These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data that you need are given in each question. The data in Question 6 are invented for practice.
Question 1 (multiple choice · foundation)
Which pair of liquids is miscible?
- (A) Water and octane, C₈H₁₈
- (B) Water and tetrachloromethane, CCl₄
- (C) Water and propanone (acetone), CH₃COCH₃
- (D) Octane and a solution of sodium chloride in water
Answer and explanation
Answer: (C). Propanone is polar: its C=O oxygen carries a partial negative charge, so water molecules can form hydrogen bonds to it, and dipole–dipole attractions also act. These propanone–water attractions are comparable to the ones lost, so the liquids mix in all proportions.
- (A) Octane is nonpolar. Mixing it with water would break water’s hydrogen bonds and give only weak dispersion attractions in return.
- (B) CCl₄ has polar C–Cl bonds, but its tetrahedral shape makes the molecule nonpolar overall, so it behaves like (A).
- (D) The aqueous layer is still mostly water, and the dissolved ions are attracted even less to octane. The liquids form two layers.
Question 2 (multiple choice · core)
Potassium chloride, KCl, dissolves readily in water but not in hexane. Which statement best explains this?
- (A) KCl reacts with water to form potassium hydroxide and hydrochloric acid.
- (B) Water molecules form strong ion–dipole attractions with K⁺ and Cl⁻ that can replace the ion–ion attractions; hexane can offer only weak dispersion attractions.
- (C) Water molecules are smaller than hexane molecules, so they fit between the ions in the crystal more easily.
- (D) Hexane molecules repel the ions because hexane is nonpolar.
Answer and explanation
Answer: (B). The ions in solid KCl attract each other strongly. They separate only if strong new attractions form, and the ion–dipole attractions to polar water molecules provide them. Nonpolar hexane has no permanent dipole to attract an ion.
- (A) Dissolving KCl gives hydrated K⁺ and Cl⁻ ions; no new substance forms. Evaporating the water gives back KCl.
- (C) Size is not the reason. Without strong attractions to the ions, small solvent molecules would not separate them either.
- (D) Hexane does not repel the ions. It simply cannot attract them strongly enough to replace the ion–ion attractions.
Question 3 (multiple choice · core)
Calcium chloride, CaCl₂, is dissolved in water. Which statement describes how water molecules are arranged around the ions?
- (A) Oxygen atoms of water point toward both Ca²⁺ and Cl⁻ ions.
- (B) Oxygen atoms point toward Ca²⁺ ions; hydrogen atoms point toward Cl⁻ ions.
- (C) Hydrogen atoms point toward Ca²⁺ ions; oxygen atoms point toward Cl⁻ ions.
- (D) The water molecules are oriented randomly, because the ions are no longer in a crystal.
Answer and explanation
Answer: (B). Opposite charges attract. The oxygen end of water is δ−, so it faces the positive Ca²⁺ ion. The hydrogen ends are δ+, so they face the negative Cl⁻ ion.
- (A) Oxygen facing Cl⁻ would put a δ− end next to a negative ion, which is repulsive.
- (C) is exactly the wrong way round for both ions.
- (D) The ion–dipole attractions are strong enough to hold the nearest water molecules in a definite orientation, even though they keep exchanging with others.
Question 4 (multiple choice · core)
Which substance is most soluble in hexane, C₆H₁₄?
- (A) Methanol, CH₃OH
- (B) Sodium chloride, NaCl
- (C) Glucose, C₆H₁₂O₆
- (D) 1-Octanol, CH₃(CH₂)₇OH
Answer and explanation
Answer: (D). 1-Octanol has one O–H group but a long eight-carbon chain. Most of the molecule interacts through dispersion forces, the same kind that hexane molecules use, so it dissolves well in hexane.
- (A) Methanol is a small molecule dominated by its O–H group. It hydrogen bonds strongly with itself, and hexane cannot replace those hydrogen bonds.
- (B) NaCl is ionic. Hexane offers nothing that can replace the strong ion–ion attractions.
- (C) Glucose has five O–H groups and forms many hydrogen bonds. It is very soluble in water and hardly soluble in hexane.
Question 5 (constructed response · core)
A student pours equal volumes of hexane (density 0.66 g/mL) and water (density 1.00 g/mL) into a separating funnel. She adds a small amount of naphthalene, C₁₀H₈, a nonpolar solid, and a small amount of sodium nitrate, NaNO₃. She shakes the funnel and lets the layers settle.
(a) State which liquid forms the top layer. (b) Predict which layer each solute ends up in. Justify each prediction in terms of intermolecular interactions. (c) Describe a particulate drawing of the layer that contains the sodium nitrate. State what your drawing must show.
Worked solution and suggested Marlbridge rubric
(a) Hexane is on top, because it is less dense than water (0.66 g/mL against 1.00 g/mL). The two liquids form separate layers because they are immiscible.
(b) Naphthalene → hexane layer. Naphthalene is nonpolar, so its molecules attract each other by dispersion forces. Hexane molecules attract it by the same kind of force. In water it would disrupt water’s hydrogen bonds and gain only weak dispersion attractions.
Sodium nitrate → water layer. NaNO₃ is ionic. Polar water molecules form strong ion–dipole attractions with Na⁺ and NO₃⁻ ions, which can replace the ion–ion attractions in the solid. Hexane cannot attract ions strongly.
