0620/22 May/June 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

40 questions · 40 marks · 45 minutes

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1. Question 1

   1 marksParticles and energy during melting
   Step 1: Melting is the change from solid to liquid and it needs energy to be supplied. Step 2: The water molecules absorb energy from the warm surroundings, which lets them break free of their fixed positions in the solid. Step 3: So as the ice melts the molecules gain energy.
   Method:

   Recall that melting requires energy to be taken in and decide what happens to the particles.
   Examiner tips

   • Melting and boiling absorb energy; freezing and condensing release it.

2. Question 2

   1 marksCooling curve through condensing and freezing
   Step 1: As the gas is cooled its temperature drops until it reaches the boiling point, where it condenses to a liquid and the temperature stays constant, giving the first plateau. Step 2: Cooling continues until the freezing point, where the liquid freezes to a solid and the temperature stays constant again, giving a second, lower plateau. Step 3: So the graph falls, has a condensing plateau, falls, has a lower freezing plateau, then falls, which is the correct cooling curve.
   Method:

   Track the cooling from gas through condensing and freezing, marking a plateau at each change of state.
   Examiner tips

   • On a cooling curve, plateaux are changes of state, ordered from higher to lower temperature.

3. Question 3

   1 marksProtons, electrons and neutrons in an atom
   Step 1: The proton number of zinc is 30, and in a neutral atom the number of electrons equals the number of protons, so there are 30 electrons. Step 2: The number of neutrons is the nucleon number minus the proton number, which is 65 − 30 = 35. Step 3: So the atom has 30 protons, 30 electrons and 35 neutrons.
   Method:

   Use the proton number for protons and electrons, then subtract it from the nucleon number for neutrons.
   Examiner tips

   • Proton number gives protons and electrons; subtract it from the mass number for neutrons.

4. Question 4

   1 marksElectronic configuration of atoms and ions
   Step 1: The configuration 2,8,8 contains 18 electrons in total. Step 2: An argon atom has 18 electrons, and a chloride ion is a chlorine atom (17 electrons) that has gained one electron to give 18, so both have 18 electrons. Step 3: The aluminium ion has 13 − 3 = 10 electrons and the sodium ion has 11 − 1 = 10 electrons, so only particles 1 and 4 have the configuration 2,8,8.
   Method:

   Count the electrons in 2,8,8, then work out the electrons in each particle including its charge.
   Examiner tips

   • For an ion, adjust the atom electron count by its charge before counting shells.

5. Question 5

   1 marksRelative atomic mass from isotopic abundances
   Step 1: The relative atomic mass is the weighted average: $(238 \times 66 + 244 \times 34) \div 100$. Step 2: This is $(15708 + 8296) \div 100 = 24004 \div 100 = 240.04$. Step 3: To three significant figures the relative atomic mass is 240.
   Method:

   Apply the weighted-average formula using the isotope masses and their percentage abundances.
   Examiner tips

   • A weighted average sits nearer the mass of the more abundant isotope.

6. Question 6

   1 marksStructure and bonding in silicon(IV) oxide
   Step 1: Silicon(IV) oxide is a giant covalent solid in which silicon and oxygen atoms are joined throughout by covalent bonds. Step 2: Each silicon atom is bonded to four oxygen atoms in a tetrahedral arrangement like diamond, so that statement is correct. Step 3: It has no delocalised electrons (so it does not conduct), it is not a simple molecule, and its arrangement is like diamond, not graphite.
   Method:

   Recall the diamond-like giant covalent structure of silicon(IV) oxide and test each statement.
   Examiner tips

   • Silicon(IV) oxide = giant covalent, diamond-like, four bonds per silicon, no conduction.

7. Question 7

   1 marksLimiting reactant in reacting masses
   Step 1: In the first experiment the 65 g of zinc exactly reacts with 32 g of sulfur to give 97 g of zinc sulfide, so this uses up all of both reactants. Step 2: Adding more sulfur (40 g) does not help because all 65 g of zinc has already reacted, so the zinc is the limiting reactant. Step 3: With no zinc left, the extra sulfur cannot react, so the mass of zinc sulfide stays at 97 g.
   Method:

   Compare the two experiments to see which reactant is completely used up and limits the product.
   Examiner tips

   • Adding more of one reactant makes no extra product once the other reactant is used up.

