot s^{-1}}$, the initial rate of decrease in $\mathrm{Y}$ would be $2 \times 2.8 \times 10^{-3} \, \mathrm{mol \cdot L^{-1} \cdot s^{-1}}$.

Calculating:

\[ 2 \times 2.8 \times 10^{-3} \, \mathrm{mol \cdot L^{-1} \cdot s^{-1}} = 5.6 \times 10^{-3} \, \mathrm{mol \cdot L^{-1} \cdot s^{-1}} \]

So, the initial rate of decrease in $\mathrm{Y}$ is $5.6 \times 10^{-3} \, \mathrm{mol \cdot L^{-1} \cdot s^{-1}}$, which corresponds to option (D).

Question22

To determine the percentage of water in a hydrated salt, a student heated a $1.2346 \mathrm{~g}$ sample of the salt for 30 minutes; when cooled to room temperature, the sample weighed $1.1857 \mathrm{~g}$. After the sample was heated for an additional $10 \mathrm{~min}-$ utes and again cooled to room temperature, the sample weighed $1.1632 \mathrm{~g}$. Which of the following should the student do next?

(A) Use the smallest mass value to calculate the percentage of water in the hydrated salt.

(B) Repeat the experiment with a new sample of the same mass and average the results.

(D) Reheat the sample until its mass is constant.

(E) Use the average of the mass values obtained after the two heatings to calculate the percentage of water in the hydrated salt.

▶️Answer/Explanation

Ans:D

To determine the percentage of water in the hydrated salt, the student should (D) Reheat the sample until its mass is constant.

Explanation:

The initial mass of the sample was 1.2346 g, and after the first heating, the mass was 1.1857 g.
After the second heating, the mass further decreased to 1.1632 g.
This indicates that the sample has not yet reached a constant mass, and more heating is required to remove all the water from the hydrated salt.
Reheating the sample until its mass is constant will give the true anhydrous mass of the salt, allowing the student to calculate the percentage of water accurately.

Question23

Which of the following statements about atoms is NOT correct?

(A) Atoms are electrically neutral because they have the same number of protons and electrons.

(B) All atoms of a given element must have the same number of protons, neutrons, and electrons.

(D) The nucleus is positively charged.

(E) Almost all of the mass of an atom is in the nucleus.

▶️Answer/Explanation

Ans:D

The statement that is NOT correct is:

(B) All atoms of a given element must have the same number of protons, neutrons, and electrons.

Explanation:

(A) is correct. Atoms are electrically neutral because they have the same number of protons (positive charge) and electrons (negative charge).

(D) is correct. The nucleus is positively charged due to the protons.

(E) is correct. Almost all of the mass of an atom is concentrated in the nucleus, which is much denser than the surrounding electron cloud.

However, (B) is not correct. Atoms of a given element must have the same number of protons (which defines the element), but the number of neutrons can vary, resulting in different isotopes of the same element. The number of electrons can also vary, resulting in different ionization states of the atom.

Question24

$
\mathrm{NH}_3(a q)+\mathrm{HCl}(a q) \rightleftarrows \mathrm{NH}_4^{+}(a q)+\mathrm{Cl}^{-}(a q)
$
. The Brønsted-Lowry bases in the reaction represented above are

(A) $\mathrm{NH}_3(a q)$ and $\mathrm{NH}_4{ }^{+}(a q)$

(B) $\mathrm{NH}_3(a q)$ and $\mathrm{Cl}^{-}(a q)$

(D) $\mathrm{HCl}(a q)$ and $\mathrm{NH}_4^{+}(a q)$

(E) $\mathrm{HCl}(a q)$ and $\mathrm{Cl}^{-}(a q)$

▶️Answer/Explanation

Ans:B

\[
\mathrm{NH}_3(a q)+\mathrm{HCl}(a q) \rightleftarrows \mathrm{NH}_4^{+}(a q)+\mathrm{Cl}^{-}(a q)
\]

In this reaction, a Brønsted-Lowry base is a substance that accepts a proton $\left(\mathrm{H}^{+}\right)$. Looking at the reaction, you can see that $\mathrm{NH}_3\left(\mathrm{NH}_3(a q)\right.$ ) accepts a proton to form $\mathrm{NH}_4^{+}(a q)$, thus $\mathrm{NH}_3$ is a Brønsted-Lowry base. Similarly, $\mathrm{Cl}^{-}(a q)$ accepts a proton to form $\mathrm{HCl}(a q)$, thus $\mathrm{Cl}^{-}(a q)$ is also a Brønsted-Lowry base. So, the correct answer is (B) $\mathrm{NH}_3(a q)$ and $\mathrm{Cl}^{-}(a q)$.

Question25

When 6.0 L of $\mathrm{He}(\mathrm{g})$ and 10. $\mathrm{L}$ of $\mathrm{N}_2(\mathrm{~g})$, both at $0^{\circ} \mathrm{C}$ and $1.0 \mathrm{~atm}$, are pumped into an evacuated $4.0 \mathrm{~L}$ rigid container, the final pressure in the container at $0^{\circ} \mathrm{C}$ is

(A) $2.0 \mathrm{~atm}$

(B) $4.0 \mathrm{~atm}$

(D) $8.8 \mathrm{~atm}$

(E) $16 \mathrm{~atm}$

▶️Answer/Explanation

Ans:B

For helium $(\mathrm{He}(g))$ :
\[
n_{\mathrm{He}}=\frac{P V}{R T}=\frac{(1.0 \mathrm{~atm})(6.0 \mathrm{~L})}{(0.0821 \mathrm{~atm} \cdot \mathrm{L} / \mathrm{mol} \cdot \mathrm{K})(273 \mathrm{~K})} \approx 0.26 \mathrm{~mol}
\]

For nitrogen $\left(\mathrm{N}_2(g)\right)$ :
\[
n_{\mathrm{N}_2}=\frac{(1.0 \mathrm{~atm})(10.0 \mathrm{~L})}{(0.0821 \mathrm{~atm} \cdot \mathrm{L} / \mathrm{mol} \cdot \mathrm{K})(273 \mathrm{~K})} \approx 0.43 \mathrm{~mol}
\]

Now, add these moles together to get the total moles of gas:
\[
n_{\text {total }}=n_{\mathrm{He}}+n_{\mathrm{N}_2} \approx 0.26 \mathrm{~mol}+0.43 \mathrm{~mol} \approx 0.69 \mathrm{~mol}
\]

Finally, use the ideal gas law to find the final pressure in the container:

\[
P_{\text {final }}=\frac{n_{\text {total }} R T}{V}=\frac{(0.69\mathrm{~mol})(0.0821 \mathrm{~atm} \cdot \mathrm{L} / \mathrm{mol} \cdot \mathrm{K})(273 \mathrm{~K})}{4.0 \mathrm{~L}} \approx 4 \mathrm{~atm}
\]

Question26

Shown above is the phase diagram of a pure substance. The substance under the conditions corresponding to point $X$ on the diagram is cooled to $40^{\circ} \mathrm{C}$ while the pressure remains constant. As the substance cools, the phase of the substance changes from

(A) gas to liquid to solid

(B) gas to solid to liquid

(D) liquid to solid to gas

(E) liquid to gas to solid

▶️Answer/Explanation

Ans:A

To determine the phase changes of the substance as it is cooled to 40°C while the pressure remains constant, we need to examine the phase diagram.

Given information:

The substance is initially at the conditions corresponding to point X on the diagram.
The substance is cooled to 40°C while the pressure remains constant.

Examining the phase diagram, we can see that as we move from right to left (decreasing temperature), the substance goes through the following phase changes:

Gas → Liquid → Solid

Question27

Oxygen is acting as an oxidizing agent in all of the following reactions EXCEPT

(A) $2 \mathrm{C}(s)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}(g)$

(B) $\mathrm{S}(s)+\mathrm{O}_2(g) \rightarrow \mathrm{SO}_2(g)$

(D) $2 \mathrm{Na}(s)+\mathrm{O}_2(g) \rightarrow \mathrm{Na}_2 \mathrm{O}_2(s)$

(E) $2 \mathrm{Mg}(s)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{MgO}(s)$

▶️Answer/Explanation

Ans:A

Oxygen is acting as an oxidizing agent in all of the following reactions EXCEPT:

(A) 2 C(s) + O₂(g) → 2 CO(g)
(B) S(s) + O₂(g) → SO₂(g)
(C) 2 F₂(g) + O₂(g) → 2 OF₂(g)
(D) 2 Na(s) + O₂(g) → Na₂O₂(s)
(E) 2 Mg(s) + O₂(g) → 2 MgO(s)

The correct answer is (A). In the reaction 2 C(s) + O₂(g) → 2 CO(g), oxygen is acting as a reducing agent, not an oxidizing agent. In all the other reactions, oxygen is acting as an oxidizing agent.

Question28

What is the maximum number of moles of $\mathrm{Al}_2 \mathrm{O}_3$ that can be produced by the reaction of $0.40 \mathrm{~mol}$ of $\mathrm{Al}$ with $0.40 \mathrm{~mol}$ of $\mathrm{O}_2$ ?

(A) $0.10 \mathrm{~mol}$

(B) $0.20 \mathrm{~mol}$

(D) $0.33 \mathrm{~mol}$

(E) $0.40 \mathrm{~mol}$

▶️Answer/Explanation

Ans:C

The balanced chemical equation is:

4 Al(s) + 3 O_₂(g) → 2 Al_₂O_₃(s)

To find the maximum moles of Al₂O₃ produced, we need to determine the limiting reactant.

The mole ratio of Al to O₂ in the balanced equation is 4:3.

Since the amounts of Al and O_₂ are equal (0.40 mol each), O_₂ is the limiting reactant.

