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 for every question: H 1.008, C 12.01, N 14.01, O 16.00, Na 22.99, Cl 35.45, K 39.10 g mol⁻¹; N_A = 6.022 × 10²³ mol⁻¹. Assume every solute dissolves completely, and that volumes add when dilute aqueous solutions are mixed.
Question 1 (multiple choice · foundation)
Which of the following is a heterogeneous mixture?
- (A) Air in a sealed flask
- (B) A brass key
- (C) Sugar that has fully dissolved in a glass of water
- (D) River water in which fine sand settles to the bottom of the jar
Answer and explanation
Answer: (D). Sand settles, so a sample from the bottom of the jar contains much more sand than a sample from the top. Its properties depend on location, which defines a heterogeneous mixture.
- (A) Air is a gaseous solution: the gases are mixed at the scale of single molecules and the composition is the same throughout the flask.
- (B) Brass is a solid solution of zinc in copper. Being solid does not stop it being homogeneous.
- (C) Fully dissolved sugar is spread evenly through the water, so every sip has the same composition.
Question 2 (multiple choice · foundation)
What is the concentration of chloride ions in a 0.15 M solution of aluminium chloride, AlCl₃?
- (A) 0.050 M
- (B) 0.15 M
- (C) 0.30 M
- (D) 0.45 M
Answer and explanation
Answer: (D). Each AlCl₃ formula unit releases one Al³⁺ and three Cl⁻ ions, so [Cl⁻] = 3 × 0.15 M = 0.45 M.
- (A) divides by 3 instead of multiplying.
- (B) is the concentration of AlCl₃ itself, and also of Al³⁺, but not of Cl⁻.
- (C) uses two Cl⁻ per formula unit, as if the compound were AlCl₂.
Question 3 (multiple choice · core)
What is the total amount of ions, in moles, in 25.0 mL of 0.400 M sodium sulfate, Na₂SO₄?
- (A) 0.0100 mol
- (B) 0.0200 mol
- (C) 0.0300 mol
- (D) 30.0 mol
Answer and explanation
Answer: (C). n(Na₂SO₄) = 0.400 mol L⁻¹ × 0.0250 L = 0.0100 mol. Each formula unit gives 2 Na⁺ and 1 SO₄²⁻, which is 3 ions, so the total is 3 × 0.0100 = 0.0300 mol.
- (A) is the amount of Na₂SO₄ formula units, not of ions.
- (B) counts only the Na⁺ ions.
- (D) uses 25.0 instead of 0.0250 L. The volume was not converted to litres.
Question 4 (calculation · core)
A student dissolves 4.50 g of glucose, C₆H₁₂O₆, in water and makes the solution up to exactly 200.0 mL in a volumetric flask.
(a) Calculate the molarity of the solution. (b) The student then pipettes a 10.0 mL portion into a test tube. How many glucose molecules does the portion contain?
Worked solution
(a)
- M(C₆H₁₂O₆) = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g mol⁻¹.
- n = 4.50 g ÷ 180.16 g mol⁻¹ = 0.024978 mol.
- M = 0.024978 mol ÷ 0.2000 L = 0.125 M.
(b)
- The portion has the same molarity, 0.1249 M (unrounded).
- n = 0.1249 mol L⁻¹ × 0.0100 L = 1.249 × 10⁻³ mol.
- N = 1.249 × 10⁻³ mol × 6.022 × 10²³ mol⁻¹ = 7.52 × 10²⁰ molecules.
Suggested mark points (3): 1 for the molar mass and moles of glucose; 1 for 0.125 M with the volume in litres; 1 for 7.52 × 10²⁰ molecules (accept 7.53 × 10²⁰ from the rounded molarity). An equally good route for (b): the portion is 10.0 ÷ 200.0 = 1/20 of the solution, so it holds 1/20 of the 0.024978 mol.
Common error: thinking the 10.0 mL portion is less concentrated than the flask. It holds fewer molecules, but the same number per litre.
Question 5 (calculation · core)
A 0.840 M stock solution of potassium chloride is diluted in two steps. First, 25.0 mL of stock is diluted to 500.0 mL (solution P). Then 10.00 mL of solution P is diluted to 100.0 mL (solution Q).
(a) Calculate the concentration of solution P. (b) Calculate [Cl⁻] in solution Q. (c) By what overall factor has the stock been diluted?
Worked solution
(a) M₁V₁ = M₂V₂, so M(P) = 0.840 M × 25.0 mL ÷ 500.0 mL = 0.0420 M.
(b) M(Q) = 0.0420 M × 10.00 mL ÷ 100.0 mL = 0.00420 M KCl. Each KCl gives one Cl⁻, so [Cl⁻] = 0.00420 M (4.20 × 10⁻³ M).
(c) 0.840 ÷ 0.00420 = 200. The steps multiply: a factor of 20 (25.0 → 500.0 mL) then a factor of 10 (10.00 → 100.0 mL).
Suggested mark points (3): 1 for each part. Accept the mole route (n = 0.840 × 0.0250 = 0.0210 mol, then ÷ 0.5000 L) for (a). For (c), accept 200 from either division or multiplying the two factors.
Common error: diluting 25.0 mL by adding 500.0 mL of water. “Diluted to 500.0 mL” means the final volume is 500.0 mL.