8. Question 8

   1 marksBalancing the thermite equation
   Step 1: Iron(III) oxide is $\text{Fe}_2\text{O}_3$ and aluminium oxide is $\text{Al}_2\text{O}_3$, so these correct formulae must appear. Step 2: Balancing then needs 2 aluminium atoms on the left and 2 iron atoms on the right. Step 3: This gives $2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$, which balances all the atoms.
   Method:

   Write the correct formulae for each compound, then balance the atoms on both sides.
   Examiner tips

   • Fix the correct formulae first, then add coefficients to balance each element.

9. Question 9

   1 marksMass from concentration and volume
   Step 1: Moles = concentration × volume in dm³ = 0.50 × (500 ÷ 1000) = 0.25 mol. Step 2: Mass = moles × relative formula mass = 0.25 × 40. Step 3: This gives a mass of 10 g of sodium hydroxide.
   Method:

   Calculate the moles from concentration and volume, then convert to a mass.
   Examiner tips

   • Always change cm³ to dm³ (divide by 1000) before using concentration.

10. Question 10

    1 marksProducts from electrolysis of molten zinc oxide
    Step 1: Molten zinc oxide contains only zinc ions and oxide ions, with no water present. Step 2: At the cathode the positive zinc ions gain electrons and form zinc metal, while at the anode the negative oxide ions lose electrons and form oxygen. Step 3: So oxygen is made at the anode and zinc at the cathode.
    Method:

    Identify the ions in the molten compound, then send the metal to the cathode and non-metal to the anode.
    Examiner tips

    • Electrolysing a molten compound gives the metal at the cathode and the non-metal at the anode.

11. Question 11

    1 marksCathode half-equation in electrolysis of dilute brine
    Step 1: At the cathode positive ions gain electrons (reduction). Step 2: In dilute sodium chloride the hydrogen ions are discharged in preference to sodium ions, because hydrogen is less reactive than sodium. Step 3: So two hydrogen ions gain two electrons to form a hydrogen molecule: $2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2$.
    Method:

    Choose the cathode reaction where positive ions gain electrons, preferring the less reactive ion.
    Examiner tips

    • Cathode = reduction = gain of electrons; the less reactive ion is discharged.

12. Question 12

    1 marksEnergy change in a hydrogen-oxygen fuel cell
    Step 1: A fuel cell uses the reaction of hydrogen and oxygen to produce a voltage, so it converts chemical energy in the fuel into electrical energy. Step 2: The overall reaction is $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$, giving only water (no carbon dioxide). Step 3: Hydrogen is not a fossil fuel, so only the statement about chemical energy becoming electrical energy is correct.
    Method:

    Recall how a hydrogen-oxygen fuel cell works and its single product, then test each statement.
    Examiner tips

    • Fuel cells make only water and convert chemical energy directly into electrical energy.

13. Question 13

    1 marksReaction pathway diagram and surroundings temperature
    Step 1: The products are higher in energy than the reactants, so energy has been taken in from the surroundings, which makes the reaction endothermic. Step 2: Because the reaction absorbs energy from the surroundings, the surroundings lose energy and their temperature falls. Step 3: So the reaction is endothermic and the temperature of the surroundings decreases.
    Method:

    Compare the energy levels of products and reactants to classify the reaction and its effect on the surroundings.
    Examiner tips

    • Endothermic: products higher on the diagram, energy absorbed, surroundings cool.

14. Question 14

    1 marksBond energy and enthalpy change of forming methane
    Step 1: Forming bonds releases energy, so making bonds gives a negative (exothermic) enthalpy change. Step 2: Each methane molecule has four C–H bonds, so the energy released per mole is 4 × 413 = 1652 kJ. Step 3: Because the bonds are being formed, the enthalpy change is −1652 kJ.
    Method:

    Count the bonds formed, multiply by the bond energy, and make the sign negative for bond making.
    Examiner tips

    • Bond forming is exothermic (negative); count every bond made.