Using the limiting reactant, we can calculate the maximum moles of Al_₂O_₃ produced:

Moles of O_₂ = 0.40 mol
Moles of Al_₂O_₃ = (0.40 mol O_₂) × (2 mol Al_₂O_₃ /3 mol O_₂) = 0.267 mol

Therefore, the correct answer is (C) 0.27 mol.

Question29

$
… \mathrm{C}_3 \mathrm{H}_8(g)+\ldots \mathrm{O}_2(g) \rightarrow \ldots \mathrm{H}_2 \mathrm{O}(g)+\ldots \mathrm{CO}_2(g)
$

When the equation for the reaction represented above is balanced and all coefficients are reduced to the lowest whole-number terms, the coefficient for $\mathrm{O}_2(\mathrm{~g})$ is

(A) 1

(B) 2

(D) 5

(E) 6

▶️Answer/Explanation

Ans:D

When the equation for the reaction represented above is balanced and all coefficients are reduced to the lowest whole-number terms, the coefficient for O₂(g) is:

The balanced equation is: C₃H₈(g) + 5 O₂(g) → 4 H₂O(g) + 3 CO₂(g)

So the coefficient for O₂(g) is 5. The correct answer is (D) 5.

Question30

A $0.1 \mathrm{M}$ solution of which of the following is colorless?

(A) $\mathrm{MgCl}_2$

(B) $\mathrm{Ni}\left(\mathrm{NO}_3\right)_2$

(D) $\mathrm{KMnO}_4$

(E) $\mathrm{CuSO}_4$

▶️Answer/Explanation

Ans:A

A 0.1 M solution of which of the following is colorless?

(A) MgCl₂ – Colorless (B) Ni(NO₃)₂ – Green (C) Na₂CrO₄ – Yellow (D) KMnO₄ – Purple (E) CuSO₄ – Blue

The correct answer is (A) MgCl₂. A 0.1 M solution of MgCl₂ is colorless.

Question31

Under which of the following conditions can an endothermic reaction be thermodynamically favorable?

(A) $\Delta G$ is positive

(B) $\Delta S$ is negative

(D) $T \Delta S=0$

(E) There are no conditions under which an endothermic reaction can be thermodynamically favorable.

▶️Answer/Explanation

Ans:C

Under which of the following conditions can an endothermic reaction be thermodynamically favorable?

(A) ΔG is positive – Incorrect, for a reaction to be thermodynamically favorable, ΔG must be negative. (B) ΔS is negative – Incorrect, for an endothermic reaction to be favorable, ΔS must be positive. (C) TΔS > ΔH – Correct. If the entropic term (TΔS) outweighs the endothermic enthalpy change (ΔH), the overall free energy change (ΔG = ΔH – TΔS) can be negative, making the reaction thermodynamically favorable. (D) TΔS = 0 – Incorrect, this would mean the reaction is not entropy-driven. (E) There are no conditions – Incorrect, as shown in option (C), there are conditions under which an endothermic reaction can be favorable.

The correct answer is (C) TΔS > ΔH.

Question32

The vapor pressure of pure water at $25^{\circ} \mathrm{C}$ is $24.0 \mathrm{~mm} \mathrm{Hg}$. What is the expected vapor pressure at $25^{\circ} \mathrm{C}$ of an ideal solution of a nonvolatile nonelectrolyte in which the mole fraction of water is 0.900 ?

(A) $1.48 \mathrm{~mm} \mathrm{Hg}$

(B) $2.40 \mathrm{~mm} \mathrm{Hg}$

(D) $24.0 \mathrm{~mm} \mathrm{Hg}$

(E) $26.7 \mathrm{~mm} \mathrm{Hg}$

▶️Answer/Explanation

Ans:C

The vapor pressure of pure water at 25°C is 24.0 mm Hg. What is the expected vapor pressure at 25°C of an ideal solution of a nonvolatile nonelectrolyte in which the mole fraction of water is 0.900?

According to Raoult’s law for ideal solutions, the vapor pressure of a solvent over an ideal solution is proportional to its mole fraction:

P = P₀ × χ₁

where P is the vapor pressure of the solution, P₀ is the vapor pressure of the pure solvent, and χ₁ is the mole fraction of the solvent.

Given:
P₀ (water) = 24.0 mm Hg
χ₁ (water) = 0.900

Plugging these values into the equation:
P = 24.0 mm Hg × 0.900 = 21.6 mm Hg

Therefore, the correct answer is (C) 21.6 mm Hg.

Question33

Which of the following salts is LEAST soluble in water?

(A) $\mathrm{NiS}$

(B) $\mathrm{MgCl}_2$

(D) $\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$

(E) $\mathrm{Pb}\left(\mathrm{NO}_3\right)_2$

▶️Answer/Explanation

Ans:A

(A) NiS – Moderately soluble
(B) MgCl₂ – Highly soluble
(C) K₂CrO₄ – Moderately soluble
(D) Al₂(SO₄)₃ – Highly soluble
(E) Pb(NO₃)₂ – Highly soluble

The salt that is least soluble in water is NiS. Nickel sulfide has a very low solubility in water compared to the other salts listed.

Therefore, the correct answer is (A) NiS.

Question34

Which of the following is the best piece of laboratory glassware for preparing $500.0 \mathrm{~mL}$ of an aqueous solution of a solid?

(A) Volumetric flask

(B) Erlenmeyer flask

(D) Graduated beaker

(E) Graduated cylinder

▶️Answer/Explanation

Ans:A

(A) Volumetric flask – Correct. A volumetric flask is the best choice for accurately preparing a specific volume of solution.
(B) Erlenmeyer flask – Not the best choice for accurately measuring volume.
(C) Test tube – Too small for preparing 500 mL of solution.
(D) Graduated beaker – Can be used, but not as accurate as a volumetric flask.
(E) Graduated cylinder – Can be used, but not as accurate as a volumetric flask.

The correct answer is (A) Volumetric flask.

Question

Questions 35-36 refer to the experiment described below.
$\mathrm{H}_2$ gas and $\mathrm{N}_2$ gas were placed in a rigid vessel and allowed to reach equilibrium in the presence of a catalyst according to the following equation.
$
3 \mathrm{H}_2(\mathrm{~g})+\mathrm{N}_2(\mathrm{~g}) \rightleftarrows 2 \mathrm{NH}_3(\mathrm{~g}) \quad \Delta H^{\circ}=-92 \mathrm{~kJ} / \mathrm{mol}_{r x n}
$

The diagram below shows how the concentrations of $\mathrm{H}_2, \mathrm{~N}_2$, and $\mathrm{NH}_3$ in this system changed over time.

Question35

Which of the following was true for the system between time $t_1$ and time $t_2$ ?

(A) The concentration of $\mathrm{N}_2$ decreased.

(B) The temperature of the system decreased.

(D) The rates of the forward and reverse reactions were equal.

(E) The rate of formation of $\mathrm{NH}_3$ molecules was equal to the rate of disappearance of $\mathrm{H}_2$ molecules.

▶️Answer/Explanation

Ans:D

Let’s analyze the given information and the graph showing the concentration changes of $ \mathrm{H}_2 $, $ \mathrm{N}_2 $, and $ \mathrm{NH}_3 $ over time:

1. The chemical equation represents the reaction:
\[3 \mathrm{H}_2(\mathrm{g}) + \mathrm{N}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{g}) \quad \Delta H^{\circ} = -92 \, \mathrm{kJ/mol}_{\text{rxn}}\]

2. The graph shows the concentrations of the three gases over time:
Concentration of $ \mathrm{N}_2 $ decreases initially and then levels off.
Concentration of $ \mathrm{H}_2 $ follows a similar pattern but at a lower overall concentration.
Concentration of $ \mathrm{NH}_3 $ increases initially and then stabilizes.

(A) The concentration of $ \mathrm{N}_2 $ decreased: True. The graph shows a decrease in $ \mathrm{N}_2 $ concentration between $ t_1 $ and $ t_2 $.
(B) The temperature of the system decreased: Not necessarily true. The graph does not provide information about temperature changes.
(C) The number of effective collisions between $ \mathrm{H}_2 $ and $ \mathrm{N}_2 $ was zero: False. Effective collisions are necessary for reactions to occur.
(D) The rates of the forward and reverse reactions were equal: True. Between $ t_1 $ and $ t_2 $, the concentrations of all species remain constant, indicating equilibrium. At equilibrium, the rates of the forward and reverse reactions are equal.
(E) The rate of formation of $ \mathrm{NH}_3 $ molecules was equal to the rate of disappearance of $ \mathrm{H}_2 $ molecules: True. At equilibrium, the rates of formation and disappearance of reactants and products are equal.

Therefore, the correct answer is(D) The rates of the forward and reverse reactions were equal.

Question36

More $\mathrm{NH}_3$ gas is added to the system at time $t_2$ while the temperature is held constant. Which of the following will most likely occur?

(A) The value of the equilibrium constant will increase.

(B) The value of the equilibrium constant will decrease.

(D) The amount of $\mathrm{N}_2$ will increase.

(E) The amount of $\mathrm{H}_2$ will decrease

▶️Answer/Explanation

Ans:D

When additional $ \mathrm{NH}_3 $ gas is added to the system at time $ t_2 $ while keeping the temperature constant, let’s consider the impact on the equilibrium:

1. The given reaction is:
\[3 \mathrm{H}_2(\mathrm{g}) + \mathrm{N}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{g}) \quad \Delta H^{\circ} = -92 \, \mathrm{kJ/mol}_{\text{rxn}}\]

2. Adding more $ \mathrm{NH}_3 $ gas will shift the equilibrium position. Let’s analyze the options:

(A) The value of the equilibrium constant will increase: False. The equilibrium constant ($ K $) depends on temperature and does not change due to changes in concentration.
(B) The value of the equilibrium constant will decrease: False. Same reason as above.
(C) The total pressure in the container will decrease: False. The total pressure remains constant because the system is rigid.
(D) The amount of $ \mathrm{N}_2 $ will increase: True. By Le Chatelier’s principle, adding more $ \mathrm{NH}_3 $ will shift the equilibrium to the left, favoring the reverse reaction. As a result, more $ \mathrm{N}_2 $ will be formed.
(E) The amount of $ \mathrm{H}_2 $ will decrease: True. Similar to the reasoning for $ \mathrm{N}_2 $, the equilibrium shift will reduce the concentration of $ \mathrm{H}_2 $.

