AP Chemistry Quiz: Representations Of Solutions

What this quiz covers

This quiz focuses on Representations Of Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Ion–dipole attractions between C$_2$H$_5$OH$^+$ ions and water because ethanol ionizes in water.
2. Hydrogen bonding between the O–H group of ethanol and water molecules, with ethanol molecules remaining intact. (correct answer)
3. Covalent bond formation between ethanol and water to produce hydrates that are new compounds in solution.
4. Electrostatic attraction between C$_2$H$_5^+$ and OH$^-$ produced by complete dissociation of ethanol in water.
5. Only London dispersion forces between ethanol and water because both molecules are nonpolar overall.

Explanation: This question tests the identification of dominant intermolecular forces in solutions of polar molecular solutes. Ethanol, C₂H₅OH, is a polar molecule that forms hydrogen bonds with water through its O-H group, where the hydrogen of ethanol can bond with oxygen in water and vice versa, while the molecules remain intact without dissociation. This interaction allows ethanol to mix thoroughly with water, as both can participate in hydrogen bonding networks. No ionization occurs because ethanol is not an acid or base in this context. A tempting distractor is choice A, which posits ionization into C₂H₅OH⁺, stemming from the misconception that all solutes with OH groups ionize like acids. To analyze solute-solvent interactions, classify the solute as molecular and identify the strongest possible intermolecular force with the solvent, such as hydrogen bonding for molecules with O-H groups.

Question 2

1. 1 Al$^{3+}$ and 3 NO$_3^-$ ions dispersed in water, each surrounded by oriented water molecules. (correct answer)
2. 1 Al(NO$_3$)$_3$(aq) unit dispersed in water, with water molecules randomly oriented around it.
3. 3 Al$^+$ and 1 NO$_3^{3-}$ ion dispersed in water, each surrounded by oriented water molecules.
4. 1 Al$^{3-}$ and 3 NO$_3^+$ ions dispersed in water, each surrounded by oriented water molecules.
5. 1 Al$^{3+}$ and 1 NO$_3^-$ ion dispersed in water because the formula reduces to AlNO$_3$ in solution.

Explanation: This question tests the ability to determine ion counts and representations from the formula of a complex ionic compound. Aluminum nitrate, Al(NO₃)₃, dissociates completely in water into one Al³⁺ ion and three NO₃⁻ ions per formula unit, as nitrate is a polyatomic ion that remains intact. Each ion is surrounded by water molecules oriented according to charge: oxygen toward Al³⁺ and hydrogen toward NO₃⁻, due to ion-dipole attractions. This assumes no further reactions like hydrolysis, as specified. A tempting distractor is choice B, which shows intact units, based on the misconception that polyatomic ions prevent full dissociation. When representing ionic solutions, dissociate the compound into its ions based on the formula coefficients, and include solvent orientation for a complete particulate view.

Question 3

1. Sucrose molecules remain intact and become uniformly dispersed as separate neutral molecules among water molecules. (correct answer)
2. Sucrose dissociates completely into C$^{+}$, H$^{+}$, and O$^{2-}$ ions that are hydrated by water.
3. Sucrose forms C$_{12}$H$_{22}$O$_{11}^{+}$ and C$_{12}$H$_{22}$O$_{11}^{-}$ ions that are hydrated by water.
4. Sucrose reacts with water to produce H$_3$O$^+$ and OH$^-$ ions in equal amounts, while the remaining solute stays as a solid.
5. Sucrose forms an ionic lattice in water, and water molecules are excluded from the region near the solute particles.

Explanation: This question tests the ability to represent particulate-level structures of molecular solutions, distinguishing between ionic and covalent solutes. Sucrose, C₁₂H₂₂O₁₁, is a molecular compound that does not dissociate into ions in water but remains as intact neutral molecules due to its covalent bonding and lack of ionization. The dissolution occurs through hydrogen bonding between the hydroxyl groups of sucrose and water molecules, allowing the sucrose to disperse uniformly without breaking into charged particles. This is why sucrose solutions are non-electrolytes and do not conduct electricity. A tempting distractor is choice B, which suggests dissociation into elemental ions, arising from the misconception that all solutes ionize like ionic compounds. When evaluating solution representations, identify if the solute is molecular or ionic, remembering that molecular solutes typically remain intact unless they are acids or bases that react with water.