(c) The drawing should show the water layer with:
- separate Na⁺ and NO₃⁻ ions, in equal numbers (1 : 1, from the formula);
- water molecules around each Na⁺ with their oxygen ends pointing toward it;
- water molecules around each NO₃⁻ with a hydrogen end pointing toward it;
- no naphthalene molecules (or at most a very few).
| Point | What earns it |
|---|---|
| 1 | Hexane on top, linked to its lower density |
| 1 | Naphthalene in hexane, with dispersion forces in both named |
| 1 | NaNO₃ in water, with ion–dipole attractions named |
| 1 | Drawing: separate ions in a 1 : 1 ratio |
| 1 | Drawing: correct orientation of water around both ions |
Do not award the naphthalene or NaNO₃ point for “like dissolves like” alone.
Question 6 (constructed response · core)
A chemist studies four compounds, W, X, Y and Z. Each has the same single polar group, which can hydrogen bond with water. Each compound has one more CH₂ group in its hydrocarbon chain than the one before it (W has the shortest chain). The data below are invented.
| Compound | Solubility in water (g/L) | Solubility in hexane (g/L) |
|---|---|---|
| W | 85 | 3 |
| X | 26 | 40 |
| Y | 7.5 | 160 |
| Z | 1.9 | 400 |
(a) Describe and explain the trend in solubility in water. (b) Explain why the trend in hexane is the opposite. (c) Compound V has one more CH₂ group than Z. Estimate its solubility in water. Show your reasoning.
Worked solution and suggested Marlbridge rubric
(a) Solubility in water falls as the chain gets longer: 85 → 26 → 7.5 → 1.9 g/L. The polar group forms hydrogen bonds with water, but it is the same in every compound. The hydrocarbon chain interacts with water only through weak dispersion attractions and disrupts water–water hydrogen bonds. As the chain lengthens, the nonpolar part becomes a larger share of the molecule, so the molecule is less like water.
(b) Hexane is nonpolar and interacts through dispersion forces. A longer chain gives larger dispersion attractions between the compound and hexane, and makes the single polar group a smaller part of the molecule. So the compound becomes more like hexane, and its solubility in hexane rises.
(c) Each extra CH₂ divides the solubility in water by about 3.3 to 3.9 (85 ÷ 26 = 3.3; 26 ÷ 7.5 = 3.5; 7.5 ÷ 1.9 = 3.9). Dividing 1.9 g/L by about 3.5 gives roughly 0.5 g/L.
| Point | What earns it |
|---|---|
| 1 | Water trend described, with the polar group’s hydrogen bonding named |
| 1 | Explanation: growing nonpolar chain disrupts water’s hydrogen bonds with only weak attractions in return |
| 1 | Hexane trend explained by increasing dispersion attractions as the chain grows |
| 1 | Estimate between 0.3 and 0.7 g/L, using the factor between neighbours |
Accept any estimate in range that is clearly based on the pattern in the data.
Question 7 (argumentation · stretch)
Hexane, C₆H₁₄, and 1-hexanol, C₆H₁₃OH, both have a six-carbon chain. Near room temperature, about 0.0095 g of hexane dissolves per litre of water, while about 5.9 g of 1-hexanol dissolves per litre of water.
(a) A student claims: “Hexane does not dissolve in water because water molecules repel hexane molecules.” Evaluate this claim. (b) Explain why 1-hexanol is several hundred times more soluble in water than hexane. (c) Explain the degree to which the like-dissolves-like model accounts for these two solubilities.
Model answer and suggested Marlbridge rubric
(a) The claim is incorrect. Water and hexane do attract each other, through weak London dispersion forces. The reason hexane stays out is that water molecules attract each other strongly through hydrogen bonds. Placing a hexane molecule among them breaks some of those hydrogen bonds and replaces them only with weak dispersion attractions, so very little hexane dissolves.
(b) 1-Hexanol has an O–H group. It can form hydrogen bonds with water, which partly replace the water–water hydrogen bonds lost when it dissolves. Hexane has no group that can do this. The ratio 5.9 ÷ 0.0095 is about 600, which is consistent with the large difference one O–H group makes. The long nonpolar chain is still the reason 1-hexanol is only slightly soluble rather than miscible.
(c) The model predicts the direction correctly: 1-hexanol is partly “like” water and hexane is not, so 1-hexanol should be more soluble. The model does not predict the size of either solubility or of the difference. It is qualitative, and 1-hexanol, with a polar head and a long nonpolar tail, fits neither “like” nor “unlike” cleanly.
| Point | What earns it |
|---|---|
| 1 | Rejects the claim: there are weak attractions between water and hexane, not repulsion |
| 1 | Identifies loss of water–water hydrogen bonds as the real reason |
| 1 | Explains 1-hexanol’s higher solubility by hydrogen bonding of its O–H group with water |
| 1 | States that the model predicts the direction of the difference |
| 1 | States that the model cannot predict the amounts, or that 1-hexanol has mixed character |
How did you do?
- Q1, Q2 or Q4 wrong: re-read “Dissolving at the particle level” and Worked example 1 in the study guide.
- Q3 or Q5(c) wrong: look again at Figure 1 and check which end of the water molecule faces each ion.
- Q6 wrong: work through Worked example 2 on chain length.
- Q7 incomplete: your answer needs the strength of water–water hydrogen bonds, and a clear statement of what the model can and cannot predict.
Then tick off the topic checklist.