2024 Cambridge IGCSE Chemistry (0620) Practice Paper with Answers

Simple distillation is used to separate a volatile solvent (water) from a non-volatile solute (copper(II) sulfate) by boiling the mixture and condensing the vapor to collect pure liquid water.

1 mark: Simple distillation. Reject: distillation, fractional distillation.

Fractional distillation uses a fractionating column to separate mixtures of liquids with different boiling points, such as the various hydrocarbon fractions found in crude oil.

1 mark: Fractional distillation. Reject: distillation, simple distillation.

Limestone (calcium carbonate, \(\text{CaCO}_3\)) undergoes thermal decomposition in the hot blast furnace to produce calcium oxide (\(\text{CaO}\)), which then reacts with acidic silicon(IV) oxide impurities to form molten slag (\(\text{CaSiO}_3\)).

1 mark: Limestone. Accept: calcium carbonate / CaCO3.

Phosphoric acid (\(\text{H}_3\text{PO}_4\)) absorbed on silica is the catalyst used in the industrial hydration of ethene with steam at \(300\ ^\circ\text{C}\) and \(60\ \text{atm}\) pressure to produce ethanol.

1 mark: Phosphoric acid. Accept: phosphoric(V) acid / H3PO4.

The fermentation of glucose uses the enzymes in yeast as a biological catalyst to convert glucose into ethanol and carbon dioxide at a moderate temperature of around \(30\ ^\circ\text{C}\) under anaerobic conditions.

1 mark: Fermentation of glucose. Accept: fermentation.

Nylon is a polyamide formed by a condensation polymerisation reaction, where dicarboxylic acid and diamine monomers react together, eliminating a small molecule (such as water or hydrogen chloride) for each amide linkage formed.

1 mark: Condensation polymerisation. Reject: addition polymerisation.

Sacrificial protection involves attaching a more reactive metal (which loses electrons more readily) to the iron or steel structure. This more reactive metal corrodes preferentially, thereby protecting the underlying iron from being oxidised.

1 mark: Sacrificial protection. Reject: barrier method.

The atomic number of potassium is 19, which represents the number of protons in its nucleus. The potassium ion has a \(1+\) charge, indicating that it has lost 1 electron. Therefore, the number of electrons is \(19 - 1 = 18\).

1 mark for 18. Reject any other values.

The number of protons is 16 (each with a positive charge) and the number of electrons is 18 (each with a negative charge). The net charge is \(16 - 18 = -2\), which is written as \(2-\).

1 mark for 2- (also accept -2).

The number of neutrons is found by subtracting the proton (atomic) number from the nucleon (mass) number:

\(\text{Number of neutrons} = 31 - 15 = 16\).

1 mark for 16.

The species is an oxygen ion (\(\text{O}^{2-}\)). The atomic number of oxygen is 8, which corresponds to the number of protons in its nucleus. The identity of the element does not change when it becomes an ion, so the proton number remains 8.

1 mark for 8.

The mass number (nucleon number) is calculated as the sum of the protons and neutrons in the nucleus:

\(\text{Mass number} = 24 + 28 = 52\).

1 mark for 52.

Sodium chloride has a giant ionic lattice structure consisting of a regular arrangement of alternating positive sodium ions and negative chloride ions. The electrostatic forces of attraction between these oppositely charged ions are extremely strong throughout the giant structure. Consequently, a large amount of thermal energy is required to overcome and break these strong ionic bonds to melt the solid.

M1: (Has a) giant ionic lattice / strong electrostatic forces of attraction between oppositely charged ions (1 mark) M2: (Requires a) large amount of energy to break / overcome these forces (1 mark)

Copper has a giant metallic structure consisting of a regular lattice of positive copper ions surrounded by a sea of mobile, delocalised electrons. Because these delocalised electrons are not fixed to any individual atom, they are free to move throughout the entire metallic structure and carry electrical charge when a voltage is applied.

M1: Reference to delocalised electrons / mobile electrons (1 mark) M2: Electrons are free to move / flow to carry charge / current (1 mark)

In graphite, carbon atoms are covalently bonded to three other carbon atoms, forming hexagonal layers. While the covalent bonds within the layers are very strong, the forces holding the different layers together are weak intermolecular forces. This lack of strong bonding between the layers allows them to slide over each other easily when physical force is applied, making the material soft and slippery.