Question 4

1. K$_2$SO$_4$ remains as intact neutral formula units dispersed in water, with no ion–dipole interactions.
2. K$_2$SO$_4$ reacts with water to form KOH(aq) and H$_2$SO$_4$(aq), which remain as molecules in solution.
3. K$_2$SO$_4$ dissociates into 1 K$^{2+}$ and 1 SO$_4^{2-}$, and water molecules surround both ions with random orientation.
4. K$_2$SO$_4$ dissociates into 2 K$^-$ and 1 SO$_4^{2+}$, and water orients H toward the cations and O toward the anions.
5. K$_2$SO$_4$ dissociates into 2 K$^+$ and 1 SO$_4^{2-}$, and water orients O toward K$^+$ and H toward SO$_4^{2-}$. (correct answer)

Explanation: This question tests the representation of polyatomic ionic solutions at the particulate level, including ion counts and solvent orientation. Potassium sulfate, K₂SO₄, dissociates completely in water into two K⁺ ions and one SO₄²⁻ ion per formula unit, as it is a soluble ionic salt. Water molecules orient with their oxygen atoms toward the positively charged K⁺ ions and hydrogen atoms toward the negatively charged SO₄²⁻ ions, facilitating hydration and stability. This ion-dipole interaction is key to understanding solubility in polar solvents. A tempting distractor is choice C, which incorrectly shows 1 K²⁺ and 1 SO₄²⁻ with random orientation, stemming from the misconception that formulas do not indicate ion ratios and that orientation is unimportant. To solve such problems, break down the ionic formula into its constituent ions and apply the principle that water dipoles align oppositely to ion charges.

Question 5

1. Dye molecules dissociate into ions that are pulled to the bottom of the beaker by gravity.
2. Dye molecules move randomly and disperse due to molecular motion, forming a uniform mixture over time. (correct answer)
3. Water molecules chemically react with dye to form a new soluble ionic compound.
4. Dye molecules remain clustered because intermolecular forces prevent any mixing with water.
5. Dye molecules spread because water molecules convert them into H$_3$O$^+$ and OH$^-$ ions.

Explanation: This question assesses the understanding of diffusion in solutions at the particulate level. The correct answer is B, as the polar dye molecules disperse throughout the water due to random molecular motion, leading to a uniform solution over time. This process is diffusion driven by kinetic energy, without needing stirring. The polarity allows interaction with water. A tempting distractor is D, claiming molecules remain clustered; this is incorrect due to the misconception that intermolecular forces prevent mixing, ignoring entropy-driven dispersion. To explain mixing, consider random particle motion and compatibility of solute-solvent interactions.

Question 6

1. PO$_4^{3-}$ remains intact as a polyatomic ion and is hydrated, while Na$^+$ ions are also hydrated. (correct answer)
2. PO$_4^{3-}$ breaks into P$^{3+}$ and four O$^{2-}$ ions that are each hydrated separately.
3. PO$_4^{3-}$ becomes neutral PO$_4$(aq) molecules, while Na$^+$ becomes NaOH(aq) molecules.
4. PO$_4^{3-}$ stays attached to Na$^+$ as Na$_3$PO$_4$ ion clusters that do not separate in water.
5. PO$_4^{3-}$ cannot exist in water, so it converts completely to P(s) and O$_2$(g).

Explanation: This question evaluates the stability of polyatomic ions in aqueous solutions. The correct answer is A, as PO43- remains an intact polyatomic ion hydrated along with Na+ ions, ignoring acid-base reactions. Phosphate does not break into monatomic ions. This maintains the solution's composition. A tempting distractor is B, suggesting decomposition; this is incorrect due to the misconception that polyatomic ions are unstable in water. Treat polyatomic ions as single units in particulate representations unless specified otherwise.

