IB DP Chemistry Structure 1.1: Introduction to the particulate nature of matter – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover Structure 1.1, Introduction to the particulate nature of matter, in IB Diploma Programme Chemistry, aligned to the IB’s Diploma Programme Chemistry guide (published February 2023, for first assessment in 2025). They test understandings Structure 1.1.1 to Structure 1.1.3, all of which are for SL and HL; the guide has no additional higher level content for this sub-topic. They suit examinations set under that guide.

For mixed questions across the whole component, try the Structure 1 practice set. Course links are in Next steps below.

Use the data given in each question. Give kelvin values using T/K = T/°C + 273.15.

Questions

1. Write a balanced equation, including state symbols, for each change.

(a) Aqueous silver nitrate reacts with aqueous sodium bromide to form a precipitate of silver bromide and a solution of sodium nitrate. [2] (b) Solid sulfur, S₈, melts. [1]

2. Name the change of state in each situation. [4]

(a) Naphthalene mothballs in a wardrobe slowly get smaller without ever becoming liquid. (b) Droplets of dew form on grass on a cool morning. (c) Molten candle wax sets on a table. (d) Purple iodine vapour forms shiny crystals on the cold base of a flask.

3. Use these data. Convert each temperature as instructed.

(a) Copper melts at 1085 °C. Give this in kelvin. [1] (b) Helium boils at 4.2 K. Give this in °C. [1] (c) Normal human body temperature, 37.0 °C, in kelvin. [1]

4. Classify each as an element, a compound, a homogeneous mixture or a heterogeneous mixture. [4]

(a) stainless steel (b) carbon monoxide (c) graphite (d) smoke, made of fine solid particles spread through air

5. Use the kinetic molecular theory to explain why a liquid keeps its volume but takes the shape of its container, whereas a gas spreads to fill any container. [3]

6. A green dye extracted from a plant is analysed by paper chromatography. The solvent front moves 8.4 cm from the base line. The dye separates into three spots, 1.26 cm, 3.78 cm and 6.05 cm from the base line. Reference substances run in the same solvent on the same paper give these Rf values: A 0.15, B 0.33, C 0.45, D 0.72, E 0.88.

(a) Determine Rf for every spot. [2] (b) Deduce which reference substances the dye contains. [1] (c) Explain why the base line is drawn in pencil. [1] (d) Explain why the reference values must come from the same solvent. [1]

7. A pure substance W is heated at a steady rate. Its temperature starts at −40 °C, rises to −7 °C and stays there from 2 to 4 minutes, rises to 59 °C and stays there from 9 to 17 minutes, then rises again.

(a) State the melting point and boiling point of W in kelvin. [2] (b) State which states of W are present at 3 minutes. [1] (c) Explain why the temperature stays constant from 9 to 17 minutes. [2] (d) Suggest why the second constant-temperature stage lasts longer than the first. [1]

8. A sample of an organic solid Y contains insoluble charcoal and a trace of a water-soluble impurity. Use these data: the solubility of Y in water is 50 g per 100 g of water at 90 °C and 2.0 g per 100 g of water at 20 °C. A student has 10.0 g of Y.

(a) Calculate the minimum mass of water at 90 °C needed to dissolve all 10.0 g of Y. [1] (b) The solution is cooled to 20 °C. Calculate the maximum mass of Y that can crystallize. [2] (c) Describe how the student should carry out the recrystallization to obtain pure, dry Y. [4] (d) Explain how a melting point measurement could show whether the product is pure. [2]

9. (a) A sealed sample of argon is heated from −73 °C to 127 °C. Determine the factor by which the average kinetic energy of the atoms changes. [2] (b) Calculate the temperature, in °C, at which the average kinetic energy of a gas sample is three times its value at 25 °C. [2]

10. Cyclohexane and methylbenzene are colourless liquids that mix completely. Use these data: cyclohexane boils at 81 °C and methylbenzene at 111 °C.

(a) State whether a cyclohexane–methylbenzene mixture is homogeneous or heterogeneous, and explain why. [2] (b) Name the method used to separate the two liquids and outline how it works. [3] (c) Convert both boiling points to kelvin. [1] (d) State and explain the thermometer reading while pure cyclohexane is being collected. [2]

11. (a) Explain, in terms of composition and separation, why air is classified as a mixture while carbon dioxide is a compound. [3] (b) A pure liquid and the same liquid containing a dissolved impurity are each cooled until solid. Describe and explain how their cooling curves differ. [3]

Answers

1. (a) AgNO₃(aq) + NaBr(aq) → AgBr(s) + NaNO₃(aq): correct balanced formulas [1]; all four state symbols correct, with (s) for the precipitate [1]. (b) S₈(s) → S₈(l) [1]. Examiner insight: a physical change still needs state symbols; S₈(s) → S₈ with no symbols does not show that melting has happened.

2. (a) sublimation [1]; (b) condensation [1]; (c) freezing [1]; (d) deposition [1]. Examiner insight: “evaporation” for the mothballs is wrong, because no liquid forms; read whether the change passes through a liquid before choosing a name.

3. (a) 1085 + 273.15 = 1358.15 K (1358 K) [1]. (b) 4.2 − 273.15 = −268.95 °C (−269.0 °C) [1]. (c) 37.0 + 273.15 = 310.15 K (310.2 K) [1]. Examiner insight: write kelvin as K with no degree sign, and check that a kelvin answer is never negative.