M1: Structure consists of layers / sheets of carbon atoms (1 mark) M2: Weak forces (of attraction) between layers allow them to slide / slip over each other (1 mark)

Sulfur dioxide has a simple molecular structure. The covalent bonds within each SO2 molecule are strong, but the forces of attraction between the individual molecules (intermolecular forces) are very weak. Because very little thermal energy is needed to overcome these weak forces, sulfur dioxide boils at a very low temperature. Note that no covalent bonds are broken during boiling.

M1: Simple molecular structure / weak intermolecular forces (or weak forces between molecules) (1 mark) M2: Little energy needed to overcome / break these forces (1 mark) [Reject: breaking covalent bonds]

Step 1: Determine the total moles of gaseous products produced per mole of copper(II) nitrate decomposed.
From the balanced equation, \(2\text{ mol}\) of \(\text{Cu(NO}_3\text{)}_2\) produces \(4\text{ mol}\) of \(\text{NO}_2\) and \(1\text{ mol}\) of \(\text{O}_2\), giving a total of \(5\text{ mol}\) of gas.
So, the ratio of reactant to gas products is \(2 : 5\).

Step 2: Calculate the moles of gas produced from \(0.050\text{ mol}\) of reactant.
\(\text{Moles of gas} = 0.050\text{ mol} \times \frac{5}{2} = 0.125\text{ mol}\)

Step 3: Calculate the volume of gas at r.t.p.
\(\text{Volume} = 0.125\text{ mol} \times 24\text{ dm}^3/\text{mol} = 3.0\text{ dm}^3\).

1 mark: Correct calculation of total moles of gas produced (0.125 mol).
1 mark: Correct calculation of the total gas volume with units (3.0 dm3 / 3 dm3).

Step 1: Calculate the moles of each element in \(100\text{ g}\) of the compound.
\(\text{Moles of P} = \frac{43.6}{31} = 1.41\text{ mol}\)
\(\text{Moles of O} = \frac{56.4}{16} = 3.53\text{ mol}\)

Step 2: Divide each mole value by the smallest value to find the simplest whole number ratio.
\(\text{Ratio of P} = \frac{1.41}{1.41} = 1\)
\(\text{Ratio of O} = \frac{3.53}{1.41} = 2.5\)

Step 3: Multiply by 2 to obtain integers, giving a ratio of \(2 : 5\).
Empirical formula is \(\text{P}_2\text{O}_5\).

1 mark: Correct calculated molar ratio of atoms (approx. 1.41 : 3.53 or 1 : 2.5).
1 mark: Correct empirical formula (P2O5).

Step 1: Calculate the formula mass (\(M_r\)) of iron(III) oxide.
\(M_r(\text{Fe}_2\text{O}_3) = (2 \times 56) + (3 \times 16) = 160\).

Step 2: Calculate the moles of \(\text{Fe}_2\text{O}_3\) reacted.
\(\text{Moles} = \frac{8.00\text{ g}}{160} = 0.050\text{ mol}\).

Step 3: Determine the moles of iron produced.
According to the stoichiometry, \(1\text{ mol}\) of \(\text{Fe}_2\text{O}_3\) yields \(2\text{ mol}\) of \(\text{Fe}\).
\(\text{Moles of Fe} = 0.050\text{ mol} \times 2 = 0.10\text{ mol}\).

Step 4: Calculate the mass of iron.
\(\text{Mass of Fe} = 0.10\text{ mol} \times 56\text{ g/mol} = 5.60\text{ g}\).

1 mark: Correct calculation of moles of Fe2O3 (0.050 mol) OR moles of Fe (0.10 mol).
1 mark: Correct final mass of iron with units (5.60 g / 5.6 g).

Step 1: Calculate the moles of sulfuric acid used.
\(\text{Moles of H}_2\text{SO}_4 = \frac{25.0}{1000} \times 0.150 = 0.00375\text{ mol}\).

Step 2: Determine moles of NaOH required from the stoichiometry (2:1 ratio).
\(\text{Moles of NaOH} = 2 \times 0.00375 = 0.00750\text{ mol}\).

Step 3: Calculate the volume of sodium hydroxide solution needed.
\(\text{Volume of NaOH} = \frac{0.00750}{0.200} = 0.0375\text{ dm}^3 = 37.5\text{ cm}^3\).

1 mark: Correct calculation of moles of NaOH required (0.00750 mol).
1 mark: Correct volume of NaOH with units (37.5 cm3).

Step 1: Calculate moles of reactants available.
\(\text{Moles of H}_2 = \frac{4.0}{2} = 2.0\text{ mol}\).
\(\text{Moles of O}_2 = \frac{24.0}{32} = 0.75\text{ mol}\).

