2024 Cambridge IGCSE Chemistry (0620) Practice Paper with Answers

Sulfuric acid is manufactured industrially on a large scale using the Contact process, which involves the catalytic oxidation of sulfur dioxide to sulfur trioxide.

Award [1] mark for 'Contact process'. Reject: 'Haber process'.

Cracking is the thermal or catalytic decomposition of longer, less useful hydrocarbon chains into smaller, highly useful molecules like alkenes and shorter alkanes.

Award [1] mark for 'cracking' or 'catalytic cracking' or 'thermal cracking'.

Chromatography (specifically paper chromatography) is used to separate and identify different soluble components, such as colored dyes, based on their relative solubility in a solvent.

Award [1] mark for 'chromatography' or 'paper chromatography'.

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

Award [1] mark for 'simple distillation' or 'distillation'. Reject: 'fractional distillation'.

Fractional distillation is used to separate liquid air into its gaseous components, such as oxygen and nitrogen, because they have different boiling points.

Award [1] mark for 'fractional distillation'.

Fermentation is the anaerobic process in which yeast enzymes catalyze the conversion of glucose into ethanol and carbon dioxide.

Award [1] mark for 'fermentation'. Accept: 'anaerobic respiration'.

Ethene monomers contain double bonds that break open and link together to form a long-chain macromolecule in addition polymerisation.

Award [1] mark for 'addition polymerisation' or 'polymerisation'. Reject: 'condensation polymerisation'.

Phosphorus has an atomic number of 15, which means a neutral atom of phosphorus has 15 protons and 15 electrons. The \(3-\). charge indicates that the ion has gained 3 additional electrons.

Number of electrons = \(15 + 3 = 18\).

18 [1]

The mass number of the iron ion is 56, and the atomic number (number of protons) is 26.

Number of neutrons = mass number - atomic number = \(56 - 26 = 30\).

30 [1]

Nickel has an atomic number of 28, meaning a neutral nickel atom contains 28 protons and 28 electrons. The \(2+\) charge indicates that the atom has lost 2 electrons.

Number of electrons = \(28 - 2 = 26\).

26 [1]

The mass number is the total number of protons and neutrons in the nucleus of an atom or ion.

Mass number = number of protons + number of neutrons = \(17 + 20 = 37\).

37 [1]

The number of protons is 34 (each with a \(1+\) charge) and the number of electrons is 36 (each with a \(1-\) charge).

Overall charge = \(34 + (-36) = -2\), which is represented as \(2-\).

2- (accept -2) [1]

In magnesium oxide, MgO, the ions present are Mg^2+ and O^2-, which have charges of +2 and -2 respectively. In sodium chloride, NaCl, the ions present are Na^+ and Cl^-, which have charges of +1 and -1 respectively. Because the charges on the ions in MgO are higher, the electrostatic forces of attraction holding the giant ionic lattice together are much stronger. Thus, more heat energy is needed to separate the ions and melt magnesium oxide.

M1: For stating that MgO has ions with higher charges (2+ and 2-) than NaCl (1+ and 1-) (1 mark). M2: For explaining that this results in stronger electrostatic forces of attraction between the ions in MgO, requiring more energy to break (1 mark).

Nitrogen is a simple molecular substance. The covalent bonds between the nitrogen atoms inside each N2 molecule (intramolecular bonds) are indeed very strong, but they do not break when nitrogen boils. Instead, boiling only involves separating the N2 molecules from each other. The forces between the molecules (intermolecular forces) are very weak, so very little thermal energy is needed to overcome them and change the state from liquid to gas.

M1: For identifying that the forces between the molecules (intermolecular forces) are weak (1 mark). M2: For stating that only these weak intermolecular forces are broken during boiling / covalent bonds do not break (1 mark).

In graphite, each carbon atom is covalently bonded to three other carbon atoms, forming hexagonal layers. However, there are no covalent bonds between the layers; they are held together only by weak intermolecular forces. Because of these weak forces, the layers can easily slide over each other, which gives graphite its soft, slippery texture and makes it an effective lubricant.