Question 7

1. Ba$^+$ and OH$^{2-}$ ions are present, with water molecules randomly oriented around them.
2. Ba$^{2+}$ and OH$^-$ ions are present, with water molecules oriented around the ions due to ion–dipole attractions. (correct answer)
3. Ba$^{2+}$ and OH$^-$ ions are present, but OH$^-$ cannot exist in water so it immediately becomes O$^{2-}$.
4. Ba(OH)$_2$ dissolves by forming covalent bonds with water to produce Ba–O–H chains.
5. Ba(OH)$_2$ remains as neutral Ba(OH)$_2$ molecules dispersed among water molecules.

Explanation: This question evaluates the particulate representation of strong base solutions. The correct answer is B, as Ba(OH)2 fully dissociates into Ba2+ and two OH- ions, with water orienting via ion-dipole attractions around each. This produces a basic, conductive solution. The hydroxide ions remain stable. A tempting distractor is A, suggesting neutral molecules; this is incorrect due to the misconception that strong bases do not ionize. Represent strong electrolytes as completely dissociated ions with proper solvent interactions.

Question 8

1. CO$_2$ dissociates completely into C$^{4+}$ and O$^{2-}$ ions that become hydrated by water.
2. CO$_2$ remains as neutral molecules dispersed in water, interacting primarily through London dispersion forces and some dipole-induced dipole. (correct answer)
3. CO$_2$ forms CO$_2^-$ and CO$_2^+$ ions that are stabilized by ion–dipole attractions in water.
4. CO$_2$ forms strong hydrogen bonds to water because CO$_2$ has O–H bonds.
5. CO$_2$ forms a lattice of alternating ions that breaks into ion pairs in solution.

Explanation: This question tests the representation of nonpolar molecular gases in water, ignoring reactions. The correct answer is B, as CO2 dissolves as neutral molecules interacting via London dispersion and dipole-induced dipole forces with water. Solubility is limited due to nonpolarity. Most CO2 remains unreacted as specified. A tempting distractor is A, claiming dissociation into ions; this is incorrect due to the misconception that gases ionize like salts. For nonpolar solutes, emphasize weak intermolecular forces in representations.

Question 9

1. CaCl$_2$ remains as neutral CaCl$_2$(aq) formula units dispersed uniformly among water molecules.
2. Ca$^{2+}$ and Cl$^-$ ions form, but each Ca$^{2+}$ stays permanently bonded to exactly one Cl$^-$ in solution.
3. Ca$^{2+}$ and 2 Cl$^-$ ions are present, each hydrated by water with appropriate orientation of water's partial charges. (correct answer)
4. Ca$^{2+}$ and Cl$^-$ ions form, but the number of Cl$^-$ ions equals the number of Ca$^{2+}$ ions.
5. CaCl$_2$ dissolves to produce Ca$^+$ and Cl$_2^-$ ions that are stabilized by water dipoles.

Explanation: This question assesses the representation of ionic solutions at the particulate level, including ion stoichiometry and hydration. The correct answer is C, as CaCl2 dissociates into one Ca2+ ion and two Cl- ions per formula unit, with each ion hydrated by water molecules oriented according to their charges—oxygen toward Ca2+ and hydrogen toward Cl-. This dissociation and hydration process allows the solid to dissolve fully, forming separated, mobile ions. The 1:2 ratio of cations to anions is crucial for accurate representation. A tempting distractor is D, which states equal numbers of Ca2+ and Cl- ions; this is incorrect due to the misconception of ignoring the subscript stoichiometry in the formula. Always verify ion ratios from the compound's formula and apply ion-dipole orientation principles for solution representations.

Question 10

1. Both solutions contain only neutral solute molecules dispersed among water molecules.
2. Both solutions contain separated ions, because any solute in water dissociates into ions.
3. Glucose remains as neutral molecules, while MgCl$_2$ forms Mg$^{2+}$ and Cl$^-$ ions that are hydrated by water. (correct answer)
4. Glucose forms C$^{6+}$ and O$^{2-}$ ions, while MgCl$_2$ remains as neutral MgCl$_2$ units.
5. Glucose and MgCl$_2$ both form ion pairs that stay together as touching units throughout the solution.