4. (a) homogeneous mixture, an alloy [1]; (b) compound [1]; (c) element, carbon only [1]; (d) heterogeneous mixture, solid particles in a gas [1]. Examiner insight: alloys are a common trap: their metallic bonding does not make them compounds, because the proportions of the metals can vary.

5. Liquid particles are close together, touching, so the volume cannot change significantly [1]. They can move past one another, so the liquid flows to take the shape of its container [1]. Gas particles are far apart with negligible attractions and move randomly in all directions, so they spread to fill the whole container, giving neither fixed shape nor fixed volume [1]. Examiner insight: answers about particles need particle words (close, far apart, move past one another, attractions); describing only what you see, such as “liquids pour”, gains little.

6. (a) Rf = spot distance ÷ solvent-front distance [1]; 1.26 ÷ 8.4 = 0.15, 3.78 ÷ 8.4 = 0.45, 6.05 ÷ 8.4 = 0.72 [1]. (b) A, C and D [1]. (c) Pencil (graphite) does not dissolve in the solvent, whereas ink would separate and mix with the spots [1]. (d) Rf depends on the solvent (and paper), so values are only comparable under identical conditions [1]. Examiner insight: measure to the solvent front, not to the top of the paper; an Rf greater than 1 shows a wrong denominator.

7. (a) Melting point −7 + 273.15 = 266.15 K [1]; boiling point 59 + 273.15 = 332.15 K [1]. (b) Solid and liquid [1]. (c) The heat taken in breaks the attractions between the particles as liquid turns to gas [1], so the average kinetic energy of the particles, and therefore the temperature, does not increase [1]. (d) Vaporization separates the particles almost completely, which needs more energy than melting [1]. Examiner insight: in part (c), “the energy is used to change state” is too vague; link it to attractions between particles and to average kinetic energy.

8. (a) 10.0 ÷ 50 × 100 = 20.0 g of water [1]. (b) At 20 °C, 20.0 g of water holds 2.0 × 20.0 ÷ 100 = 0.40 g of Y [1]; maximum mass crystallizing = 10.0 − 0.40 = 9.60 g [1]. (c) Dissolve the solid in the minimum volume of hot (90 °C) water [1]. Filter while hot to remove the charcoal [1]. Cool the filtrate so Y crystallizes while the trace impurity stays dissolved [1]. Collect the crystals by filtration, rinse them with a small volume of cold water, then dry them [1]. (d) Measure the melting point and compare it with the known value for pure Y [1]; a pure sample melts sharply at that value, while an impure one melts lower and over a range [1]. Examiner insight: the word “minimum” carries the reasoning in recrystallization; an answer that just says “dissolve in hot water” misses why so much Y crystallizes in part (b).

9. (a) Convert: −73 °C = 200.15 K and 127 °C = 400.15 K [1]; 400.15 ÷ 200.15 = 2.00, so the average kinetic energy doubles [1]. (b) 25 °C = 298.15 K, and 3 × 298.15 = 894.45 K [1]; 894.45 − 273.15 = 621.3 °C (621 °C) [1]. Examiner insight: tripling the Celsius value (to 75 °C) is the classic wrong answer; proportionality holds only on the kelvin scale.

10. (a) Homogeneous [1]: the liquids are miscible, so the mixture is a single uniform phase [1]. (b) Fractional distillation [1]. The vapour rises up a fractionating column, repeatedly condensing and re-vaporizing [1]. Cyclohexane, with the lower boiling point, reaches the top and distils over first [1]. (c) 354.15 K and 384.15 K [1]. (d) About 81 °C (354 K), staying constant [1], because cyclohexane is pure and a pure substance boils at a single fixed temperature [1]. Examiner insight: “simple distillation” is not enough for two liquids whose boiling points are this close; name the column and explain that the lower-boiling liquid distils first.

11. (a) Air contains several substances (nitrogen, oxygen, argon and others) in a variable ratio, not chemically bonded together [1], so they can be separated physically, for example by fractional distillation of liquid air [1]. Carbon dioxide contains carbon and oxygen atoms chemically bonded in a fixed 1:2 ratio and can only be broken down by a chemical reaction [1]. (b) The pure liquid gives a flat plateau: it freezes at one fixed temperature [1]. The impure liquid starts to freeze at a lower temperature and the temperature keeps falling while it freezes, so there is no sharp plateau [1]. This is because a mixture does not have a fixed composition, so it has no single freezing point [1]. Examiner insight: in part (a), give a reason for each substance; a definition of a mixture alone leaves the carbon dioxide half unanswered.

Where marks are usually lost

• Leaving state symbols off equations for physical changes, or using (aq) for a pure liquid.
• Calling a gas-to-solid change “sublimation” instead of deposition.
• Classifying alloys, or molecules of one element such as O₃, as compounds.
• Leaving out “minimum” and “hot filtration” from a recrystallization method.
• Measuring Rf to the end of the paper rather than to the solvent front.
• Saying temperature rises during a change of state, or not linking the plateau to attractions between particles.
• Working out kinetic energy ratios from Celsius values.
• Writing °K, or reporting a negative kelvin value.

Next steps

• Go back over weak areas in the Structure 1.1 revision notes.
• Reread full explanations in the Structure 1.1 study guide.
• See every resource for the course on the IB DP Chemistry hub.
• Tick off understandings on the IB DP Chemistry checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme Chemistry guide, first assessment 2025 (published February 2023): Structure 1.1, Introduction to the particulate nature of matter.