15. Question 15

    1 marksEffect of a catalyst on a reaction
    Step 1: A catalyst provides an alternative pathway with a lower activation energy, so it lowers $E_a$, it does not raise it. Step 2: A catalyst does not change the energies of the reactants or products, so the enthalpy change $\Delta H$ stays the same. Step 3: A catalyst does not lower the collision frequency or raise the particles’ kinetic energy, so the only correct statement is that $\Delta H$ is unchanged.
    Method:

    Recall what a catalyst does and does not change, then test each statement.
    Examiner tips

    • A catalyst changes the rate by lowering Ea, but ΔH for the reaction is unaffected.

16. Question 16

    1 marksConditions for the Contact process
    Step 1: The Contact process uses vanadium(V) oxide as the catalyst, not iron (iron is used in the Haber process). Step 2: The key step is carried out at a temperature of about 450 °C and only a low pressure of about 2 atmospheres, because a higher pressure gives little extra yield and is expensive. Step 3: So the correct conditions are a vanadium(V) oxide catalyst, a low pressure and about 450 °C.
    Method:

    Recall the standard Contact process conditions and pick the matching row.
    Examiner tips

    • Match the catalyst to the process: vanadium(V) oxide = Contact process, iron = Haber process.

17. Question 17

    1 marksIdentifying a reducing agent in redox equations
    Step 1: A reducing agent is the species that is itself oxidised, that is, it loses electrons (its oxidation state increases). Step 2: In equation 2 the iron goes from 0 in Fe to +2 in $\text{FeSO}_4$, and in equation 3 the iron goes from +2 in $\text{Fe}^{2+}$ to +3 in $\text{Fe}^{3+}$, so in both the iron is oxidised and acts as the reducing agent. Step 3: In equation 1 the iron in $\text{Fe}_2\text{O}_3$ is reduced (it gains electrons, so it is the oxidising agent), and in equation 4 the iron stays +2 throughout (no redox), so only equations 2 and 3 show the iron as a reducing agent.
    Method:

    Assign oxidation states to the iron in each equation and find where it increases.
    Examiner tips

    • Track the oxidation state of the iron: increasing means oxidised, so it is the reducing agent.

18. Question 18

    1 marksReleasing ammonia from an ammonium salt
    Step 1: A gas that turns damp red litmus paper blue is alkaline, and the common alkaline gas is ammonia. Step 2: Ammonia is released when an ammonium salt is warmed with a base, such as a basic oxide. Step 3: So the two substances are a basic oxide and an ammonium salt; potassium salts and acidic oxides do not release ammonia.
    Method:

    Identify the alkaline gas as ammonia, then choose the base plus ammonium salt that releases it.
    Examiner tips

    • Ammonia (turns red litmus blue) is produced when an ammonium salt meets a base.

19. Question 19

    1 marksDissociation of strong and weak acids
    Step 1: A strong acid is fully dissociated, so its equation uses a single forward arrow, while a weak acid is only partly dissociated, so its equation uses a reversible arrow. Step 2: Hydrochloric acid is strong, so $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$ has a forward arrow, and ethanoic acid is weak, so $\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$ has a reversible arrow. Step 3: So the correct row gives HCl a full arrow and ethanoic acid a reversible arrow.
    Method:

    Match each acid to its strength, then choose the forward or reversible arrow accordingly.
    Examiner tips

    • Use → for a strong acid and ⇌ for a weak acid to show the extent of dissociation.

20. Question 20

    1 marksIdentifying acidic oxides
    Step 1: Acidic oxides are formed by non-metals, while metal oxides such as calcium oxide and sodium oxide are basic. Step 2: Carbon dioxide, nitrogen dioxide and sulfur dioxide are all non-metal oxides, so all three are acidic. Step 3: Any set containing CaO or $\text{Na}_2\text{O}$ includes a basic oxide, so only $\text{CO}_2$, $\text{NO}_2$ and $\text{SO}_2$ are all acidic.
    Method:

    Classify each oxide as metal (basic) or non-metal (acidic) and pick the all-non-metal set.
    Examiner tips

    • Spot the metal oxides (basic) and remove any set that contains one.