Explanation: This question assesses comparative particulate representations of molecular and ionic solutions. The correct answer is C, as glucose is a molecular solute that remains as neutral C6H12O6 molecules dispersed in water, while MgCl2 dissociates into hydrated Mg2+ and Cl- ions. The ionic dissociation in MgCl2 leads to conductivity, unlike the non-ionic glucose solution. Both solutions are at the same temperature and volume, highlighting the difference in solute behavior. A tempting distractor is B, suggesting both dissociate into ions; this is incorrect due to the misconception that all solutes ionize in water, failing to distinguish molecular from ionic compounds. To compare solutions, classify solutes as ionic or molecular and represent their dissociation accordingly.

Question 11

1. Ethanol dissociates into $\text{CH}_3\text{CH}_2\text{O}^-$ and $\text{H}^+$ ions that are strongly hydrated by water.
2. Ethanol molecules disperse; the $\text{O–H}$ group can hydrogen-bond with water while the hydrocarbon portion interacts mostly by dispersion. (correct answer)
3. Ethanol remains as a separate liquid layer because polar molecules cannot mix with water molecules.
4. Ethanol forms an ionic lattice in water, producing $\text{CH}_3\text{CH}_2\text{OH}^+$ and $\text{OH}^-$ ions.
5. Ethanol molecules disperse; water molecules orient with H atoms toward the ethanol hydrocarbon end due to negative charge there.

Explanation: This question tests the skill of representing polar molecular compound dissolution and multiple types of intermolecular forces. Ethanol (CH₃CH₂OH) is a molecular compound that remains intact when dissolved in water, dispersing as whole molecules rather than ionizing. The ethanol molecule has two distinct regions: a polar -OH group that can form hydrogen bonds with water (both as donor and acceptor), and a nonpolar hydrocarbon portion (CH₃CH₂-) that interacts with water primarily through weaker London dispersion forces. This dual nature makes ethanol miscible with water while maintaining its molecular structure. Choice A incorrectly suggests that ethanol would ionize by losing its hydrogen as H⁺, but the O-H bond in alcohols is not acidic enough to ionize appreciably in water. When representing organic molecules in water, consider both the polar functional groups (which hydrogen bond) and nonpolar regions (which interact through dispersion forces).

Question 12

1. Both solutes produce hydrated ions because any solute separates into cations and anions when mixed with water.
2. $\text{CO}_2$ dissolves primarily as intact molecules, whereas $\text{NaBr}$ dissolves as separated $\text{Na}^+$ and $\text{Br}^-$ ions. (correct answer)
3. $\text{CO}_2$ dissolves primarily as $\text{C}^{4+}$ and $\text{O}^{2-}$ ions, whereas $\text{NaBr}$ dissolves as intact molecules.
4. $\text{CO}_2$ dissolves as $\text{CO}_2^{2-}$ ions, whereas $\text{NaBr}$ dissolves as $\text{Na}^{2+}$ and $\text{Br}^-$ ions.
5. Both solutes remain as undissolved solids because water cannot separate particles in ionic or molecular substances.

Explanation: This question tests the skill of distinguishing between molecular and ionic compound dissolution at the particulate level. Carbon dioxide (CO₂) is a molecular compound that dissolves in water primarily as intact CO₂ molecules, with only a tiny fraction reacting to form carbonic acid; the CO₂ molecules are held in solution by weak dipole-induced dipole forces and some hydrogen bonding with the small amount of H₂CO₃ formed. In contrast, sodium bromide (NaBr) is an ionic compound that completely dissociates into Na⁺ and Br⁻ ions when dissolved, with these ions stabilized by strong ion-dipole interactions with water molecules. Choice C incorrectly suggests that CO₂ would break into atomic ions (C⁴⁺ and O²⁻), which would require breaking the strong covalent bonds within the molecule—something that doesn't occur during simple dissolution. To distinguish dissolution types, remember that ionic compounds separate into their constituent ions, while molecular compounds typically remain as intact molecules.

