Properties of Substances and Mixtures: Unit Diagnostic (Chemistry Unit 3)

What this diagnostic is for

Use this page to find which Unit 3 topics to revisit. There is one question per topic, and two for Topic 3.1, which the rest of the unit builds on. Answer every question before opening any answer.

These are original Marlbridge practice questions, not past exam questions. They are not calibrated and give no predicted score: a wrong answer points to a topic, not a grade.

Calculator and data. A scientific calculator is allowed. R = 0.08206 L atm mol⁻¹ K⁻¹; K = °C + 273.15; h = 6.626 × 10⁻³⁴ J s; c = 2.998 × 10⁸ m s⁻¹; N_A = 6.022 × 10²³ mol⁻¹. Molar masses: H 1.008, He 4.003, C 12.01 g mol⁻¹. Treat gases as ideal.

Question 1 (multiple choice · 3.1)

Between the molecules of which substance do only London dispersion forces act?

(A) Dichloromethane, CH₂Cl₂
(B) Carbon dioxide, CO₂
(C) Hydrogen sulfide, H₂S
(D) Methylamine, CH₃NH₂

Answer and explanation

Answer: (B). CO₂ is linear, so its two C=O bond dipoles cancel: nonpolar, dispersion only.

(A) In CH₂Cl₂ the C–Cl and C–H dipoles do not cancel.
(C) H₂S is bent, so polar.
(D) H on N: hydrogen bonding.

If you missed this: read “Forces that need a permanent dipole” in the Topic 3.1 study guide.

Question 2 (multiple choice · 3.1)

Which hydrogen halide has the highest boiling point?

(A) HI
(B) HBr
(C) HCl
(D) HF

Answer and explanation

Answer: (D). Only HF molecules hydrogen bond (H on F). Boiling points: HF about 20 °C, HCl −85 °C, HBr −67 °C, HI −35 °C.

(A) is the “most electrons” trap. Dispersion explains HCl < HBr < HI, but hydrogen bonding in HF outweighs it.
(B) and (C) have fewer electrons than HI and no hydrogen bonding.

If you missed this: read “Hydrogen bonding” in the Topic 3.1 study guide.

Question 3 (multiple choice · 3.2)

Why does a crystal of an ionic solid shatter when struck with a hammer?

(A) Only intermolecular forces hold neighbouring layers together.
(B) Ionic bonds are weak, so a small force breaks them.
(C) A shift of one layer lines up ions of the same charge, and they repel.
(D) The blow releases mobile electrons, which push the layers apart.

Answer and explanation

Answer: (C). Each ion is surrounded by opposite charges. A small shift brings like charges face to face, and they repel.

(A) An ionic solid has no molecules; it is held by ion–ion attractions.
(B) Ionic attractions are strong, which is why melting points are high.
(D) Ionic solids have no mobile electrons, which is why they do not conduct.

If you missed this: read “Ionic solids” in the Topic 3.2 study guide.

Question 4 (multiple choice · 3.3)

A gas can be squeezed into a much smaller volume, a liquid hardly at all. Why?

(A) Gas particles are far apart with empty space between them; liquid particles already touch.
(B) Gas particles are smaller than liquid particles of the same substance.
(C) Liquid particles are joined by covalent bonds; gas particles are not.
(D) Gas particles move more slowly, so they are easier to push together.

Answer and explanation

Answer: (A). Compressing a gas removes empty space; a liquid has almost none.

(B) The molecules are the same in both phases.
(C) Liquid molecules are held by intermolecular forces, not bonds between molecules.
(D) Speed does not set compressibility.

If you missed this: read “Why solids and liquids have similar molar volumes” in the Topic 3.3 study guide.

Question 5 (short answer · 3.4)

A rigid 1.50 L flask at 27.0 °C contains 0.0600 mol of N₂ and some helium. The total pressure is 1.64 atm.

(a) Calculate the partial pressure of N₂. (b) Calculate the amount of helium, in moles. (c) Calculate the mole fraction of helium.

Worked answer

(a) T = 300.15 K. P(N₂) = nRT/V = 0.0600 × 0.08206 × 300.15 ÷ 1.50 = 0.985 atm.

(b) P(He) = 1.64 − 0.985 = 0.655 atm, so n(He) = PV/RT = 0.655 × 1.50 ÷ (0.08206 × 300.15) = 0.0399 mol.