Question 6 (constructed response · core)
A student needs 250.0 mL of 0.100 M potassium nitrate, KNO₃. The student writes this plan: “Weigh the correct mass of KNO₃, tip it into a beaker containing 250.0 mL of water, and stir until it dissolves.”
(a) Calculate the mass of KNO₃ needed. (b) Explain why the student’s method would not give a concentration of exactly 0.100 M. (c) Describe a better method, naming the key piece of glassware. (d) The student pours 50.0 mL of the correctly made solution into a beaker. State the molarity of the solution in the beaker and calculate the mass of KNO₃ it contains.
Worked solution and suggested Marlbridge rubric
(a) n = 0.100 mol L⁻¹ × 0.2500 L = 0.0250 mol. M(KNO₃) = 39.10 + 14.01 + 3(16.00) = 101.11 g mol⁻¹. m = 0.0250 × 101.11 = 2.53 g.
(b) Molarity is moles per litre of solution, not per litre of water. Adding a solid to 250.0 mL of water changes the volume, so the final volume is not 250.0 mL. For example, if it ended at 251.2 mL the molarity would be 0.0250 ÷ 0.2512 = 0.0995 M, not 0.100 M. (A beaker is also not precise enough to measure 250.0 mL.)
(c) Dissolve the 2.53 g in a smaller volume of water in a beaker. Transfer the solution to a 250.0 mL volumetric flask, rinsing the beaker into the flask so that no solute is lost. Add water until the bottom of the meniscus sits on the mark, then stopper and invert to mix.
(d) The beaker holds 0.100 M, the same as the flask, because the solution is homogeneous. n = 0.100 × 0.0500 = 0.00500 mol, so m = 0.00500 × 101.11 = 0.506 g of KNO₃.
| Point | What earns it |
|---|---|
| 1 | 2.53 g with working (moles and molar mass shown) |
| 1 | Explains that molarity uses the volume of solution and the solid changes the volume |
| 1 | Volumetric flask, filled to the mark after the solid has dissolved |
| 1 | Molarity stays 0.100 M, with the reason (homogeneous solution) |
| 1 | 0.506 g in the 50.0 mL portion |
Do not award the point in (b) for “the student did not measure accurately” without the link to the volume of solution.
Question 7 (calculation and explanation · stretch)
40.0 mL of 0.250 M potassium sulfate, K₂SO₄, is mixed with 60.0 mL of 0.100 M potassium chloride, KCl. No reaction occurs.
(a) Calculate [K⁺], [SO₄²⁻] and [Cl⁻] in the mixture. (b) A classmate says: “[K⁺] is the average of 0.500 M and 0.100 M, so it is 0.300 M.” Explain the error. (c) Explain, at the particle level, why a 5 mL sample taken from any part of the well-stirred mixture has the same [K⁺].
Worked solution and suggested Marlbridge rubric
(a)
- K₂SO₄: n = 0.250 × 0.0400 = 0.0100 mol, giving 0.0200 mol K⁺ and 0.0100 mol SO₄²⁻.
- KCl: n = 0.100 × 0.0600 = 0.00600 mol, giving 0.00600 mol K⁺ and 0.00600 mol Cl⁻.
- Total volume = 0.0400 + 0.0600 = 0.1000 L.
- [K⁺] = (0.0200 + 0.00600) ÷ 0.1000 = 0.260 M; [SO₄²⁻] = 0.0100 ÷ 0.1000 = 0.100 M; [Cl⁻] = 0.00600 ÷ 0.1000 = 0.0600 M.
Charge check: positive 0.260 M; negative 2(0.100) + 0.0600 = 0.260 M. Neutral.
(b) A simple average treats the two solutions as equal volumes. Here 60.0 mL of the weaker solution is mixed with only 40.0 mL of the stronger one, so the result is pulled below 0.300 M. The correct method adds moles of K⁺ and divides by the total volume.
(c) The mixture is a solution. After stirring, the K⁺, SO₄²⁻ and Cl⁻ ions are spread evenly among the water molecules, one ion at a time, and they stay spread out because they do not settle. Every 5 mL therefore holds the same number of K⁺ ions, so the concentration is the same everywhere.
| Point | What earns it |
|---|---|
| 1 | Uses 2 K⁺ per K₂SO₄ (0.0200 mol K⁺ from that solution) |
| 1 | Adds moles and divides by 0.1000 L to get [K⁺] = 0.260 M |
| 1 | Correct [SO₄²⁻] and [Cl⁻] |
| 1 | Identifies unequal volumes as the flaw in averaging |
| 1 | Particle-level reason: ions evenly distributed in a homogeneous mixture, so equal volumes hold equal numbers of K⁺ |
Common error: forgetting the factor of 2 gives [K⁺] = 0.160 M.
How did you do?
- Q1 or Q7(c) wrong: re-read “Homogeneous and heterogeneous mixtures” and Figure 1 in the study guide.
- Q2, Q3 or Q7(a) wrong: practise counting ions from the formula (“Ionic solutes” in the guide).
- Q4 or Q5 wrong: check your mL → L conversions and use moles as the middle step (Figure 2).
- Q6 incomplete: your answer needs the why: molarity uses litres of solution.
Then tick off the topic checklist.