Question 13

1. Intact $\text{CaCl}_2$ units remain clustered, with water molecules randomly oriented around the clusters.
2. $\text{Ca}^{2+}$ and $\text{Cl}^-$ ions are separated; water's O atoms face $\text{Ca}^{2+}$ and water's H atoms face $\text{Cl}^-$. (correct answer)
3. $\text{Ca}^{2+}$ and $\text{Cl}^-$ ions are separated; water's H atoms face $\text{Ca}^{2+}$ and water's O atoms face $\text{Cl}^-$.
4. $\text{Ca}^{2+}$ ions form, but chloride remains as neutral $\text{Cl}_2(aq)$ molecules dispersed in water.
5. $\text{CaCl}_2$ dissolves as neutral molecules, each surrounded by water with no preferred orientation.

Explanation: This question tests the skill of representing ionic compound dissolution and ion-dipole interactions at the particulate level. When calcium chloride (CaCl₂) dissolves in water, it completely dissociates into Ca²⁺ cations and Cl⁻ anions according to the equation CaCl₂(s) → Ca²⁺(aq) + 2Cl⁻(aq). Water molecules, being polar with a partial negative charge on oxygen and partial positive charges on hydrogen, orient specifically around these ions: the negative oxygen atoms of water molecules face toward the positive Ca²⁺ ions, while the positive hydrogen atoms face toward the negative Cl⁻ ions. This arrangement maximizes the attractive ion-dipole interactions and stabilizes the dissolved ions in solution. Choice C incorrectly reverses the water orientation, suggesting that hydrogen atoms would face the positive calcium ions, which would create repulsion rather than attraction. To correctly represent dissolved ionic compounds, remember that water always orients with opposite charges facing each other: O toward cations, H toward anions.

Question 14

1. NaCl remains as intact neutral formula units dispersed in water, with water molecules randomly oriented around each unit.
2. NaCl dissociates into Na+ and Cl− ions that are separated, and water molecules orient with O atoms toward Na+ and H atoms toward Cl−. (correct answer)
3. NaCl dissociates into Na− and Cl+ ions that are separated, and water molecules orient with H atoms toward Na− and O atoms toward Cl+.
4. NaCl dissolves as Na+ and Cl− ions, but the ions remain paired together as NaCl(aq) with no consistent water orientation.
5. NaCl dissolves as individual Na and Cl atoms, and water molecules form covalent bonds to each atom to stabilize them.

Explanation: This question tests understanding of ionic compound dissolution and ion-dipole interactions in aqueous solutions. When NaCl dissolves in water, it dissociates completely into Na+ cations and Cl− anions, which become separated and surrounded by water molecules. The polar water molecules orient themselves specifically around each ion: the partially negative oxygen atoms of water point toward the positive Na+ ions, while the partially positive hydrogen atoms point toward the negative Cl− ions. This orientation occurs because of ion-dipole attractions between the charged ions and the polar water molecules. Choice A incorrectly suggests NaCl remains as neutral formula units, which is a common misconception that ionic compounds don't dissociate in water. To solve problems about ionic dissolution, remember that ionic compounds separate into individual ions in water, and water molecules orient based on charge attractions: O toward cations, H toward anions.

Question 15

1. NaCl remains as neutral NaCl(aq) units dispersed, with water molecules randomly oriented around each unit.
2. Na$^+$ and Cl$^-$ ions are separated, with water's O atoms oriented toward Na$^+$ and water's H atoms oriented toward Cl$^-$. (correct answer)
3. Na$^+$ and Cl$^-$ ions are separated, with water's H atoms oriented toward Na$^+$ and water's O atoms oriented toward Cl$^-$.
4. NaCl forms Na$^+$ and Cl$^-$, but the ions remain paired as touching ion pairs throughout the solution.
5. NaCl dissolves by forming covalent bonds with water to make H–Cl and Na–OH molecules.