Therefore, the correct answers are (D) The amount of $ \mathrm{N}_2 $ will increase and (E) The amount of $ \mathrm{H}_2 $ will decrease.

Question37

When heated, metallic carbonates generally produce

(A) metallic peroxide $+\mathrm{CO}$

(B) metal $+\mathrm{CO}+\mathrm{O}_2$

(D) metallic oxalate $+\mathrm{O}_2$

(E) metallic oxide $+\mathrm{CO}_2$

▶️Answer/Explanation

Ans:E

When heated, metallic carbonates generally decompose into metallic oxide and carbon dioxide. So, the correct option is (E) metallic oxide $+\mathrm{CO}_2$.

Question38

$
\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftarrows \mathrm{Fe}(\mathrm{SCN})^{2+}(a q)
$

38 For the reaction represented above, the value of the equilibrium constant, $K_{e q}$, is 240 at $25^{\circ} \mathrm{C}$. From this information, correct deductions about the reaction at $25^{\circ} \mathrm{C}$ include which of the following?

I. The reaction is quite rapid.

II. The product is favored over the reactants at equilibrium.

III. The reaction is endothermic.

(A) I only

(B) II only

(D) II and III only

(E) I, II, and III

▶️Answer/Explanation

Ans:B

The equilibrium constant $K_{eq}$ indicates the ratio of products to reactants at equilibrium. A high value of $K_{eq}$ (such as 240) indicates that at equilibrium, the products are favored over the reactants. Since $K_{eq}$ is greater than 1, it indicates that the product side is favored. Therefore, deduction II is correct. Deduction I about the rate of reaction cannot be made solely based on the value of $K_{eq}$. Deduction III about the reaction being endothermic cannot be made solely based on the value of $K_{eq}$. So, the correct answer is (B) II only.

Question39

The volume of water that must be added in order to dilute $40 \mathrm{~mL}$ of $9.0 \mathrm{M} \mathrm{HCl}$ to a concentration of $6.0 \mathrm{M}$ is closest to

(A) $10 \mathrm{~mL}$

(B) $20 \mathrm{~mL}$

(D) $40 \mathrm{~mL}$

(E) $60 \mathrm{~mL}$

▶️Answer/Explanation

Ans:B

To dilute a solution, we use the formula:

\[
M_1V_1 = M_2V_2
\]

Where:
$M_1$ and $V_1$ are the initial concentration and volume,
$M_2$ and $V_2$ are the final concentration and volume.

Given $M_1 = 9.0 \mathrm{M}$, $V_1 = 40 \mathrm{~mL}$, and $M_2 = 6.0 \mathrm{M}$, we can solve for $V_2$:

\[
(9.0 \mathrm{M})(40 \mathrm{~mL}) = (6.0 \mathrm{M})(V_2)
\]
\[
V_2 = \frac{(9.0 \mathrm{M})(40 \mathrm{~mL})}{6.0 \mathrm{M}} = 60 \mathrm{~mL}
\]

So, the volume of water that must be added is $20 \mathrm{~mL}$ to get $60 \mathrm{~mL}$

Question40

Which of the following statements best explains why an increase in temperature of 5-10 Celsius degrees can substantially increase the rate of a chemical reaction?
(A) The activation energy for the reaction is lowered.

(B) The number of effective collisions between reactant particles is increased.

(D) $\Delta H$ for the reaction is lowered.

(E) $\Delta G$ for the reaction becomes more positive.

▶️Answer/Explanation

Ans:B

An increase in temperature can substantially increase the rate of a chemical reaction because it increases the kinetic energy of the reactant molecules, leading to more frequent and energetic collisions. Among the given options, the best explanation for this phenomenon is (B) The number of effective collisions between reactant particles is increased.

Question41

$
2 \mathrm{KClO}_3(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_2(g)
$

What is the percentage yield of $\mathrm{O}_2$ if $12.3 \mathrm{~g}$ of $\mathrm{KClO}_3$ (molar mass $123 \mathrm{~g}$ ) is decomposed to produce $3.2 \mathrm{~g}$ of $\mathrm{O}_2$ (molar mass $32 \mathrm{~g}$ ) according to the equation above?

(A) $100 \%$

(B) $67 \%$

(D) $33 \%$

(E) $10 \%$

▶️Answer/Explanation

Ans:B

\[ 2 \mathrm{KClO}_3 \rightarrow 2 \mathrm{KCl} + 3 \mathrm{O}_2 \]

Clearly , 2 moles of $\mathrm{KClO}_3$ produce 3 moles of $\mathrm{O}_2$.

Mass of $\mathrm{KClO}_3$ = 12.3 g
Molar mass of $\mathrm{KClO}_3$ = 123 g/mol

\[ \text{moles of } \mathrm{KClO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{12.3 \, \text{g}}{123 \, \text{g/mol}} = 0.1 \, \text{mol} \]

According to the stoichiometry of the reaction, 0.1 moles of $\mathrm{KClO}_3$ will produce $0.1 \times \frac{3}{2} = 0.15$ moles of $\mathrm{O}_2$.

Molar mass of $\mathrm{O}_2$ = 32 g/mol

theoretical yield of $\mathrm{O}_2$:

\[ \text{mass of } \mathrm{O}_2 = \text{moles} \times \text{molar mass} = 0.15 \, \text{mol} \times 32 \, \text{g/mol} = 4.8 \, \text{g} \]

\[ \text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 \]
\[ \text{Percentage yield} = \frac{3.2 \, \text{g}}{4.8 \, \text{g}} \times 100 \]
\[ \text{Percentage yield} = \frac{2}{3} \times 100 \]
\[ \text{Percentage yield} = 66.67\% \]

Rounded to the nearest whole number, the percentage yield is $67\%$. So, the correct answer is (B) $67\%$.

Question42

When a strong acid is titrated with a strong base using phenolphthalein as an indicator, the color changes abruptly at the endpoint of the titration and can be switched back and forth by the addition of only one drop of acid or base. The reason for the abruptness of this color change is that

(A) a large change in $\mathrm{pH}$ occurs near the endpoint of the titration

(B) a buffer solution exists at the endpoint of the titration

(D) the $\mathrm{pH}$ of water is very resistant to change

(E) phenolphthalein is much more sensitive to the $\mathrm{pH}$ of a solution than most other indicators

▶️Answer/Explanation

Ans:A

The abruptness of the color change in a titration using phenolphthalein as an indicator occurs because phenolphthalein transitions between its colorless and pink forms in a very narrow pH range, close to the equivalence point of the titration. This happens because phenolphthalein undergoes a pH-dependent structural change from its colorless to pink form as the solution transitions from acidic to slightly basic conditions. Therefore, the reason for the abruptness of the color change is (A) a large change in pH occurs near the endpoint of the titration.

Question43

A 1 mol sample of zinc can reduce the greatest number of moles of which of the following ions?

(A) $\mathrm{Al}^{3+}$

(B) $\mathrm{Pb}^{2+}$

(D) $\mathrm{Cl}^{-}$

(E) $\mathrm{N}^{3-}$

▶️Answer/Explanation

Ans:C

To determine which ion can be reduced by a 1 mol sample of zinc, we need to consider the reduction potentials of the ions listed. Zinc will reduce ions with reduction potentials lower than its own. Among the ions listed, $\mathrm{Ag}^+$ has the lowest reduction potential. Therefore, a 1 mol sample of zinc can reduce the greatest number of moles of $\mathrm{Ag}^+$. So, the correct answer is (C) $\mathrm{Ag}^+$.

Question44

At $298 \mathrm{~K}$ and $1 \mathrm{~atm}$, bromine is a liquid with a high vapor pressure, whereas chlorine is a gas. This provides evidence that, under these conditions, the

(A) forces among $\mathrm{Br}_2$ molecules are greater than those among $\mathrm{Cl}_2$ molecules

(B) forces among $\mathrm{Br}_2$ molecules are weaker than the $\mathrm{Br}-\mathrm{Br}$ bond

(D) $\mathrm{Br}-\mathrm{Br}$ bond is stronger than the $\mathrm{Cl}-\mathrm{Cl}$ bond

(E) $\mathrm{Br}-\mathrm{Br}$ bond is weaker than the $\mathrm{Cl}-\mathrm{Cl}$ bond

▶️Answer/Explanation

Ans:A

The fact that bromine ($\mathrm{Br}_2$) is a liquid with a high vapor pressure at $298 \, \text{K}$ and $1 \, \text{atm}$, while chlorine ($\mathrm{Cl}_2$) is a gas under the same conditions, indicates that the forces among $\mathrm{Br}_2$ molecules are greater than those among $\mathrm{Cl}_2$ molecules. In other words, bromine molecules experience stronger intermolecular forces, such as van der Waals forces, compared to chlorine molecules. So, the correct answer is (A) forces among $\mathrm{Br}_2$ molecules are greater than those among $\mathrm{Cl}_2$ molecules.

Question45

The value of $K_{s p}$ for $\mathrm{PbCl}_2$ is $1.6 \times 10^{-5}$. What is the lowest concentration of $\mathrm{Cl}^{-}(a q)$ that would be needed to begin precipitation of $\mathrm{PbCl}_2(s)$ in $0.010 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_3\right)_2$ ?

