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. The diagrams are described in words; where a question asks you to draw, a description of what you would draw with the same features earns the same credit. Water is polar: O is δ−, each H is δ+. Assume ionic solutes dissolve completely.
Question 1 (multiple choice · foundation)
Which description matches a correct particle diagram of magnesium chloride solution, MgCl₂(aq)?
- (A) Several MgCl₂ units, each a Mg atom joined to two Cl atoms, spread among water molecules
- (B) Separate Mg²⁺ and Cl⁻ ions, twice as many Cl⁻ as Mg²⁺, each ion surrounded by water molecules
- (C) Separate Mg²⁺ and Cl⁻ ions in equal numbers, each surrounded by water molecules
- (D) Mg atoms and Cl₂ molecules spread among water molecules
Answer and explanation
Answer: (B). MgCl₂ is ionic, so its ions separate when it dissolves. The formula gives 2 Cl⁻ for every Mg²⁺, which also balances the charges: +2 and 2 × (−1).
- (A) shows the compound as intact units. Dissolved ionic compounds do not stay as pairs or groups.
- (C) has the wrong ratio, so the charges in the box would not balance.
- (D) shows neutral metal atoms and chlorine molecules: dissolving does not turn ions back into elements.
Question 2 (multiple choice · foundation)
In a solution of potassium bromide, how are the water molecules nearest to a K⁺ ion arranged?
- (A) Each points a hydrogen atom towards the K⁺ ion.
- (B) Each points its oxygen atom towards the K⁺ ion.
- (C) They point in random directions, because K⁺ does not attract water.
- (D) They are joined to the K⁺ ion by covalent bonds.
Answer and explanation
Answer: (B). K⁺ is positive and attracts the partially negative oxygen end of each water molecule (an ion–dipole attraction).
- (A) is the arrangement around an anion such as Br⁻. Next to K⁺, the δ+ hydrogens would be repelled.
- (C) ignores the ion–dipole attraction. K⁺ does attract water; this attraction is what holds water molecules in an ordered shell around the ion.
- (D) The attraction is an intermolecular (ion–dipole) force, not a shared electron pair.
Question 3 (multiple choice · core)
Diagram P represents 0.10 L of a solution and shows 8 solute particles. Diagram Q represents 0.20 L of a solution of the same solute and shows 12 solute particles. Which statement is correct?
- (A) Q is more concentrated, because it contains more solute particles.
- (B) P is more concentrated, because it has more solute particles per unit volume.
- (C) P and Q have the same concentration, because both contain the same solute.
- (D) The concentrations cannot be compared without the molar mass of the solute.
Answer and explanation
Answer: (B). P has 8 ÷ 0.10 L = 80 particles per litre; Q has 12 ÷ 0.20 L = 60 particles per litre. P is more concentrated.
- (A) compares particle counts without dividing by volume.
- (C) The identity of the solute does not fix its concentration.
- (D) Concentration in a diagram is particles per volume; molar mass is not needed to compare.
Question 4 (calculation · core)
A particle diagram represents 0.200 L of calcium nitrate solution, Ca(NO₃)₂(aq). It shows 4 Ca²⁺ ions and 8 NO₃⁻ ions. Each ion drawn represents 0.00500 mol of that ion.
(a) Show that the diagram is consistent with the formula. (b) Calculate [Ca²⁺] and [NO₃⁻]. (c) Water is added until the solution represented has a volume of 0.400 L. State how many of each ion the new diagram should show, and give the new concentrations.
Worked solution
(a) Ca(NO₃)₂ gives 1 Ca²⁺ : 2 NO₃⁻. The diagram has 4 : 8 = 1 : 2. Charge: 4(+2) + 8(−1) = 0. Consistent.
(b) n(Ca²⁺) = 4 × 0.00500 = 0.0200 mol, so [Ca²⁺] = 0.0200 ÷ 0.200 = 0.100 M. n(NO₃⁻) = 8 × 0.00500 = 0.0400 mol, so [NO₃⁻] = 0.0400 ÷ 0.200 = 0.200 M.
(c) Dilution does not add or remove ions: the diagram should still show 4 Ca²⁺ and 8 NO₃⁻, now spread through a box representing 0.400 L. [Ca²⁺] = 0.0200 ÷ 0.400 = 0.0500 M; [NO₃⁻] = 0.0400 ÷ 0.400 = 0.100 M.
Suggested mark points (4): 1 for the ratio or charge argument in (a); 1 for both concentrations in (b); 1 for the unchanged particle counts in (c); 1 for both halved concentrations. Accept “each concentration halves because the volume doubles” for the last point if the values are given.
Common error: drawing 2 Ca²⁺ and 4 NO₃⁻ after dilution. That would represent removing solute, not adding water.
Question 5 (constructed response · core)
Methanol, CH₃OH, mixes with water in all proportions.
(a) Describe a particle diagram of a dilute solution of methanol in water, showing the interactions between methanol and water. (b) A student draws methanol in water as CH₃⁺ and OH⁻ ions. Explain why this is wrong. (c) Explain how your diagram for (a) would differ from one for sodium chloride solution.
Model answer and suggested Marlbridge rubric
(a) A few whole CH₃OH molecules spread evenly among many water molecules. Dashed lines show hydrogen bonds: from the H on methanol’s O–H to the O of a nearby water molecule, and from an H of a water molecule to methanol’s O atom. The CH₃ group does not form hydrogen bonds.