Explanation: This question tests the ability to represent solutions at the particulate level, focusing on the dissociation and hydration of ionic solutes in water. The correct answer is B, as NaCl dissociates into separate Na+ and Cl- ions when dissolved, with water molecules orienting their partially negative oxygen atoms toward the positive Na+ ions and partially positive hydrogen atoms toward the negative Cl- ions. This orientation is due to ion-dipole attractions, which stabilize the ions and allow the solid to dissolve into a clear solution. The stirring ensures uniform distribution of these hydrated ions throughout the water. A tempting distractor is A, which suggests NaCl remains as neutral units; this is incorrect due to the misconception that ionic compounds do not dissociate in polar solvents like water. When analyzing solution representations, identify whether the solute is ionic or molecular and apply principles of ion-dipole interactions for hydration.

Question 16

1. Solution X has more total ions because Na$^+$ is smaller and dissociates more completely than Ca$^{2+}$.
2. Solution Y has more total ions because each Ca(NO$_3$)$_2$ unit produces three ions, while each NaNO$_3$ unit produces two ions. (correct answer)
3. Solution X and Solution Y have the same total ions because both contain nitrate ions and have the same volume.
4. Solution X has more total ions because each NaNO$_3$ unit produces three ions, while each Ca(NO$_3$)$_2$ unit produces two ions.
5. Solution Y has fewer total ions because Ca(NO$_3$)$_2$ stays mostly undissociated in water due to its higher molar mass.

Explanation: This question tests the comparison of total ion concentrations in solutions of different ionic compounds with varying dissociation products. Both solutions have equal moles of solute and the same volume, but Ca(NO₃)₂ dissociates into three ions (one Ca²⁺ and two NO₃⁻) per formula unit, while NaNO₃ dissociates into two ions (one Na⁺ and one NO₃⁻) per unit. Therefore, Solution Y has 1.5 times more total ions than Solution X, assuming complete dissociation for both soluble salts. This affects properties like colligative effects or conductivity. A tempting distractor is choice D, which reverses the ion counts, arising from the misconception of miscounting the ions from each formula. For comparing ion numbers, calculate the van't Hoff factor (number of ions per formula unit) and multiply by the number of moles, keeping volume constant.

Question 17

1. Both solutions contain only ions (H$^+$ and the conjugate base) because both acids fully ionize in water.
2. HCl(aq) contains mostly H$_3$O$^+$ and Cl$^-$, whereas CH$_3$COOH(aq) contains mostly CH$_3$COOH molecules with some H$_3$O$^+$ and CH$_3$COO$^-$. (correct answer)
3. Both solutions contain the same ratio of ions to molecules because they have the same molarity.
4. Both solutions contain mostly undissociated acid molecules because acids remain molecular in water.
5. HCl(aq) contains mostly HCl molecules, whereas CH$_3$COOH(aq) contains mostly ions because acetate is a strong base.

Explanation: This question tests the distinction between strong and weak acids in particulate representations of solutions. HCl is a strong acid that fully dissociates into H₃O⁺ and Cl⁻ ions in water, resulting in mostly ions with negligible undissociated molecules. In contrast, CH₃COOH is a weak acid that only partially ionizes, so the solution contains mostly intact CH₃COOH molecules along with small amounts of H₃O⁺ and CH₃COO⁻ ions. This difference arises from the equilibrium constants, where strong acids have complete ionization and weak acids do not. A tempting distractor is choice A, which claims both fully ionize, based on the misconception that all acids behave identically regardless of strength. When comparing acid solutions, recall that strength determines the extent of ionization, and represent weak acids with predominant molecular species.