Step 2: Determine the limiting reactant.
According to the equation, \(0.75\text{ mol}\) of \(\text{O}_2\) requires \(2 \times 0.75 = 1.5\text{ mol}\) of \(\text{H}_2\).
Since we have \(2.0\text{ mol}\) of \(\text{H}_2\), oxygen is the limiting reactant.

Step 3: Calculate moles of water produced from the limiting reactant.
\(\text{Moles of H}_2\text{O} = 2 \times \text{moles of O}_2 = 2 \times 0.75 = 1.50\text{ mol}\).

Step 4: Calculate mass of water.
\(\text{Mass of H}_2\text{O} = 1.50\text{ mol} \times 18\text{ g/mol} = 27.0\text{ g}\).

1 mark: Identification of O2 as the limiting reactant (or showing 0.75 mol of O2 limits production).
1 mark: Correct calculation of water mass with units (27.0 g / 27 g).

Step 1: Calculate the theoretical moles of ethyl ethanoate from ethanol.
\(\text{Moles of ethanol} = \frac{11.5}{46} = 0.250\text{ mol}\).
Since the reaction is 1:1, theoretical moles of ethyl ethanoate = \(0.250\text{ mol}\).

Step 2: Calculate the theoretical maximum mass of ethyl ethanoate.
\(\text{Theoretical mass} = 0.250\text{ mol} \times 88\text{ g/mol} = 22.0\text{ g}\).

Step 3: Calculate the percentage yield.
\(\text{Percentage yield} = \left( \frac{16.5\text{ g}}{22.0\text{ g}} \right) \times 100 = 75.0\%\).

1 mark: Correct calculation of the theoretical mass of ethyl ethanoate (22.0 g).
1 mark: Correct calculation of the percentage yield (75% / 75.0%).

1. Identify the effect of temperature on yield at a constant pressure: At 20 atm, when the temperature is raised from 300 \(^{\circ}C\) to 500 \(^{\circ}C\), the percentage yield of product \(X\) decreases from 45\\% to 15\\%.
2. Relate to the equilibrium shift: A decrease in yield indicates that the equilibrium position has shifted to the left (towards reactants).
3. Apply Le Chatelier's principle: An increase in temperature always shifts the equilibrium position in the direction of the endothermic reaction to absorb the added thermal energy. Since the shift is to the left, the backward reaction is endothermic.
4. Deduce the enthalpy change of the forward reaction: If the backward reaction is endothermic, the forward reaction must be exothermic.

Award marks as follows:
- Deduce that the forward reaction is exothermic [1 mark]
- State that increasing temperature (from 300 \(^{\circ}C\) to 500 \(^{\circ}C\)) decreases the yield of \(X\) (using data from the table at a constant pressure) [1 mark]
- Explain that an increase in temperature shifts equilibrium in the endothermic direction, meaning the backward reaction is endothermic / forward reaction is exothermic [1 mark]

1. State the trend shown in the table: As pressure increases from 50 atm to 150 atm at a constant temperature of 250 \(^{\circ}C\), the percentage yield of methanol increases from 58\\% to 82\\%.
2. Relate this to the number of moles of gas: The left-hand side of the equation has 3 moles of gaseous reactants (1 \(\text{CO}\) + 2 \(\text{H}_2\)), while the right-hand side has only 1 mole of gaseous product (\(\text{CH}_3\text{OH}\)).
3. Apply Le Chatelier's principle: Increasing pressure shifts the equilibrium position to the side with fewer moles of gas (the right-hand side), thereby increasing the yield of methanol.

Award marks as follows:
- State that increasing pressure increases the yield of methanol (supported by the table values: 58\\% at 50 atm increases to 82\\% at 150 atm) [1 mark]
- State that there are fewer moles of gas on the product side (1 mole) than on the reactant side (3 moles) [1 mark]
- Explain that increasing pressure shifts the equilibrium position to the side with fewer moles of gas / shifts to the right [1 mark]

For (a):
- Compare Experiment 1 and Experiment 2 at 400 \(^{\circ}C\).
- The yield is identical at 98\\% for both, meaning the catalyst does not alter the final position of equilibrium.
- However, the time taken decreases significantly from 3600 s to 45 s, showing that the catalyst increases the rate of reaction.