21. Question 21

    1 marksPredicting properties of a noble gas from trends
    Step 1: Going down the noble gases both the density and the boiling point increase, so krypton’s values must lie between those of argon and xenon. Step 2: The density must be between 0.9 and 5.9 g/dm³, and the boiling point between −186 and −108 °C. Step 3: A density of 3.8 g/dm³ and a boiling point of −153 °C both fall in these ranges, so they are the correct predictions.
    Method:

    Use the increasing trend down the group to place krypton between argon and xenon.
    Examiner tips

    • Interpolate: a value between two neighbours must lie between their two values.

22. Question 22

    1 marksPhysical properties of a Group I metal
    Step 1: Like other metals, sodium is malleable, meaning it can be bent or hammered into shape. Step 2: As a Group I metal sodium is also a good conductor of electricity, has a low melting point and a low density. Step 3: So of the options given, only being malleable is a correct physical property of sodium.
    Method:

    Recall the physical properties of Group I metals and find the correct one.
    Examiner tips

    • Remember Group I metals are soft, low-density and low-melting, yet still malleable conductors.

23. Question 23

    1 marksPredicting the state of a Group VII element
    Step 1: Down Group VII the elements change from gases (fluorine, chlorine) to a liquid (bromine) to solids (iodine, astatine). Step 2: Tennessine lies below astatine, which is already a solid, so the trend predicts tennessine is also a solid. Step 3: So at room temperature and pressure tennessine is expected to be a solid.
    Method:

    Follow the trend in state down Group VII to predict the state of the lowest element.
    Examiner tips

    • Continue the gas-liquid-solid trend down the group to predict the heaviest member.

24. Question 24

    1 marksDisplacement reaction of halogens
    Step 1: Chlorine is more reactive than bromine because reactivity decreases down Group VII. Step 2: The more reactive chlorine takes the place of the less reactive bromine, forming chloride ions and releasing bromine: $\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2$. Step 3: So chlorine (the element) displaces bromine (the element) from the sodium bromide.
    Method:

    Use the reactivity order of the halogens to decide which element displaces which.
    Examiner tips

    • Write it as element displaces element: chlorine displaces bromine.

25. Question 25

    1 marksComparing the hardness of brass and copper
    Step 1: An alloy contains atoms of different sizes, which disrupt the regular layers of atoms in the metal. Step 2: The different-sized atoms in brass stop the layers from sliding over each other easily, so brass is harder than pure copper. Step 3: Brass is a mixture (not a compound) and still conducts, copper conducts because of its free electrons (not free atoms), and a pure metal is softer, so only the statement about brass being harder is correct.
    Method:

    Use the idea of mixed-size atoms blocking layers to compare the hardness of an alloy and pure metal.
    Examiner tips

    • Alloying distorts the layers, making alloys harder than the pure metal.

26. Question 26

    1 marksOrdering metals by reactivity
    Step 1: The given equations show magnesium > zinc > copper > silver and that tin > copper, but the position of tin relative to zinc is still unknown. Step 2: If zinc reacts with $\text{Sn}^{2+}$ ions, then zinc is more reactive than tin, which fixes tin between zinc and copper. Step 3: This completes the order magnesium > zinc > tin > copper > silver, so the useful new information is that zinc reacts with $\text{Sn}^{2+}$ ions.
    Method:

    Work out the order from the given reactions, then choose the test that fixes the only missing comparison.
    Examiner tips

    • Identify the one pair whose order is still missing, then find the experiment that compares them.

27. Question 27

    1 marksGalvanising and rust prevention
    Step 1: The zinc coating keeps out water and oxygen, so galvanising is a barrier method, making statement 1 correct. Step 2: Even if the zinc is scratched, the more reactive zinc still corrodes in preference to the iron, so the iron is protected and does not rust quickly, making statement 2 wrong and statement 3 (sacrificial protection) correct. Step 3: Zinc, being more reactive, loses (donates) electrons more readily rather than accepting them, so statement 4 is wrong, leaving 1 and 3.
    Method:

    Judge each statement using how a more reactive coating protects iron physically and sacrificially.
    Examiner tips

    • Zinc protects iron in two ways: a physical barrier and sacrificial (it corrodes first).