Question 18

1. Only NH$_3$ molecules because molecular solutes do not form ions in water.
2. Mostly NH$_3$ molecules, with some NH$_4^+$ and OH$^-$ ions formed by reaction with water. (correct answer)
3. Mostly NH$_4^+$ and O$^{2-}$ ions because OH$^-$ is not stable in water.
4. Mostly N$^{3-}$ and H$^+$ ions because NH$_3$ dissociates into its elements in water.
5. Only NH$_4^+$ and OH$^-$ ions because NH$_3$ is a strong base and fully reacts with water.

Explanation: This question tests the particulate representation of weak base solutions, including partial reaction with solvent. Ammonia, NH₃, is a weak base that partially reacts with water to form NH₄⁺ and OH⁻ ions, but most remains as undissociated NH₃ molecules due to its small base dissociation constant. The equilibrium favors the molecular form, with only a small fraction ionizing. This is why ammonia solutions are weakly basic and conduct electricity poorly. A tempting distractor is choice A, which claims full ionization, stemming from the misconception that all bases are strong like NaOH. To represent weak electrolytes, depict mostly intact molecules with minor ionized products, and recall that strength determines the extent of dissociation in water.

Question 19

1. NaCl remains as intact neutral NaCl units dispersed in water, and water molecules are randomly oriented around each unit.
2. NaCl dissociates into Na$^+$ and Cl$^-$ ions, and water molecules orient with O atoms toward Na$^+$ and H atoms toward Cl$^-$. (correct answer)
3. NaCl dissociates into Na$^+$ and Cl$^-$ ions, and water molecules orient with H atoms toward Na$^+$ and O atoms toward Cl$^-$.
4. NaCl reacts with water to form HCl(aq) and NaOH(aq), which then remain as neutral molecules dispersed in water.
5. NaCl forms Na$^{2+}$ and Cl$^{2-}$ ions in water, and water molecules orient with O atoms toward the anions and H atoms toward the cations.

Explanation: This question tests the understanding of particulate-level representations of ionic solutions, focusing on dissociation and solvent-solute interactions. Sodium chloride, NaCl, is an ionic compound that fully dissociates in water into Na⁺ cations and Cl⁻ anions due to the polar nature of water overcoming the lattice energy. Water molecules, being polar, orient themselves with the oxygen atom (partial negative charge) facing the Na⁺ ions and the hydrogen atoms (partial positive charge) facing the Cl⁻ ions, forming ion-dipole attractions that stabilize the ions in solution. This hydration shell around each ion is crucial for the solubility of ionic compounds in polar solvents like water. A tempting distractor is choice C, which reverses the water molecule orientation, stemming from the misconception that hydrogen is more electronegative than oxygen in water. To approach similar problems, always recall that ionic compounds dissociate completely in water, and polar solvents orient their negative end toward cations and positive end toward anions.

Question 20

1. Mostly MgCl$_2$(aq) units with a small amount of Mg$^+$ and Cl$^-$ ions from partial dissociation.
2. Only Mg$^{2+}$ ions are present because chloride ions are converted to HCl(aq) in water.
3. Equal numbers of Mg$^{2+}$ ions and Cl$^-$ ions because the compound is electrically neutral overall.
4. Twice as many Cl$^-$ ions as Mg$^{2+}$ ions because each formula unit produces 2 chloride ions. (correct answer)
5. Twice as many Mg$^{2+}$ ions as Cl$^-$ ions because Mg has a $2+$ charge.

Explanation: This question tests the comprehension of ion ratios in dissociated ionic solutions based on chemical formulas. Magnesium chloride, MgCl₂, is a soluble ionic compound that completely dissociates in water into one Mg²⁺ ion and two Cl⁻ ions per formula unit, reflecting the 1:2 ratio in its formula. This results in twice as many chloride ions as magnesium ions in the solution to maintain electrical neutrality overall. Ignoring water autoionization, no other particles are present, and the relative amounts are directly proportional to the stoichiometry of dissociation. A tempting distractor is choice B, which assumes equal numbers of Mg²⁺ and Cl⁻ ions, based on the misconception that neutrality requires a 1:1 ion ratio regardless of charges. For similar questions, use the chemical formula to determine the number and type of ions produced upon dissociation, ensuring the total charge balances to zero.