(c) X(He) = 0.0399 ÷ 0.0999 = 0.399. Check: 0.655 ÷ 1.64 = 0.399.

Common slip: using 27.0 instead of 300.15 K.

If you missed this: read “Mixtures of gases: partial pressures” in the Topic 3.4 study guide.

Question 6 (multiple choice · 3.5)

Helium is at 200 K. At about what temperature would methane, CH₄, molecules have the same average speed as the helium atoms?

(A) 50 K
(B) 200 K
(C) 800 K
(D) 3200 K

Answer and explanation

Answer: (C). Average kinetic energy, ½mv², is proportional to Kelvin temperature. For equal speeds, T must scale with mass: 16.04 ÷ 4.003 = 4.01, so T ≈ 800 K.

(A) divides by the mass ratio; a heavier molecule needs a higher temperature.
(B) gives equal kinetic energy, not equal speed.
(D) squares the mass ratio.

If you missed this: read “Comparing speeds of different gases” in the Topic 3.5 study guide.

Question 7 (multiple choice · 3.6)

At very high pressure, a gas occupies a larger volume than the ideal gas law predicts. Which assumption of the kinetic molecular theory has failed?

(A) Collisions between particles are elastic.
(B) The particles take up a negligible share of the container’s volume.
(C) Average kinetic energy is proportional to Kelvin temperature.
(D) The particles move randomly in straight lines between collisions.

Answer and explanation

Answer: (B). Closely packed particles fill a real share of the space, so the sample needs more than the ideal volume.

(A), (C) and (D) do not explain an oversized volume.

If you missed this: read “Effect 2: particle volume raises the volume (or pressure)” in the Topic 3.6 study guide.

Question 8 (multiple choice · 3.7)

What volume of 0.250 M sodium phosphate, Na₃PO₄, contains 0.0450 mol of Na⁺?

(A) 11.3 mL
(B) 180 mL
(C) 540 mL
(D) 60.0 mL

Answer and explanation

Answer: (D). Each Na₃PO₄ gives 3 Na⁺: n(Na₃PO₄) = 0.0150 mol, and V = 0.0150 ÷ 0.250 = 0.0600 L.

(A) multiplies moles by molarity.
(B) ignores the 3 Na⁺ per formula unit.
(C) multiplies by 3 instead of dividing.

If you missed this: read “Ionic solutes: each ion has its own concentration” in the Topic 3.7 study guide.

Question 9 (multiple choice · 3.8)

A particle diagram represents 0.20 L of Al₂(SO₄)₃(aq) and shows 4 Al³⁺ ions. Each ion drawn stands for 0.010 mol. How many sulfate ions should it show, and what is [SO₄²⁻]?

(A) 6; 0.060 M
(B) 6; 0.30 M
(C) 4; 0.20 M
(D) 12; 0.60 M

Answer and explanation

Answer: (B). Al³⁺ : SO₄²⁻ is 2 : 3, so 6 sulfate ions (charge check: 4(+3) + 6(−2) = 0). 0.060 mol ÷ 0.20 L = 0.30 M.

(A) gives moles, not moles per litre.
(C) draws equal numbers, as if the charges matched.
(D) uses 3 sulfate per Al³⁺, leaving a net charge of −12.

If you missed this: read “Reading concentration from a diagram” in the Topic 3.8 study guide.

Question 10 (multiple choice · 3.9)

Propan-1-ol, CH₃CH₂CH₂OH, and its isomer methoxyethane, CH₃OCH₂CH₃, are mixed and distilled. Which statement about the first distillate is correct?

(A) It is rich in methoxyethane: its molecules cannot hydrogen bond to each other, so its vapor pressure is higher.
(B) It is rich in propan-1-ol, because its O–H bonds break easily when heated.
(C) It matches the mixture, because the isomers have the same molar mass.
(D) It is rich in methoxyethane, because its molecules have fewer electrons.

Answer and explanation

Answer: (A). Propan-1-ol hydrogen bonds (O–H); methoxyethane has no H on O. Its weaker attractions let more molecules escape, so the vapor is rich in it.

(B) Distillation separates whole molecules; no covalent bonds break.
(C) and (D) Both have 34 electrons and similar dispersion forces; hydrogen bonding makes the difference.

If you missed this: read “Distillation: using vapor pressure” in the Topic 3.9 study guide.