28. Question 28

    1 marksReaction in the blast furnace
    Step 1: In the blast furnace, carbon monoxide is the reducing agent that removes oxygen from the iron(III) oxide. Step 2: The carbon monoxide reduces the iron(III) oxide to iron, forming carbon dioxide: $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$. Step 3: Calcium is not used to reduce the ore, so the equations involving calcium do not represent the extraction of iron.
    Method:

    Identify the reduction of iron(III) oxide by carbon monoxide as the extraction reaction.
    Examiner tips

    • The key blast-furnace reaction is reduction of Fe2O3 by CO to give iron and CO2.

29. Question 29

    1 marksColour changes when water is added to anhydrous salts
    Step 1: Anhydrous copper(II) sulfate is white and turns blue when water is added, which is the standard test for water. Step 2: Anhydrous cobalt(II) chloride is blue and turns pink when water is added, the other test for water. Step 3: So copper(II) sulfate goes white to blue and cobalt(II) chloride goes blue to pink.
    Method:

    Recall the two standard tests for water and the direction of each colour change.
    Examiner tips

    • These two colour changes are the chemical tests for the presence of water.

30. Question 30

    1 marksRemoving pollutant gases in a catalytic converter
    Step 1: A catalytic converter removes the harmful carbon monoxide and nitrogen monoxide from the exhaust. Step 2: It does this by reacting them together so that carbon monoxide is oxidised to carbon dioxide and nitrogen monoxide is reduced to nitrogen: $2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2$. Step 3: The other equations make harmful gases or carbon monoxide rather than removing them.
    Method:

    Choose the equation that turns the harmful exhaust gases into harmless carbon dioxide and nitrogen.
    Examiner tips

    • The converter pairs CO with NO to give harmless CO2 and N2.

31. Question 31

    1 marksCarbon dioxide and global warming
    Step 1: Carbon dioxide is a greenhouse gas that absorbs thermal energy radiated from the Earth, so statement 1 is correct, and burning fossil fuels has raised its atmospheric level, so statement 3 is correct. Step 2: By absorbing this energy it reduces the thermal energy lost to space, so statement 2 is wrong. Step 3: Other gases such as methane also cause global warming, so statement 4 is wrong, leaving statements 1 and 3.
    Method:

    Use what greenhouse gases do to judge each statement about carbon dioxide.
    Examiner tips

    • Greenhouse gases trap heat (reduce loss to space); CO2 is one of several, not the only one.

32. Question 32

    1 marksStructural formula of butan-2-ol
    Step 1: Butan-2-ol has a chain of four carbon atoms (but-) and an –OH group (-ol) on carbon number 2. Step 2: Writing a four-carbon chain with the –OH on the second carbon gives $\text{CH}_3\text{CH}_2\text{CH(OH)CH}_3$. Step 3: So this is the structural formula of butan-2-ol.
    Method:

    Build the four-carbon chain and place the –OH on carbon 2 to match the name.
    Examiner tips

    • Count the carbons for the stem name and check which carbon carries the –OH group.

33. Question 33

    1 marksDisplayed formula of an ester
    Step 1: An ester forms when the –OH of the alcohol reacts with the –COOH of the carboxylic acid, losing water and making a C(=O)–O ester linkage. Step 2: Ethanoic acid provides $\text{CH}_3\text{CO}{-}$ and methanol provides $-\text{OCH}_3$, which join to give methyl ethanoate, $\text{CH}_3{-}\text{C(=O)}{-}\text{O}{-}\text{CH}_3$. Step 3: So the correct displayed formula is the methyl ethanoate structure with the ester linkage in the right place.
    Method:

    Combine the acid and alcohol parts through an ester linkage to build the correct displayed formula.
    Examiner tips

    • Methanol + ethanoic acid → methyl ethanoate, CH3COOCH3, plus water.

34. Question 34

    1 marksUses of petroleum fractions
    Step 1: Kerosene is the aircraft fuel fraction, so it is correctly used as jet fuel. Step 2: Bitumen is used to surface roads (not as a lubricant), naphtha is mainly a feedstock for making chemicals (not home heating), and refinery gas is used as a fuel for cooking and heating (not making chemicals). Step 3: So only kerosene matched with jet fuel is correct.
    Method:

    Recall the standard use of each petroleum fraction and find the correct pairing.
    Examiner tips

    • Learn each fraction with its main use; kerosene = jet fuel is a key one.