For (b):
- Compare Experiment 2 and Experiment 3.
- When temperature increases from 400 \(^{\circ}C\) to 500 \(^{\circ}C\) (with Pt catalyst), the yield of \(\text{SO}_3\) decreases from 98\\% to 91\\%.
- Since a temperature rise decreases the yield of product, the equilibrium shifts to the left.
- An increase in temperature always shifts equilibrium in the endothermic direction, so the backward reaction is endothermic.
- Therefore, the forward reaction must be exothermic, which corresponds to a negative enthalpy change (\(\Delta H < 0\)).

Award marks as follows:
- (a) Explains that the catalyst has no effect on the position of equilibrium/yield (remains 98\\%) [1 mark]
- (a) Explains that the catalyst increases the reaction rate / decreases the time taken to reach equilibrium (from 3600 s to 45 s) [1 mark]
- (b) States that \(\Delta H\) is negative / exothermic AND refers to the temperature increase causing a yield decrease (98\\% to 91\\%) [1 mark]

To obtain the copper(II) carbonate: 1. Add distilled water to the solid mixture in a beaker and stir thoroughly. The sodium chloride dissolves while the copper(II) carbonate remains insoluble. 2. Filter the mixture using a filter funnel and filter paper. The copper(II) carbonate is collected as the residue on the filter paper, while the sodium chloride solution passes through as the filtrate. 3. Wash the residue of copper(II) carbonate with a small amount of distilled water to remove any remaining sodium chloride solution. 4. Dry the copper(II) carbonate residue by pressing it gently between sheets of filter paper or by placing it in a warm oven.

1 mark: Add water and stir to dissolve the sodium chloride. 1 mark: Filter the mixture to obtain copper(II) carbonate as the residue. 0.5 mark: Wash the residue with distilled water and dry it.

The preparation of an insoluble salt is done via precipitation: 1. Mix equal volumes of aqueous barium chloride and aqueous sodium sulfate in a beaker to form a precipitate of barium sulfate. 2. Filter the mixture to separate the insoluble barium sulfate precipitate (residue) from the soluble sodium chloride solution (filtrate). 3. Wash the residue on the filter paper with distilled water to remove any soluble impurities. 4. Dry the precipitate in a warm oven or leave it to dry between filter papers.

1 mark: Mix the two aqueous solutions together. 1 mark: Filter the mixture to collect the precipitate as the residue. 0.5 mark: Wash the residue with distilled water and dry it.

1. The Rf value is calculated by dividing the distance traveled by the solute by the distance traveled by the solvent front: Rf = 4.2 \/ 6.0 = 0.70. 2. The baseline must be drawn in pencil because graphite (pencil lead) is insoluble in chromatography solvents. If ink from a pen were used, the ink dyes would dissolve in the solvent and run up the paper, interfering with the results and contaminating the chromatogram.

1 mark: Correct calculation of Rf = 0.70. 1 mark: State that pencil lead is insoluble in the solvent. 0.5 mark: Explain that ink contains soluble dyes which would run and interfere with the chromatogram.

1. The two essential pieces of apparatus are a gas syringe (or an inverted measuring cylinder filled with water) to measure the volume of gas, and a stopwatch (or timer) to measure time. 2. A gas syringe is preferred because downward delivery in air only collects gas in a container but does not allow the volume of gas to be measured continuously or quantitatively over time. Furthermore, hydrogen is extremely light and would easily escape or mix with air during downward delivery.

1 mark: Identify gas syringe (or inverted measuring cylinder) AND a timer (0.5 marks each). 1 mark: Explain that downward delivery does not allow continuous quantitative measurement of gas volume. 0.5 mark: Mention that hydrogen has a low density or that gas syringe prevents gas loss.

1. Find the percentage of oxygen by mass:
\(100\% - 56.0\% = 44.0\%\)

2. Calculate the number of moles of each element in a \(100\text{ g}\) sample:
Moles of \(\text{V} = \frac{56.0}{51} = 1.10\text{ mol}\)
Moles of \(\text{O} = \frac{44.0}{16} = 2.75\text{ mol}\)

3. Determine the simplest whole-number ratio of atoms:
Divide both by the smaller value (\(1.10\)):
\(\text{V} = \frac{1.10}{1.10} = 1.0\)
\(\text{O} = \frac{2.75}{1.10} = 2.5\)

Multiply both values by 2 to obtain whole numbers:
\(\text{V} = 2\)
\(\text{O} = 5\)

4. Write the final empirical formula: \(\text{V}_2\text{O}_5\).

[1] Calculates the percentage of oxygen as \(44.0\%\) (or shows calculation \(100 - 56.0\)).
[1] Divides percentages by relative atomic masses: \(\frac{56.0}{51}\) and \(\frac{44.0}{16}\) (moles of \(\text{V} = 1.10\), moles of \(\text{O} = 2.75\)).
[1] Identifies the simplest ratio as \(1 : 2.5\) or \(2 : 5\).
[1] Deduces the correct empirical formula: \(\text{V}_2\text{O}_5\).