(A) $1.6 \times 10^{-7} \mathrm{M}$

(B) $4.0 \times 10^{-4} \mathrm{M}$

(D) $2.6 \times 10^{-3} \mathrm{M}$

(E) $4.0 \times 10^{-2} \mathrm{M}$

▶️Answer/Explanation

Ans:E

To find the lowest concentration of $\mathrm{Cl}^-$ that would be needed to begin precipitation of $\mathrm{PbCl}_2(s)$, we set up the following equilibrium expression for the dissociation of $\mathrm{PbCl}_2$:
\[ \mathrm{PbCl}_2 \rightleftharpoons \mathrm{Pb}^{2+} + 2\mathrm{Cl}^- \]

The $K_{sp}$ expression for this equilibrium is:
\[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2 \]

Given $K_{sp} = 1.6 \times 10^{-5}$ and the initial concentration of $\mathrm{Pb}^{2+}$ as $0.010 \, \mathrm{M}$ (from $0.010 \, \mathrm{M} \, \mathrm{Pb(NO}_3)_2$), we can assume that initially, there is no $\mathrm{PbCl}_2$ present, so the concentration of $\mathrm{Pb}^{2+}$ comes from the dissociation of $\mathrm{Pb(NO}_3)_2$. Therefore, the concentration of $\mathrm{Pb}^{2+}$ is $0.010 \, \mathrm{M}$. We can rearrange the $K_{sp}$ expression to solve for $\mathrm{Cl}^-$:
\[ [\mathrm{Cl}^-] = \sqrt{\frac{K_{sp}}{[\mathrm{Pb}^{2+}]}} \]
\[ [\mathrm{Cl}^-] = \sqrt{\frac{1.6 \times 10^{-5}}{0.010}} \]
\[ [\mathrm{Cl}^-] = \sqrt{1.6 \times 10^{-3}} \]
\[ [\mathrm{Cl}^-] \approx 0.040 \, \mathrm{M} \]

Therefore, the lowest concentration of $\mathrm{Cl}^-$ needed to begin precipitation of $\mathrm{PbCl}_2(s)$ is $0.040 \, \mathrm{M}$. So, the closest answer choice is (E) $4.0 \times 10^{-2} \, \mathrm{M}$.

Question46

Which of the following aqueous solutions has the lowest freezing point?

(A) $0.2 \mathrm{~m} \mathrm{NaCl}$

(B) $0.2 \mathrm{~m} \mathrm{CaCl}$

(D) $0.2 \mathrm{~m} \mathrm{NH}_3$

(E) $0.2 \mathrm{~m} \mathrm{Al}\left(\mathrm{NO}_3\right)_3$

▶️Answer/Explanation

Ans:E

To determine which aqueous solution has the lowest freezing point, we need to identify the solute that will provide the maximum number of particles when dissolved. This is because the extent of freezing point depression is directly proportional to the concentration of solute particles.

Let’s analyze the options:

(A) $0.2 \, \text{m} \, \mathrm{NaCl}$: When NaCl dissolves in water, it dissociates into one Na⁺ ion and one Cl⁻ ion, so it provides two particles.

(B) $0.2 \, \text{m} \, \mathrm{CaCl}_2$: When CaCl₂ dissolves in water, it dissociates into one Ca²⁺ ion and two Cl⁻ ions, so it provides three particles.

(C) $0.2 \, \text{m} \, \mathrm{H}_2\mathrm{SO}_4$: When H₂SO₄ dissociates in water, it dissociates into two H⁺ ions and one SO₄²⁻ ion, so it provides three particles.

(D) $0.2 \, \text{m} \, \mathrm{NH}_3$: Ammonia ($ \mathrm{NH}_3 $) does not dissociate significantly in water, so it provides only one particle.

(E) $0.2 \, \text{m} \, \mathrm{Al}(\mathrm{NO}_3)_3$: When $ \mathrm{Al}(\mathrm{NO}_3)_3 $ dissolves in water, it dissociates into one Al³⁺ ion and three NO₃⁻ ions, so it provides four particles.

Therefore, among the given options, the aqueous solution of $0.2 \, \text{m} \, \mathrm{Al}(\mathrm{NO}_3)_3$ (option E) will have the lowest freezing point.

Question47

Step 1: $\quad \mathrm{NO}(g)+\mathrm{O}_3(g) \rightarrow \mathrm{NO}_2(g)+\mathrm{O}_2(g)$
Step 2: $\quad \mathrm{NO}_2(g)+\mathrm{O}(g) \rightarrow \mathrm{NO}(g)+\mathrm{O}_2(g)$
A reaction mechanism for the destruction of ozone, $\mathrm{O}_3(\mathrm{~g})$, is represented above. In the overall reaction, $\mathrm{NO}(g)$ is best described as
(A) an inhibitor

(B) a catalyst

(D) an intermediate

(E) a product

▶️Answer/Explanation

Ans:D

In the given reaction mechanism:

Step 1: $ \mathrm{NO}(\mathrm{g}) + \mathrm{O}_3(\mathrm{g}) \rightarrow \mathrm{NO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) $

Step 2: $ \mathrm{NO}_2(\mathrm{g}) + \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) $

The overall reaction is the combination of these two steps, and $ \mathrm{NO}(\mathrm{g}) $ appears as both a reactant and a product.

However, it’s important to note that in the context of the overall reaction, $ \mathrm{NO}(\mathrm{g}) $ is not consumed to produce other products directly. Instead, it is regenerated in the second step, making it an intermediate species.

Therefore, the best description for $ \mathrm{NO}(\mathrm{g}) $ in the overall reaction is **(D) an intermediate**.

Question48

When a buret is rinsed before a titration, which of the techniques below is the best procedure?
(A) Rinse the buret one time with some of the titrant solution.

(B) Rinse the buret one time with some of the titrant solution and then dry the buret in an oven.

(D) Rinse the buret two times: each time with some of the titrant solution.

(E) Rinse the buret two times: each time with distilled water.

▶️Answer/Explanation

Ans:C

The best procedure for rinsing a buret before a titration is to ensure that the buret is clean and free from any contaminants that could affect the accuracy of the titration results.

Option (C) Rinse the buret two times: once with some of the titrant solution, then once with distilled water, is the most appropriate technique.

Rinsing the buret once with the titrant solution helps to remove any residual particles or contaminants from the previous use of the buret, ensuring that the titration solution will not be contaminated.

Following this, rinsing the buret with distilled water helps to further clean the buret and remove any remaining traces of the titrant solution, ensuring that the next titration is not influenced by any leftover solution from the previous rinsing.

Therefore, the correct answer is (C) Rinse the buret two times: once with some of the titrant solution, then once with distilled water.

Question49

$
\mathrm{Sr}^{2+}(a q)+\mathrm{F}^{-}(a q) \rightleftarrows \operatorname{SrF}^{+}(a q)
$

At $25^{\circ} \mathrm{C}$, the equilibrium constant for the reaction represented above has a value of 1.3 . At $50^{\circ} \mathrm{C}$, the value of the equilibrium constant is less than 1.3. Based on this information, which of the following must be correct?

(A) The reaction rate decreases as the temperature is increased.

(B) The reaction is thermodynamically favorable only at temperatures above $25^{\circ} \mathrm{C}$.

(D) At $25^{\circ} \mathrm{C}, \Delta S^{\circ}$ for the reaction is positive.

(E) At $25^{\circ} \mathrm{C}, \Delta H^{\circ}$ for the reaction is negative.

▶️Answer/Explanation

Ans:E

Equilibrium constant ($K_{\text{eq}}$) decreases as the temperature is increased from $25^\circ \text{C}$ to $50^\circ \text{C}$, it implies that the reaction shifts in the direction of the reactants as the temperature increases.

(A) The reaction rate decreases as the temperature is increased.
This statement is false because according to Le Chatelier’s principle, an increase in temperature favors the endothermic direction of the reaction, which would increase the reaction rate.

(B) The reaction is thermodynamically favorable only at temperatures above $25^\circ \text{C}$.
This statement is false because the equilibrium constant being greater than 1 at $25^\circ \text{C}$ indicates that the reaction is thermodynamically favorable at that temperature as well.

(C) At $25^\circ \text{C}, \Delta G^\circ$ for the reaction is positive.
This statement is false because a positive value of $ \Delta G^\circ $ would indicate a non-spontaneous reaction, which is not the case when $ K_{\text{eq}} > 1 $.

(D) At $25^\circ \text{C}, \Delta S^\circ$ for the reaction is positive.
This statement cannot be determined solely based on the given information about the equilibrium constant and temperature.

(E) At $25^\circ \text{C}, \Delta H^\circ$ for the reaction is negative.
This statement is true because when the equilibrium constant is greater than 1 ($K_{\text{eq}} > 1$), it indicates that the reaction is exothermic ($ \Delta H^\circ < 0$).

Therefore, the correct answer is (E) At $25^\circ \text{C}, \Delta H^\circ$ for the reaction is negative.

Question50

$
\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_3(a q) \rightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}(a q)
$

The reaction represented above is best classified as

(A) a Lewis acid-base reaction

(B) a Brønsted-Lowry acid-base reaction

(D) an oxidation-reduction reaction

(E) a precipitation reaction

▶️Answer/Explanation

Ans:A

The reaction represented is a coordination complex formation reaction, where the nickel ion ($\mathrm{Ni}^{2+}$) reacts with ammonia ($\mathrm{NH}_3$) ligands to form a complex ion $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$. In this complex, the nickel ion is surrounded by six ammonia molecules as ligands.

This reaction does not involve the transfer of protons, so it is not an acid-base reaction. It also does not involve the transfer of electrons between reactants, so it is not an oxidation-reduction reaction. Additionally, it does not result in the formation of a precipitate, so it is not a precipitation reaction.