(b) Methanol is a molecular compound held together by covalent bonds. It dissolves as intact molecules; its C–O bond does not break, so no CH₃⁺ or OH⁻ ions form.
(c) For NaCl(aq) the solute is drawn as separate Na⁺ and Cl⁻ ions in a 1 : 1 ratio. The interactions are ion–dipole attractions, so water’s O faces each Na⁺ and an H faces each Cl⁻. In methanol solution there are no ions; the interactions are hydrogen bonds between neutral molecules.
| Point | What earns it |
|---|---|
| 1 | Methanol shown as whole molecules spread among water molecules |
| 1 | Hydrogen bond correctly placed between the O–H of one molecule and an O atom of the other |
| 1 | (b) Methanol is molecular / covalently bonded, so it does not form ions on dissolving |
| 1 | (c) NaCl drawn as separate ions with ion–dipole orientation, contrasted with hydrogen bonding between molecules |
Do not award the second point for a hydrogen bond drawn to the H atoms of the CH₃ group.
Question 6 (constructed response · core)
A student is asked to draw a particle diagram of sodium sulfate solution, Na₂SO₄(aq). The student’s diagram shows:
- 3 Na⁺ ions;
- 3 separate S atoms and 12 separate O atoms;
- water molecules around each Na⁺ with their H atoms pointing towards the Na⁺.
(a) Identify three errors in the diagram. (b) Describe a corrected diagram that uses 3 sulfate units.
Model answer and suggested Marlbridge rubric
(a)
- Sulfate broken apart. SO₄²⁻ is a polyatomic ion held together by covalent bonds; it stays as one unit with a 2− charge.
- Wrong ratio / charges do not balance. Na₂SO₄ needs 2 Na⁺ per SO₄²⁻. With 3 sulfate units there must be 6 Na⁺. As drawn, 3 Na⁺ against 3 sulfate units would leave a total charge of 3 − 6 = −3.
- Water reversed around Na⁺. The δ− oxygen of each water molecule should face the positive Na⁺ ion.
(b) 6 Na⁺ and 3 SO₄²⁻ ions, all separate. Each SO₄²⁻ is drawn as one unit (an S joined to four O, with an overall 2− label). Water around each Na⁺ has O facing inwards; water around each SO₄²⁻ points an H towards the ion. Charge: 6(+1) + 3(−2) = 0.
| Point | What earns it |
|---|---|
| 1 | Sulfate must stay as a whole SO₄²⁻ ion |
| 1 | Ratio must be 2 Na⁺ : 1 SO₄²⁻ (or charge does not balance) |
| 1 | Water’s O, not H, should face Na⁺ |
| 1 | Corrected diagram: 6 Na⁺ and 3 SO₄²⁻ with correct orientation around both ions |
Question 7 (calculation and explanation · stretch)
Diagram A represents 0.10 L of calcium chloride solution and shows 3 Ca²⁺ and 6 Cl⁻ ions. Diagram B represents 0.10 L of sodium chloride solution and shows 4 Na⁺ and 4 Cl⁻ ions. Each ion drawn represents 0.010 mol. The two solutions are mixed, and no reaction occurs.
(a) State the number of each ion in a diagram of the whole mixture, and the volume it represents. (b) Calculate [Ca²⁺], [Na⁺] and [Cl⁻] in the mixture. (c) Compare [Cl⁻] in the mixture with [Cl⁻] in A and in B, and explain the result.
Worked solution and suggested Marlbridge rubric
(a) 3 Ca²⁺, 4 Na⁺ and 6 + 4 = 10 Cl⁻ ions, in a box representing 0.10 + 0.10 = 0.20 L.
(b)
- [Ca²⁺] = (3 × 0.010) ÷ 0.20 = 0.15 M
- [Na⁺] = (4 × 0.010) ÷ 0.20 = 0.20 M
- [Cl⁻] = (10 × 0.010) ÷ 0.20 = 0.50 M
Charge check: 2(0.15) + 0.20 = 0.50 M positive; 0.50 M negative.
(c) Before mixing, [Cl⁻] was 0.60 M in A (6 × 0.010 ÷ 0.10) and 0.40 M in B (4 × 0.010 ÷ 0.10). The mixture’s 0.50 M lies between them. Because equal volumes were mixed, the result is the simple average here; the chloride ions from both solutions are now spread through twice the volume. Each of Ca²⁺ and Na⁺, which came from only one solution, has its concentration halved.
| Point | What earns it |
|---|---|
| 1 | Correct particle counts and 0.20 L |
| 1 | All three concentrations correct |
| 1 | Correct [Cl⁻] for A and B before mixing |
| 1 | Explains that Cl⁻ from both solutions is spread through the combined volume (and why the value lies between) |
Common error: keeping the volume at 0.10 L gives [Cl⁻] = 1.0 M, which is higher than either starting solution and so cannot be right.
How did you do?
- Q1, Q5 or Q6 wrong: re-read “Drawing ionic solutes” and “Drawing molecular solutes” in the study guide.
- Q2 wrong: look again at Figure 1 and ask “do these ends attract?”
- Q3, Q4 or Q7 wrong: practise “Reading concentration from a diagram”: always divide by the volume represented.
Then tick off the topic checklist.