1. Calculate the relative formula mass (\(M_r\)) of \(\text{Li}_2\text{CO}_3\):
\(M_r = (2 \times 7) + 12 + (3 \times 16) = 14 + 12 + 48 = 74\)

2. Calculate the number of moles of \(\text{Li}_2\text{CO}_3\) reacted:
\(\text{Moles} = \frac{3.70\text{ g}}{74\text{ g/mol}} = 0.050\text{ mol}\)

3. Determine the moles of \(\text{CO}_2\) produced:
According to the stoichiometry of the equation, the ratio of \(\text{Li}_2\text{CO}_3\) to \(\text{CO}_2\) is \(1:1\).
Therefore, moles of \(\text{CO}_2 = 0.050\text{ mol}\)

4. Calculate the volume of \(\text{CO}_2\) gas at r.t.p.:
\(\text{Volume} = \text{moles} \times 24.0\text{ dm}^3/\text{mol} = 0.050 \times 24.0 = 1.20\text{ dm}^3\).

[1] Calculates the \(M_r\) of \(\text{Li}_2\text{CO}_3\) as \(74\).
[1] Calculates the moles of \(\text{Li}_2\text{CO}_3\) as \(0.050\text{ mol}\) (allow ecf from incorrect \(M_r\)).
[1] Uses the \(1:1\) mole ratio to state that moles of \(\text{CO}_2 = 0.050\text{ mol}\).
[1] Calculates the volume of \(\text{CO}_2\) gas as \(1.20\text{ dm}^3\) (or \(1.2\text{ dm}^3\); accept \(1200\text{ cm}^3\) if unit is clearly stated).

1. Calculate the moles of \(\text{H}_2\text{SO}_4\) used:
\(\text{Moles} = \text{concentration} \times \text{volume in dm}^3\)
\(\text{Moles} = 0.0500\text{ mol/dm}^3 \times \frac{18.5}{1000}\text{ dm}^3 = 9.25 \times 10^{-4}\text{ mol}\) (or \(0.000925\text{ mol}\))

2. Determine the moles of \(\text{NaOH}\) that reacted:
From the balanced equation, \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\).
\(\text{Moles of NaOH} = 2 \times 9.25 \times 10^{-4}\text{ mol} = 1.85 \times 10^{-3}\text{ mol}\) (or \(0.00185\text{ mol}\))

3. Calculate the concentration of the \(\text{NaOH}\) solution:
\(\text{Concentration} = \frac{\text{moles}}{\text{volume in dm}^3}\)
\(\text{Concentration} = \frac{1.85 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0740\text{ mol/dm}^3\).

[1] Calculates the moles of sulfuric acid as \(9.25 \times 10^{-4}\text{ mol}\) (or \(0.000925\text{ mol}\)).
[1] Uses the \(1:2\) stoichiometric ratio to find moles of \(\text{NaOH} = 1.85 \times 10^{-3}\text{ mol}\) (or \(0.00185\text{ mol}\)).
[1] Divides the moles of \(\text{NaOH}\) by the volume of \(\text{NaOH}\) in \(\text{dm}^3\) (\(0.0250\text{ dm}^3\)).
[1] Obtains the correct concentration: \(0.0740\text{ mol/dm}^3\) (or \(0.074\text{ mol/dm}^3\)).

In the blast furnace, iron(III) oxide (hematite) reacts with carbon monoxide gas at high temperatures to produce molten iron and carbon dioxide gas. The balanced equation is \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\).

1 mark: correct reactant and product formulae: \(\text{Fe}_2\text{O}_3\), \(\text{CO}\), \(\text{Fe}\), \(\text{CO}_2\).
1 mark: correct balancing: \(1:3 \rightarrow 2:3\).

Calcium carbonate decomposes at high temperatures to produce calcium oxide and carbon dioxide: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\). The calcium oxide (which is a basic metal oxide) then reacts with the acidic silicon(IV) oxide impurity to form slag (calcium silicate): \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\).