The formation of coordination complexes involves the interaction of metal ions with ligands to form complex ions. These reactions are best classified as Lewis acid-base reactions, where the metal ion acts as a Lewis acid by accepting electron pairs from the ligands (Lewis bases). Therefore, the correct answer is (A) a Lewis acid-base reaction.

Question51

$
\begin{array}{ll}
\mathrm{Cu}^{2+}(a q)+2 e^{-} \rightarrow \mathrm{Cu}(s) & E^{\circ}=0.34 \mathrm{~V} \\
\mathrm{Cr}^{3+}(a q)+e^{-} \rightarrow \mathrm{Cr}^{2+}(a q) & E^{\circ}=-0.41 \mathrm{~V}
\end{array}
$

According to the half-reactions represented above, which of the following occurs in aqueous solutions under standard conditions?

(A) $\mathrm{Cu}^{2+}(a q)+\mathrm{Cr}^{3+}(a q) \rightarrow \mathrm{Cu}(s)+\mathrm{Cr}^{2+}(a q)$

(B) $\mathrm{Cu}^{2+}(a q)+2 \mathrm{Cr}^{2+}(a q) \rightarrow \mathrm{Cu}(s)+2 \mathrm{Cr}^{3+}(a q)$

(D) $\mathrm{Cu}(s)+\mathrm{Cr}^{3+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cr}^{2+}(a q)$

(E) $2 \mathrm{Cu}^{2+}(a q)+\mathrm{Cr}^{3+}(a q) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Cr}^{2+}(a q)$

▶️Answer/Explanation

Ans:E

To determine which of the given reactions occurs under standard conditions, we need to compare the standard reduction potentials ($E^\circ$) for the two half-reactions provided.

Given:

1. $ \mathrm{Cu}^{2+}(aq) + 2e^- \rightarrow \mathrm{Cu}(s) \quad E^\circ = 0.34 \, \text{V} $
2. $ \mathrm{Cr}^{3+}(aq) + e^- \rightarrow \mathrm{Cr}^{2+}(aq) \quad E^\circ = -0.41 \, \text{V} $

The reaction that occurs spontaneously is the one where the reduction potential of the cathode (the species being reduced) is greater than the reduction potential of the anode (the species being oxidized).

Since $ \mathrm{Cu}^{2+}(aq) $ has a higher reduction potential ($E^\circ = 0.34 \, \text{V}$) compared to $ \mathrm{Cr}^{3+}(aq) $ ($E^\circ = -0.41 \, \text{V}$), the reaction where $ \mathrm{Cu}^{2+}(aq) $ is reduced and $ \mathrm{Cr}^{3+}(aq) $ is oxidized is the one that occurs spontaneously.

The correct reaction is option (E) $2 \mathrm{Cu}^{2+}(aq) + \mathrm{Cr}^{3+}(aq) \rightarrow 2 \mathrm{Cu}(s) + \mathrm{Cr}^{2+}(aq)$.

Question52

$
\ldots \mathrm{Zn}(s)+\ldots \mathrm{H}^{+}(a q)+\ldots \mathrm{NO}_3^{-}(a q) \rightarrow \ldots \mathrm{Zn}^{2+}(a q)+\ldots \mathrm{NH}_4^{+}(a q)+\ldots \mathrm{H}_2 \mathrm{O}(l)
$

When the equation above is balanced and all coefficients are reduced to lowest whole number terms, the coefficient for $\mathrm{Zn}(s)$ is

(A) 2

(B) 4

(D) 10

(E) 14

▶️Answer/Explanation

Ans:B

Oxidation half reaction
$\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})$
$\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$
[To balance charges]

\[
\begin{aligned}
& \text { Reduction half-reaction } \\
& \mathrm{NO}_3^{-}(\mathrm{aq}) \longrightarrow \mathrm{NH}_4^{-}(\mathrm{aq}) \\
& \mathrm{NO}_3^{-}(\mathrm{aq}) \longrightarrow \mathrm{NH}_4^{+}(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
& {\left[\text { To balance } \mathrm{O}^{-} \text {atom] } \ldots(2)\right.} \\
& \mathrm{NO}_3^{-}(\mathrm{aq})+10 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{NH}_4^{+}(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(1)
\end{aligned}
\]
[To balance $\mathrm{H}^{+1}$ atoms]
\[
\mathrm{NO}_3^{-}(\mathrm{aq})+10 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{NH}_4^{+}(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(1)
\]

Multiply (1) equation by 4 to equalize the no. of electrons in both. Add both half reactions
\[
4 \mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+8 \mathrm{e}^{-}
\]
then $(3)+(2)$ is
\[
4 \mathrm{Zn}(\mathrm{s})+\mathrm{NO}_3^{-}(\mathrm{aq})+10 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 4 \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{NH}_4^{+}(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})
\]

Question53

$
4 \mathrm{NH}_3(g)+3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{~N}_2(g)+6 \mathrm{H}_2 \mathrm{O}(g)
$

If the standard molar heats of formation of ammonia, $\mathrm{NH}_3(\mathrm{~g})$, and gaseous water, $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$, are $-46 \mathrm{~kJ} / \mathrm{mol}$ and $-242 \mathrm{~kJ} / \mathrm{mol}$, respectively, what is the value of $\Delta H_{298}^{\circ}$ for the reaction represented above?

(A) $-190 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

(B) $-290 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

(D) $-1,270 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

(E) $-1,640 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

▶️Answer/Explanation

Ans:D

The standard heat of formation ($\Delta H_f^\circ$) of a compound is the change in enthalpy that results when one mole of the compound is formed from its elements in their standard states. The standard enthalpy change ($\Delta H_{298}^\circ$) for a reaction is the sum of the standard heats of formation of the products minus the sum of the standard heats of formation of the reactants.

Given the balanced chemical equation:
\[ 4 \mathrm{NH}_3(g) + 3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{N}_2(g) + 6 \mathrm{H}_2\mathrm{O}(g) \]

We can calculate $\Delta H_{298}^\circ$ using the standard heats of formation ($\Delta H_f^\circ$) of the reactants and products:
\[ \Delta H_{298}^\circ = \sum \Delta H_f^\circ(\text{products}) – \sum \Delta H_f^\circ(\text{reactants}) \]

Given:
\[ \Delta H_f^\circ(\mathrm{NH}_3) = -46 \, \text{kJ/mol} \]
\[ \Delta H_f^\circ(\mathrm{H}_2\mathrm{O}) = -242 \, \text{kJ/mol} \]

Using these values:
\[ \Delta H_{298}^\circ = [2 \cdot \Delta H_f^\circ(\mathrm{N}_2) + 6 \cdot \Delta H_f^\circ(\mathrm{H}_2\mathrm{O})] – [4 \cdot \Delta H_f^\circ(\mathrm{NH}_3) + 3 \cdot \Delta H_f^\circ(\mathrm{O}_2)] \]
\[ \Delta H_{298}^\circ = [2 \cdot 0 + 6 \cdot (-242)] – [4 \cdot (-46) + 3 \cdot 0] \]
\[ \Delta H_{298}^\circ = (-1452) – (-184) \]
\[ \Delta H_{298}^\circ = -1452 + 184 \]
\[ \Delta H_{298}^\circ = -1268 \, \text{kJ/mol}_{\text{rxn}} \]

So, the correct answer is (D) $-1,270 \, \text{kJ/mol}_{\text{rxn}}$.

Question54

When a magnesium wire is dipped into a solution of lead(II) nitrate, a black deposit forms on the wire. Which of the following can be concluded from this observation?

(A) The standard reduction potential, $E^{\circ}$, for $\mathrm{Pb}^{2+}(a q)$ is greater than that for $\mathrm{Mg}^{2+}(a q)$.

(B) $\mathrm{Mg}(s)$ is less easily oxidized than $\mathrm{Pb}(s)$.

(D) The magnesium wire will be the cathode of a $\mathrm{Mg} / \mathrm{Pb}$ cell.

(E) $\mathrm{Pb}(s)$ can spontaneously displace $\mathrm{Mg}^{2+}(a q)$ from solution.

▶️Answer/Explanation

Ans:A

The formation of a black deposit on the magnesium wire indicates that magnesium is displacing lead from the lead(II) nitrate solution. This suggests that magnesium is more reactive than lead, and therefore, the standard reduction potential ($E^\circ$) for $\mathrm{Pb}^{2+}(aq)$ is greater than that for $\mathrm{Mg}^{2+}(aq)$. Thus, the correct answer is (A) The standard reduction potential, $E^\circ$, for $\mathrm{Pb}^{2+}(aq)$ is greater than that for $\mathrm{Mg}^{2+}(aq)$.

Question55

Which of the molecules represented below contains carbon with $s p^2$ hybridization?

(A) $\mathrm{CH}_4$

(B) $\mathrm{CH}_2 \mathrm{Cl}_2$

(D) $\mathrm{C}_2 \mathrm{H}_2 \mathrm{Cl}_2$

(E) $\mathrm{C}_2 \mathrm{H}_4 \mathrm{Cl}_2$

▶️Answer/Explanation

Ans:D

Carbon with $sp^2$ hybridization forms trigonal planar geometries. Among the molecules listed, option (D) $\mathrm{C}_2\mathrm{H}_2\mathrm{Cl}_2$ has a trigonal planar geometry due to the presence of a double bond between the two carbon atoms, causing $sp^2$ hybridization in each carbon atom. Therefore, the correct answer is (D) $\mathrm{C}_2\mathrm{H}_2\mathrm{Cl}_2$.