Question 11 (short answer · 3.10)

Ammonia, NH₃, and methane, CH₄, both have 10 electrons. Ammonia is very soluble in water; methane is only slightly soluble.

(a) Name the strongest attraction between water and (i) NH₃, (ii) CH₄. (b) Explain the difference in solubility using the attractions broken and formed.

Worked answer

(a) (i) Hydrogen bonding: NH₃ donates (N–H) and accepts (lone pair on N). (ii) Dipole–induced dipole, since CH₄ is nonpolar.

(b) Dissolving breaks some water–water hydrogen bonds. NH₃ replaces them with similar hydrogen bonds to water; CH₄ offers only much weaker attractions, so little dissolves.

If you missed this: read “Dissolving at the particle level” in the Topic 3.10 study guide.

Question 12 (multiple choice · 3.11)

For one molecule, which order of energy gaps runs from smallest to largest?

(A) rotational < electronic < vibrational
(B) vibrational < rotational < electronic
(C) electronic < vibrational < rotational
(D) rotational < vibrational < electronic

Answer and explanation

Answer: (D). Rotational gaps match microwave photons, vibrational gaps infrared, and electronic gaps visible and ultraviolet.

(A) UV and visible photons carry more energy than infrared ones.
(B) Microwaves have longer wavelengths than infrared, so lower photon energy.
(C) reverses the order: the “longer wavelength, more energy” error.

If you missed this: read “Matching the region to the transition” in the Topic 3.11 study guide.

Question 13 (short answer · 3.12)

An excited atom drops to a lower level and emits light of wavelength 512 nm.

(a) Calculate the frequency. (b) Calculate the energy of one photon, and of one mole of photons in kJ mol⁻¹. (c) What does the photon energy tell you about the atom?

Worked answer

(a) ν = c/λ = 2.998 × 10⁸ ÷ 5.12 × 10⁻⁷ = 5.86 × 10¹⁴ s⁻¹.

(b) E = hν = 6.626 × 10⁻³⁴ × 5.856 × 10¹⁴ = 3.88 × 10⁻¹⁹ J. × 6.022 × 10²³ = 234 kJ mol⁻¹.

(c) It equals the gap between the two energy levels.

If you missed this: read “The two equations” in the Topic 3.12 study guide.

Question 14 (multiple choice · 3.13)

A student makes calibration standards from a dye stock solution that had partly evaporated, so each standard is more concentrated than its label says. The concentration she then finds for an unknown will be

(A) correct, because the calibration curve corrects for errors in the standards.
(B) too high, because the standards absorb more light.
(C) too low, because the standards read too high for their labels, so the line is too steep.
(D) too high, because evaporation increases the molar absorptivity.

Answer and explanation

Answer: (C). Each standard is plotted at its label but absorbs for its true, higher concentration, so the slope (εb) is too large and c = A ÷ slope too low.

(A) A calibration curve is only as good as its standards.
(B) gets the direction wrong.
(D) ε belongs to the dye at that wavelength; it does not change with concentration.

If you missed this: read “Calibration curves” and Worked example 3 in the Topic 3.13 study guide.

Your next step

Topic	Question(s)	If you missed it, read
3.1 Intermolecular forces	1, 2	3.1 study guide
3.2 Properties of Solids	3	3.2 study guide
3.3 Solids, liquids, gases	4	3.3 study guide
3.4 Ideal Gas Law	5	3.4 study guide
3.5 Kinetic Molecular Theory	6	3.5 study guide
3.6 Non-ideal gases	7	3.6 study guide
3.7 Solutions and Mixtures	8	3.7 study guide
3.8 Representing solutions	9	3.8 study guide
3.9 Separation	10	3.9 study guide
3.10 Solubility	11	3.10 study guide
3.11 Spectroscopy	12	3.11 study guide
3.12 Photons	13	3.12 study guide
3.13 Beer-Lambert Law	14	3.13 study guide

How to use your result

Look at topics, not the total. Count a guess as a miss.
Fix gaps in order. Topic 3.1 feeds 3.2, 3.9 and 3.10; 3.4 feeds 3.5 and 3.6; 3.7 feeds 3.8 and 3.13; 3.11 feeds 3.12. Start with the earliest topic you missed.
Then practise. Try that topic’s practice questions, then the Unit 3 mixed review, where every question combines topics.