1 mark: stating that calcium carbonate decomposes to form calcium oxide (or that calcium oxide reacts with the acidic oxide in a neutralisation reaction).
1 mark: identifying the final waste product as slag / calcium silicate.

Carbon is more reactive than zinc, so it can reduce zinc oxide by displacing zinc from the compound. However, aluminium is more reactive than carbon, meaning carbon is not a strong enough reducing agent to displace aluminium from its oxide.

1 mark: stating aluminium is more reactive than carbon (or has a stronger affinity for oxygen than carbon).
1 mark: stating zinc is less reactive than carbon (or carbon is more reactive than zinc, allowing reduction to occur).

In pure copper, the atoms are all of the same size and arranged in neat, regular layers which can easily slide over one another when a force is applied. In brass, the introduction of zinc atoms, which have a different atomic size, disrupts and distorts this regular arrangement. This prevents the layers of atoms from sliding past one another easily, making the alloy much harder.

1 mark: different size of zinc atoms disrupts/distorts the regular layers/arrangement of copper atoms.
1 mark: prevents/stops layers from sliding over each other easily.

Aluminium oxide has an extremely high melting point of about 2000 \(^{\circ}\text{C}\). Dissolving it in molten cryolite lowers the operating temperature of the cell to around 950 \(^{\circ}\text{C}\), saving a vast amount of thermal energy. Additionally, molten cryolite improves the overall electrical conductivity of the electrolyte mixture, allowing electrolysis to proceed more efficiently.

1 mark: lowers the melting point of the electrolyte/aluminium oxide mixture (or reduces energy costs).
1 mark: improves/increases the electrical conductivity of the electrolyte mixture.

(a) Structural isomers are defined as compounds that share the exact same molecular formula but have different arrangements of atoms, resulting in different structural formulae.

(b) The two straight-chain alkene isomers of \(\text{C}_4\text{H}_8\) are:
- but-1-ene, which has the double bond between the first and second carbon atoms: \(\text{CH}_2\text{=CH-CH}_2\text{-CH}_3\)
- but-2-ene, which has the double bond between the second and third carbon atoms: \(\text{CH}_3\text{-CH=CH-CH}_3\)

M1: (Same molecular formula but) different structural formulae / different structures / different arrangement of atoms [1]

M2: but-1-ene [1]

M3: but-2-ene [1]

(a) Polymer **Y** is a polyester, containing ester links (\(\text{-O-CO-}\)) in its backbone. Polyesters are formed when monomer units join together with the elimination of a small molecule (such as water). This process is known as condensation polymerisation.

(b) To synthesise a polyester, the monomers required must be a dicarboxylic acid (which contains carboxylic acid / carboxyl groups, \(\text{-COOH}\)) and a diol (which contains alcohol / hydroxyl groups, \(\text{-OH}\)).

M1: Condensation (polymerisation) [1]

M2: Carboxylic acid / carboxyl group (accept \(\text{-COOH}\)) [1]

M3: Alcohol / hydroxyl group / diol (accept \(\text{-OH}\)) [1]

(a) Esters with three carbon atoms (molecular formula \(\text{C}_3\text{H}_6\text{O}_2\)) can only be made from:
1. Methanol and ethanoic acid, which yields the ester methyl ethanoate (\(\text{CH}_3\text{COOCH}_3\)).
2. Ethanol and methanoic acid, which yields the ester ethyl methanoate (\(\text{HCOOCH}_2\text{CH}_3\)).

(b) Methyl ethanoate is synthesized from the alcohol methanol and the carboxylic acid ethanoic acid.

M1: methyl ethanoate [1]

M2: ethyl methanoate [1]

M3: ethanoic acid [1]

(a) By looking at the repeating unit of the polymer, the main carbon chain contains two carbon atoms, and one of those carbons has a methyl (\(\text{-CH}_3\)) branch. Thus, the monomer is propene (\(\text{CH}_3\text{-CH=CH}_2\)), which is an alkene.

(b) Alkenes are unsaturated hydrocarbons, whereas addition polymers like poly(propene) are saturated. To distinguish them, bromine water (aqueous bromine) is used. Propene will rapidly react with bromine in an addition reaction, causing the orange-brown solution to become colourless. The saturated polymer will not react and the bromine water remains orange-brown.

M1: propene [1]

M2: Bromine water / aqueous bromine / \(\text{Br}_2\text{(aq)}\) [1]

M3: (With monomer) decolourises / turns colourless AND (with polymer) remains orange / brown / yellow / no change [1]