Question56

A chemical supply company sells a concentrated solution of aqueous $\mathrm{H}_2 \mathrm{SO}_4$ (molar mass $98 \mathrm{~g} \mathrm{~mol}^{-1}$ ) that is 50 . percent $\mathrm{H}_2 \mathrm{SO}_4$ by mass. At $25^{\circ} \mathrm{C}$, the density of the solution is $1.4 \mathrm{~g} \mathrm{~mL}^{-1}$. What is the molarity of the $\mathrm{H}_2 \mathrm{SO}_4$ solution at $25^{\circ} \mathrm{C}$ ?

(A) $1.8 \mathrm{M}$

(B) $3.6 \mathrm{M}$

(D) $7.1 \mathrm{M}$

(E) $14 \mathrm{M}$

▶️Answer/Explanation

Ans:D

To find the molarity of the $\mathrm{H}_2\mathrm{SO}_4$ solution, we need to first calculate the number of moles of $\mathrm{H}_2\mathrm{SO}_4$ present in the solution and then divide by the volume of the solution in liters.

The solution is 50% $\mathrm{H}_2\mathrm{SO}_4$ by mass.
The molar mass of $\mathrm{H}_2\mathrm{SO}_4$ is $98 \, \text{g/mol}$.
The density of the solution is $1.4 \, \text{g/mL}$.

Let’s assume we have $100 \, \text{g}$ of the solution. Since the solution is 50% $\mathrm{H}_2\mathrm{SO}_4$ by mass, we have $50 \, \text{g}$ of $\mathrm{H}_2\mathrm{SO}_4$.

Now, let’s find the number of moles of $\mathrm{H}_2\mathrm{SO}_4$:
\[
\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{50 \, \text{g}}{98 \, \text{g/mol}} \approx 0.51 \, \text{mol}
\]

Next, let’s convert the volume of the solution from grams to liters:
\[
\text{volume} = \frac{\text{mass}}{\text{density}} = \frac{100 \, \text{g}}{1.4 \, \text{g/mL}} \approx 71.43 \, \text{mL} \approx 0.07143 \, \text{L}
\]

Now, we can find the molarity ($M$) using the formula:
\[
M = \frac{\text{moles}}{\text{volume in liters}} = \frac{0.51 \, \text{mol}}{0.07143 \, \text{L}} \approx 7.14 \, \text{M}
\]

So, the molarity of the $\mathrm{H}_2\mathrm{SO}_4$ solution at $25^\circ \text{C}$ is approximately $7.14 \, \text{M}$. The closest option is (D) $7.1 \, \text{M}$.

Question57

A reaction produces a colorless gas, which is collected by water displacement. A glowing splint inserted into a bottle full of the gas is extinguished. The gas could be

(A) $\mathrm{N}_2$

(B) $\mathrm{NO}_2$

(D) $\mathrm{Br}_2$

(E) $\mathrm{Cl}_2$

▶️Answer/Explanation

Ans:A

The fact that a glowing splint inserted into a bottle full of the gas is extinguished suggests that the gas is non-combustible. Among the options provided, only option (A) $\mathrm{N}_2$ (nitrogen gas) is colorless, non-combustible, and would extinguish a flame. Therefore, the correct answer is (A) $\mathrm{N}_2$.

Question58

A solution of methanol, $\mathrm{CH}_3 \mathrm{OH}$, in water is prepared by mixing together $128 \mathrm{~g}$ of methanol and $108 \mathrm{~g}$ of water. The mole fraction of methanol in the solution is closest to

(A) 0.80

(B) 0.60

(D) 0.40

(E) 0.20

▶️Answer/Explanation

Ans:D

The mole fraction ($X$) of methanol in the solution is given by the ratio of the number of moles of methanol to the total number of moles of solute (methanol) and solvent (water).

Moles of methanol ($\mathrm{CH}_3\mathrm{OH}$):

\[
\text{moles of methanol} = \frac{\text{mass}}{\text{molar mass}} = \frac{128 \, \text{g}}{32 \, \text{g/mol}} = 4 \, \text{mol}
\]

Moles of water ($\mathrm{H}_2\mathrm{O}$):
\[
\text{moles of water} = \frac{\text{mass}}{\text{molar mass}} = \frac{108 \, \text{g}}{18 \, \text{g/mol}} = 6 \, \text{mol}
\]

\[
\text{total moles} = \text{moles of methanol} + \text{moles of water} = 4 \, \text{mol} + 6 \, \text{mol} = 10 \, \text{mol}
\]

Now, we can calculate the mole fraction of methanol ($X_{\text{methanol}}$):
\[
X_{\text{methanol}} = \frac{\text{moles of methanol}}{\text{total moles}} = \frac{4 \, \text{mol}}{10 \, \text{mol}} = 0.4
\]

So, the mole fraction of methanol in the solution is $0.4$.

Question59

A sample of a compound contains $3.21 \mathrm{~g}$ of sulfur and $11.4 \mathrm{~g}$ of fluorine. Which of the following represents the empirical formula of the compound?

(A) $\mathrm{SF}_2$

(B) $\mathrm{SF}_3$

(D) $\mathrm{SF}_5$

(E) $\mathrm{SF}_6$

▶️Answer/Explanation

Ans:E

To find the empirical formula of the compound, we need to determine the simplest whole number ratio of atoms present in the compound.
Mass of sulfur ($\text{S}$) = $3.21 \, \text{g}$
Mass of fluorine ($\text{F}$) = $11.4 \, \text{g}$

First, let’s find the number of moles of each element:
\[
\text{moles of S} = \frac{\text{mass of S}}{\text{molar mass of S}} = \frac{3.21 \, \text{g}}{32.06 \, \text{g/mol}} \approx 0.100 \, \text{mol}
\]
\[
\text{moles of F} = \frac{\text{mass of F}}{\text{molar mass of F}} = \frac{11.4 \, \text{g}}{19.00 \, \text{g/mol}} \approx 0.600 \, \text{mol}
\]

Next, we’ll find the ratio of moles of each element. Since sulfur is present in the smaller amount, we’ll divide the moles of fluorine by the moles of sulfur:
\[
\text{ratio} = \frac{\text{moles of F}}{\text{moles of S}} = \frac{0.600 \, \text{mol}}{0.100 \, \text{mol}} = 6
\]

This means that the ratio of sulfur to fluorine is $1 : 6$. Thus, the empirical formula of the compound is $\text{SF}_6$. Therefore, the correct answer is (E) $\text{SF}_6$.

Question60

Of the following, the best explanation for the fact that most gases are easily compressed is that the molecules in a gas

(A) are in constant motion

(B) are relatively far apart

(D) have a real, nonzero volume

(E) move slower as temperature decreases

▶️Answer/Explanation

Ans:B

The fact that most gases are easily compressed can be best explained by option (B) – “are relatively far apart.”

In a gas, the molecules are not tightly packed together like in a liquid or solid. Instead, they are spread out and move freely. This means that there is a significant amount of empty space between gas molecules. When pressure is applied to a gas, the gas molecules can be pushed closer together because of the space between them. As a result, the volume of the gas can be reduced, leading to compression.

Therefore, the ability of gases to be easily compressed is primarily due to the relatively large distances between gas molecules compared to the sizes of the molecules themselves.

Question61

Given that the density of $\mathrm{Hg}(l)$ at $0^{\circ} \mathrm{C}$ is about $14 \mathrm{~g} \mathrm{~mL}^{-1}$, which of the following is closest to the volume of one mole of $\mathrm{Hg}(l)$ at this temperature?

(A) $0.070 \mathrm{~mL}$

(B) $0.14 \mathrm{~mL}$

(D) $14 \mathrm{~mL}$

(E) $28 \mathrm{~mL}$

▶️Answer/Explanation

Ans:D

Density of liquid mercury $(\mathrm{Hg}(l))$ at $0^{\circ} \mathrm{C}=14 \mathrm{~g} / \mathrm{mL}$

Now, we need to find the mass of one mole of mercury $(\mathrm{Hg})$. The molar mass of mercury is $200.59 \mathrm{~g} / \mathrm{mol}$ (rounded to two decimal places).

Using this information, we can calculate the volume of one mole of mercury:
\[
\begin{aligned}
& V=\frac{m_{\mathrm{Hg}}}{d}=\frac{200.59 \mathrm{~g} / \mathrm{mol}}{14 \mathrm{~g} / \mathrm{mL}} \\
& V=\frac{200.59}{14} \mathrm{~mL} / \mathrm{mol} \approx 14.33 \mathrm{~mL} / \mathrm{mol}
\end{aligned}
\]

Therefore, the volume of one mole of liquid mercury $(\mathrm{Hg}(l))$ at $0^{\circ} \mathrm{C}$ is approximately $14.33 \mathrm{~mL} / \mathrm{mol}$.

Among the given options, the closest one is (D) $14 \mathrm{~mL}$.

Question62

A sample of an unknown gas from a cylinder is collected over water in the apparatus shown above. After all the gas sample has been collected, the water levels inside and outside the gas collection tube are made the same. Measurements that must be made to calculate the molar mass of the gas include all of the following EXCEPT

(A) atmospheric pressure

(B) temperature of the water

(D) initial and final mass of the gas cylinder

(E) mass of the water in the apparatus

▶️Answer/Explanation

Ans:E

To calculate the molar mass of the gas collected over water, we need to consider the following measurements:

Atmospheric pressure: The atmospheric pressure affects the pressure of the gas collected. We need this information to apply the ideal gas law.
Temperature of the water: The temperature affects the volume of the gas collected. We use the ideal gas law, which includes temperature.
Volume of gas in the gas-collection tube: This is essential for calculating the volume of the gas. The ideal gas law involves volume.
Initial and final mass of the gas cylinder: These measurements allow us to determine the mass of the gas collected. Mass is crucial for calculating molar mass.
Mass of the water in the apparatus: This measurement is not necessary for calculating the molar mass of the gas. The water does not directly impact the gas’s molar mass.

Therefore, the measurement that is not required to calculate the molar mass of the gas is (E) mass of the water in the apparatus.