35. Question 35

    1 marksSubstitution reaction of ethane with chlorine
    Step 1: Ethane is a saturated alkane, so with chlorine and ultraviolet light it undergoes substitution, where a chlorine atom replaces one hydrogen atom. Step 2: One hydrogen of ethane is replaced by a chlorine, giving chloroethane, $\text{C}_2\text{H}_5\text{Cl}$, and the displaced hydrogen joins a chlorine to form hydrogen chloride, HCl. Step 3: So the products are $\text{C}_2\text{H}_5\text{Cl}$ and HCl.
    Method:

    Apply the substitution rule for an alkane and chlorine to find the chloroalkane and HCl.
    Examiner tips

    • Alkane + Cl2 (UV) → chloroalkane + HCl by substitution.

36. Question 36

    1 marksComparing methods of making ethanol
    Step 1: The catalytic addition of steam to ethene is a fast, continuous process that gives pure ethanol. Step 2: Compared with fermentation, two clear advantages are that it produces a purer product and is a faster reaction. Step 3: Using renewable raw materials is an advantage of fermentation (ethene comes from finite petroleum), and using more energy is a disadvantage, so the correct pair is a purer product and a faster reaction.
    Method:

    Compare the two ethanol methods and pick the two genuine advantages of hydrating ethene.
    Examiner tips

    • Hydration of ethene: fast, continuous, pure product; fermentation: renewable but slow and impure.

37. Question 37

    1 marksMonomers of a polyamide
    Step 1: The repeat unit contains amide linkages, –N(H)–C(=O)–, so it is a polyamide made from a diamine and a dicarboxylic acid. Step 2: The –N(H)–CH₂–CH₂–N(H)– part comes from a two-carbon diamine, $\text{H}_2\text{N}{-}\text{CH}_2\text{CH}_2{-}\text{NH}_2$, and the –C(=O)–CH₂–CH₂–C(=O)– part comes from a dicarboxylic acid, $\text{HOOC}{-}\text{CH}_2\text{CH}_2{-}\text{COOH}$. Step 3: So the correct monomers are this diamine and dicarboxylic acid, matching the chain lengths in the repeat unit.
    Method:

    Recognise the amide linkage, then split the repeat unit into the matching diamine and diacid monomers.
    Examiner tips

    • Polyamide = diamine + dicarboxylic acid; match the carbon chains to the repeat unit.

38. Question 38

    1 marksApparatus for measuring an exact volume
    Step 1: A volume given to one decimal place, such as 26.3 cm³, needs apparatus with a fine, accurate scale. Step 2: A burette has a precise graduated scale and a tap, so it can deliver any accurate variable volume like 26.3 cm³. Step 3: A measuring cylinder is far less precise, and beakers and conical flasks only hold liquids, so the burette is the correct choice.
    Method:

    Choose the apparatus whose fine graduations allow an exact variable volume to be measured.
    Examiner tips

    • For an accurate, variable volume to 0.1 cm³, use a burette.

39. Question 39

    1 marksResidue and filtrate in filtration
    Step 1: During filtration, the insoluble solid is trapped on the filter paper and is called the residue, so X is the residue. Step 2: The liquid that passes through the filter paper is called the filtrate, so Y is the filtrate. Step 3: The filtrate is the aqueous salt solution, not pure water, so X is the residue and Y is the filtrate.
    Method:

    Use the filtration terms: residue for the solid on the paper, filtrate for the liquid collected.
    Examiner tips

    • In filtration, name the trapped solid the residue and the collected liquid the filtrate.

40. Question 40

    1 marksEffect of an impurity on melting and boiling points
    Step 1: Dissolving an impurity in a substance lowers its melting point and raises its boiling point. Step 2: So the melting point falls below −114 °C, to −116 °C, and the boiling point rises above 78 °C, to 79 °C. Step 3: So the impure ethanol melts at −116 °C and boils at 79 °C.
    Method:

    Apply the effect of an impurity to shift the melting point down and the boiling point up.
    Examiner tips

    • Impurity: melting point goes down, boiling point goes up, and the ranges broaden.