Question63

Addition of sulfurous acid (a weak acid) to barium hydroxide (a strong base) results in the formation of a precipitate. The net ionic equation for this reaction is

(A) $2 \mathrm{H}^{+}(a q)+2 \mathrm{OH}^{-}(a q) \rightleftarrows 2 \mathrm{H}_2 \mathrm{O}(l)$

(B) $\mathrm{H}_2 \mathrm{SO}_3(a q)+\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \rightleftarrows \mathrm{BaSO}_3(s)+2 \mathrm{H}_2 \mathrm{O}(l)$

(C) $2 \mathrm{H}^{+}(a q)+\mathrm{SO}_3{ }^{2-}(a q)+\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \rightleftarrows \mathrm{BaSO}_3(s)+2 \mathrm{H}_2 \mathrm{O}(l)$

(D) $\mathrm{H}_2 \mathrm{SO}_3(a q)+\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \rightleftarrows \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_3{ }^{2-}(a q)+2 \mathrm{H}_2 \mathrm{O}(l)$

(E) $\mathrm{H}_2 \mathrm{SO}_3(a q)+\mathrm{Ba}(\mathrm{OH})_2(a q) \rightleftarrows \mathrm{BaSO}_3(s)+2 \mathrm{H}_2 \mathrm{O}(l)$

▶️Answer/Explanation

Ans:B

To write the net ionic equation, we first need to break down the reaction into its ionic components and then eliminate the spectator ions.

The balanced chemical equation is:

\[ \text{H}_2\text{SO}_3(aq) + \text{Ba}^{2+}(aq) + 2\text{OH}^-(aq) \rightleftharpoons \text{BaSO}_3(s) + 2\text{H}_2\text{O}(l) \]

To write the net ionic equation, we remove the spectator ions, which are the ions that appear on both sides of the equation without undergoing a chemical change.

The ions that do not participate in the formation of the solid precipitate are $ \text{Ba}^{2+}(aq) $ and $ \text{OH}^-(aq) $. Therefore, we remove them from both sides of the equation:

\[ \text{H}_2\text{SO}_3(aq) + 2\text{OH}^-(aq) \rightleftharpoons \text{BaSO}_3(s) + 2\text{H}_2\text{O}(l) \]

So, the net ionic equation for the reaction between sulfurous acid and barium hydroxide is:

\[ \text{H}_2\text{SO}_3(aq) + 2\text{OH}^-(aq) \rightleftharpoons \text{BaSO}_3(s) + 2\text{H}_2\text{O}(l) \]

This equation represents the chemical change occurring during the reaction, highlighting the species that are directly involved in forming the precipitate.

Question64

$
\begin{aligned}
& \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(s) \rightarrow \mathrm{HI}(g) \quad \Delta H=26 \mathrm{~kJ} / \mathrm{mol}_{r x n} \\
& \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \rightarrow \mathrm{HI}(g) \quad \Delta H=-5.0 \mathrm{~kJ} / \mathrm{mol}_{r x n} \\
&
\end{aligned}
$
Based on the information above, what is the enthalpy change for the sublimation of iodine, represented below?
$
\mathrm{I}_2(s) \rightarrow \mathrm{I}_2(g)
$
(A) $15 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

(B) $21 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

(D) $42 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

(E) $62 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

▶️Answer/Explanation

Ans:C

To find the enthalpy change for the sublimation of iodine ($ \mathrm{I}_2(s) \rightarrow \mathrm{I}_2(g) $), we can use Hess’s Law, which states that the total enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps.

We can write the desired sublimation reaction as the combination of the given reactions:

1. $ \frac{1}{2} \mathrm{H}_2(g) + \frac{1}{2} \mathrm{I}_2(g) \rightarrow \mathrm{HI}(g) $ with $ \Delta H = -5.0 \, \text{kJ/mol}_{rxn} $
2. $ \frac{1}{2} \mathrm{H}_2(g) + \frac{1}{2} \mathrm{I}_2(s) \rightarrow \mathrm{HI}(g) $ with $ \Delta H = 26 \, \text{kJ/mol}_{rxn} $

To obtain the desired sublimation reaction, we need to subtract the second reaction from the first:

\[ \left( \frac{1}{2} \mathrm{H}_2(g) + \frac{1}{2} \mathrm{I}_2(g) \rightarrow \mathrm{HI}(g) \right) – \left( \frac{1}{2} \mathrm{H}_2(g) + \frac{1}{2} \mathrm{I}_2(s) \rightarrow \mathrm{HI}(g) \right) \]

This gives us:

\[ \frac{1}{2} \mathrm{I}_2(s) \rightarrow \frac{1}{2} \mathrm{I}_2(g) \]

To find the enthalpy change for this reaction, we subtract the second reaction’s enthalpy from the first:

\[ -5.0 \, \text{kJ/mol}_{rxn} – 26 \, \text{kJ/mol}_{rxn} = -31 \, \text{kJ/mol}_{rxn} \]

So, the enthalpy change for the sublimation of iodine is $ -31 \, \text{kJ/mol}_{rxn} $.

Therefore, the correct answer is (C) $ 31 \, \text{kJ/mol}_{rxn} $

Question65

Which of the following graphs correctly shows the relationship between potential energy and internuclear separation for two hydrogen atoms?

▶️Answer/Explanation

Ans:D

Question66

Which of the following compounds is LEAST likely to exist?

(A) $\mathrm{PCl}_5$

(B) $\mathrm{PBr}_3$

(D) $\mathrm{NI}_5$

(E) $\mathrm{SbF}_5$

▶️Answer/Explanation

Ans:D

To determine which compound is least likely to exist, we should consider the expected valence electron configuration and the stability of the central atom in each compound.

Let’s analyze each compound:

(A) $\mathrm{PCl}_5$: Phosphorus can expand its octet and accommodate five chlorine atoms around it due to its empty 3d orbitals. This compound exists and is known.

(B) $\mathrm{PBr}_3$: Similar to $\mathrm{PCl}_5$, phosphorus can accommodate three bromine atoms around it, and this compound is also known.

(C) $\mathrm{NF}_3$: Nitrogen typically forms three covalent bonds and a lone pair, consistent with the $sp^2$ hybridization. This compound is known and exists.

(D) $\mathrm{NI}_5$: Iodine can potentially accommodate five iodine atoms around it due to its large size and availability of d orbitals. However, this compound is highly reactive and unstable and hasn’t been synthesized.

(E) $\mathrm{SbF}_5$: Antimony, like phosphorus, can expand its octet and accommodate five fluorine atoms around it due to empty d orbitals. This compound is known and exists.

So, among the given options, $\mathrm{NI}_5$ (D) is the least likely to exist.

Question67

The graph above shows the titration curve that resulted when a sample of $0.1 \mathrm{M}$ monoprotic acid was titrated with a solution of $\mathrm{NaOH}$. Based on the graph, the $\mathrm{p} K_a$ of the acid is closest to

(A) 3.0

(B) 4.0

(D) 8.0

(E) 12.0

▶️Answer/Explanation

Ans:B

The pH at the half-equivalence point corresponds to the pKa of the acid. Let’s analyze the graph:

The half-equivalence point occurs when exactly half of the acid has reacted with the base (i.e., moles of acid = moles of base).
At this point, the pH is equal to the pKa of the acid.

From the graph, we can see that the half-equivalence point occurs around a pH of 4.0. Therefore, the pKa of the acid is closest to 4.0.

The correct answer is (B) 4.0

Question68

The rate law for the reaction of nitrogen dioxide and chlorine is found to be rate $=k\left[\mathrm{NO}_2\right]^2\left[\mathrm{Cl}_2\right]$. By what factor does the rate of the reaction change when the concentrations of both $\mathrm{NO}_2$ and $\mathrm{Cl}_2$ are doubled?

(A) 2

(B) 3

(D) 6

(E) 8

▶️Answer/Explanation

Ans:E

The rate law given is: rate $=k\left[\mathrm{NO}_2\right]^2\left[\mathrm{Cl}_2\right]$.
If both $\left[\mathrm{NO}_2\right]$ and $\left[\mathrm{Cl}_2\right]$ are doubled, the new rate will be $k(2[\mathrm{NO}_2])^2(2[\mathrm{Cl}_2])$.
Simplifying, we get a factor of $2^2 \times 2 = 8$.
So, the rate of the reaction changes by a factor of 8, which matches option (E).

Question69

When a student prepares an aqueous solution containing the five cations $\mathrm{Ag}^{+}(a q), \mathrm{Hg}_2{ }^{2+}(a q)$, $\mathrm{Cu}^{2+}(a q), \mathrm{Mn}^{2+}(a q)$, and $\mathrm{Ba}^{2+}(a q)$, the student observes that no precipitates form in the solution. Which of the following could be the identity of the anion in the solution?

(A) $\mathrm{Cl}^{-}(a q)$

(B) $\mathrm{CO}_3{ }^{2-}(a q)$

(D) $\mathrm{NO}_3^{-}(a q)$

(E) $\mathrm{SO}_4{ }^{2-}(a q)$

▶️Answer/Explanation

Ans:D

The absence of precipitates suggests that none of the cations react with the anion to form insoluble compounds.
Among the given options, $\mathrm{NO}_3^{-}$ and $\mathrm{Cl}^{-}$ ions usually form soluble salts with cations, thus not likely to cause precipitation.
$\mathrm{CO}_3^{2-}$, $\mathrm{CrO}_4^{2-}$, and $\mathrm{SO}_4^{2-}$ ions can form insoluble salts with some cations. However, if no precipitate forms, it suggests that the anion is unlikely to be any of these three.

Therefore, option (D) $\mathrm{NO}_3^{-}(aq)$ is the most suitable choice.

Question70

What is the molarity of $\mathrm{I}^{-}(a q)$ in a solution that contains $34 \mathrm{~g}$ of $\mathrm{SrI}_2$ (molar mass $341 \mathrm{~g}$ ) in $1.0 \mathrm{~L}$ of the solution?

(A) $0.034 \mathrm{M}$

(B) $0.068 \mathrm{M}$

(D) $0.20 \mathrm{M}$

(E) $0.68 \mathrm{M}$

▶️Answer/Explanation

Ans:D

The molar mass of $\mathrm{SrI}_2$ is $341 \mathrm{~g/mol}$, and $34 \mathrm{~g}$ of $\mathrm{SrI}_2$ is dissolved in $1.0 \mathrm{~L}$ of solution.

This means there are $34 \mathrm{~g}/341 \mathrm{~g/mol} = 0.1 \mathrm{~mol}$ of $\mathrm{SrI}_2$.
Since $\mathrm{SrI}_2$ dissociates into $2$ moles of $I^{-}$ ions per mole of $\mathrm{SrI}_2$, we have $0.1 \mathrm{~mol} \times 2 = 0.2 \mathrm{~mol}$ of $I^{-}$ ions.
The volume of the solution is $1.0 \mathrm{~L}$, so the molarity ($M$) of $I^{-}$ ions is $0.2 \mathrm{~mol}/1.0 \mathrm{~L} = 0.20 \mathrm{~M}$, which matches option (D).

Question71

Which of the following compounds contains both ionic and covalent bonds?

(A) $\mathrm{SO}_3$

(B) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

(D) $\mathrm{H}_2 \mathrm{~S}$

(E) $\mathrm{NH}_4 \mathrm{Cl}$

▶️Answer/Explanation

Ans:E

The compound that contains both ionic and covalent bonds is (E) $\mathrm{NH}_4 \mathrm{Cl}$.

In this compound, the bond between ammonium (NH4+) and chloride (Cl-) ions is predominantly ionic, as ammonium donates a proton to chloride. However, within the ammonium ion itself, there are covalent bonds holding together the nitrogen and hydrogen atoms. So, it exhibits both types of bonding.

Question72

Some pollutant gases in the atmosphere act as contributors to the formation of acid rain, a serious environmental problem. An example of such a gas is

(A) $\mathrm{N}_2$

(B) $\mathrm{O}_2$

(D) $\mathrm{NO}_2$

(E) $\mathrm{CH}_4$

▶️Answer/Explanation

Ans:D

Acid rain is formed when sulfur dioxide ($\mathrm{SO}_2$) and nitrogen oxides ($\mathrm{NO}_x$) react with water vapor in the atmosphere to produce sulfuric acid ($\mathrm{H}_2 \mathrm{SO}_4$) and nitric acid ($\mathrm{HNO}_3$), respectively. Among the given options, sulfur dioxide is represented by $\mathrm{SO}_2$ (D), so the correct answer is (D) $\mathrm{NO}_2$, which is a nitrogen oxide and contributes to the formation of acid rain.

Question73

$
\begin{aligned}
& 2 \mathrm{~S}(s)+2 \mathrm{O}_2(g) \rightleftarrows 2 \mathrm{SO}_2(g) \quad K_1=2 \times 10^{105} \\
& 2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftarrows 2 \mathrm{SO}_3(g) \quad K_2=7 \times 10^{24}
\end{aligned}
$

Given the value of the equilibrium constants $K_1$ and $K_2$ for the reactions represented above, what is the value of the equilibrium constant, $K_3$, for the following reaction?
$
2 \mathrm{~S}(s)+3 \mathrm{O}_2(g) \rightleftarrows 2 \mathrm{SO}_3(g) \quad K_3=?
$
(A) $1 \times 10^{130}$

(B) $3 \times 10^{80}$

(D) $2 \times 10^{40}$

(E) $7 \times 10^{24}$

▶️Answer/Explanation

Ans:A

To find the equilibrium constant $K_3$ for the given reaction, we can use the equilibrium constants $K_1$ and $K_2$ for the two individual reactions.

The overall reaction $2 \mathrm{~S}(s) + 3 \mathrm{O}_2(g) \rightleftarrows 2 \mathrm{SO}_3(g)$ is the combination of the two given reactions. We can obtain this overall reaction by multiplying the two reactions and canceling out the intermediates:

\[
\begin{aligned}
2 \mathrm{~S}(s) + 2 \mathrm{O}_2(g) &\rightleftarrows 2 \mathrm{SO}_2(g) \quad \text{(1st reaction)} \\
2 \mathrm{SO}_2(g) + \mathrm{O}_2(g) &\rightleftarrows 2 \mathrm{SO}_3(g) \quad \text{(2nd reaction)}
\end{aligned}
\]

By multiplying these reactions, we get:

\[
2 \mathrm{~S}(s) + 3 \mathrm{O}_2(g) \rightleftarrows 2 \mathrm{SO}_3(g)
\]

To find $K_3$, we multiply the equilibrium constants for the individual reactions:

\[
K_3 = K_1 \times K_2 = (2 \times 10^{105}) \times (7 \times 10^{24})
\]

Now, multiply the coefficients of $K_1$ and $K_2$ to find the value of $K_3$:

\[
K_3 = 2 \times 7 \times 10^{105 + 24} = 14 \times 10^{129}
\]

This value can be expressed as $1.4 \times 10^{130}$.

So, the closest option to $K_3$ is (A) $1 \times 10^{130}$.

Question74

Which of the following molecules is nonpolar but has polar covalent bonds?

(A) $\mathrm{N}_2$

(B) $\mathrm{H}_2 \mathrm{O}_2$

(D) $\mathrm{CCl}_4$

(E) $\mathrm{CH}_2 \mathrm{Cl}_2$

▶️Answer/Explanation

Ans:D

A molecule can be nonpolar overall if its polar bonds are arranged symmetrically such that the polarities cancel out. Let’s analyze each option:

(A) $\mathrm{N}_2$ – Nitrogen gas ($\mathrm{N}_2$) consists of two nitrogen atoms, which have the same electronegativity. The bond between them is nonpolar because the two nitrogen atoms share the electrons equally, and the molecule is symmetrical. Therefore, $\mathrm{N}_2$ is nonpolar with nonpolar covalent bonds.

(B) $\mathrm{H}_2 \mathrm{O}_2$ – Hydrogen peroxide ($\mathrm{H}_2 \mathrm{O}_2$) has polar covalent bonds due to the electronegativity difference between hydrogen and oxygen. However, the molecule is not symmetrical, so it is polar overall.

(C) $\mathrm{H}_2 \mathrm{O}$ – Water ($\mathrm{H}_2 \mathrm{O}$) has polar covalent bonds due to the electronegativity difference between hydrogen and oxygen. It is also a bent molecule, which makes it polar overall.

(D) $\mathrm{CCl}_4$ – Carbon tetrachloride ($\mathrm{CCl}_4$) has polar covalent bonds between carbon and chlorine, but the molecule is symmetrical, with the four chlorine atoms arranged tetrahedrally around the central carbon atom. This arrangement cancels out the polarity of the individual bonds, resulting in a nonpolar molecule.

(E) $\mathrm{CH}_2 \mathrm{Cl}_2$ – Dichloromethane ($\mathrm{CH}_2 \mathrm{Cl}_2$) has polar covalent bonds between carbon and chlorine, but the molecule is not symmetrical. The chlorine atoms are on one side of the molecule, resulting in an overall dipole moment and making it polar overall.

Therefore, the correct answer is (D) $\mathrm{CCl}_4$, which is nonpolar but has polar covalent bonds.

Question75

A $0.10 \mathrm{M}$ solution of which of the following salts is most basic?

(A) $\mathrm{LiNO}_3$

(B) $\mathrm{Na}_2 \mathrm{SO}_4$

(D) $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$

(E) $\mathrm{K}_2 \mathrm{CO}_3$

▶️Answer/Explanation

Ans:E

The basicity of a salt can be determined by analyzing the ions it dissociates into in solution. The ions that contribute to basicity are those that can accept protons ($H^+$) from water, resulting in the formation of hydroxide ions ($OH^-$). Let’s analyze each option:

(A) $\mathrm{LiNO}_3$ – Lithium nitrate dissociates into lithium ions ($\mathrm{Li}^+$) and nitrate ions ($\mathrm{NO}_3^-$). Neither of these ions can accept protons from water to form hydroxide ions. Therefore, the solution is not basic.

(B) $\mathrm{Na}_2 \mathrm{SO}_4$ – Sodium sulfate dissociates into sodium ions ($\mathrm{Na}^+$) and sulfate ions ($\mathrm{SO}_4^{2-}$). None of these ions can accept protons from water to form hydroxide ions. Therefore, the solution is not basic.

(C) $\mathrm{CaCl}_2$ – Calcium chloride dissociates into calcium ions ($\mathrm{Ca}^{2+}$) and chloride ions ($\mathrm{Cl}^-$). Chloride ions do not react with water to form hydroxide ions. Therefore, the solution is not basic.

(D) $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ – Aluminum nitrate dissociates into aluminum ions ($\mathrm{Al}^{3+}$) and nitrate ions ($\mathrm{NO}_3^-$). None of these ions can accept protons from water to form hydroxide ions. Therefore, the solution is not basic.

(E) $\mathrm{K}_2 \mathrm{CO}_3$ – Potassium carbonate dissociates into potassium ions ($\mathrm{K}^+$) and carbonate ions ($\mathrm{CO}_3^{2-}$). Carbonate ions can accept protons from water to form bicarbonate ions ($\mathrm{HCO}_3^-$), and then further accept another proton to form hydroxide ions. Therefore, the solution is basic.

Therefore, the correct answer is (E) $\mathrm{K}_2 \mathrm{CO}_3$, which is the most basic among the given salts.