M1: For stating that graphite has a layered structure with weak forces between the layers (1 mark). M2: For explaining that these layers can slide over each other easily (1 mark).

Copper consists of a regular giant lattice of positive copper ions surrounded by a sea of delocalised electrons. The ions are arranged in neat, orderly layers. When the metal is hammered or shaped, these layers of ions can slide past each other into new positions without breaking the metallic bond, because the mobile delocalised electrons can move to adjust and keep holding the ions together.

M1: For stating that the copper ions are arranged in regular layers that can slide over each other (1 mark). M2: For explaining that the delocalised electrons can move to maintain the metallic bonding / prevent structure from shattering (1 mark).

1. Calculate the formula mass of anhydrous cobalt(II) chloride, \(\text{CoCl}_2\):
\(M_r(\text{CoCl}_2) = 59 + (35.5 \times 2) = 130\)

2. Determine the mass of the water of crystallisation:
\(238 - 130 = 108\)

3. Divide by the relative formula mass of water (\(\text{H}_2\text{O} = 18\)):
\(x = \frac{108}{18} = 6\)

M1: Calculation of anhydrous \(\text{CoCl}_2\) mass as 130 or mass of water as 108 [1]
M2: Correct calculation of \(x = 6\) [1]

1. Calculate the moles of \(\text{CO}_2\) gas produced:
\(\text{Moles of CO}_2 = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{/mol}} = 0.01\text{ mol}\)

2. Determine the mass of pure \(\text{FeCO}_3\) that reacted:
Since \(\text{FeCO}_3\) reacts in a 1:1 ratio with \(\text{CO}_2\):
\(\text{Moles of FeCO}_3 = 0.01\text{ mol}\)
\(\text{Mass of pure FeCO}_3 = 0.01\text{ mol} \times 116\text{ g/mol} = 1.16\text{ g}\)

3. Calculate the percentage purity:
\(\text{Percentage purity} = \frac{1.16\text{ g}}{1.45\text{ g}} \times 100 = 80\%\)

M1: Correct calculation of moles of \(\text{CO}_2\) (\(0.01\text{ mol}\)) or mass of pure \(\text{FeCO}_3\) (\(1.16\text{ g}\)) [1]
M2: Correct percentage purity calculation of \(80\%\) [1]

1. Find the ratio of moles of atoms of each element:
\(\text{Moles of P} = \frac{27.9}{31} = 0.90\text{ mol}\)
\(\text{Moles of S} = \frac{72.1}{32} = 2.25\text{ mol}\)

2. Divide each by the smaller number of moles (0.90):
\(\text{P} = \frac{0.90}{0.90} = 1\)
\(\text{S} = \frac{2.25}{0.90} = 2.5\)

3. Convert to whole numbers by multiplying by 2:
\(\text{P} = 2\), \(\text{S} = 5\)
Empirical formula is \(\text{P}_2\text{S}_5\).

M1: Correct calculation of mole ratio (approx. \(0.90 : 2.25\)) or dividing by the smallest value to get \(1 : 2.5\) [1]
M2: Correct whole-number empirical formula \(\text{P}_2\text{S}_5\) [1]

1. Calculate the number of moles of \(\text{Fe}_2\text{O}_3\):
\(\text{Moles of Fe}_2\text{O}_3 = \frac{8.0\text{ g}}{160\text{ g/mol}} = 0.05\text{ mol}\)

2. Determine the moles of iron produced:
From the stoichiometry of the equation, \(1\text{ mol}\) of \(\text{Fe}_2\text{O}_3\) produces \(2\text{ mol}\) of \(\text{Fe}\).
\(\text{Moles of Fe} = 0.05\text{ mol} \times 2 = 0.10\text{ mol}\)

3. Calculate the mass of iron:
\(\text{Mass of Fe} = 0.10\text{ mol} \times 56\text{ g/mol} = 5.6\text{ g}\)

M1: Calculation of moles of \(\text{Fe}_2\text{O}_3\) as \(0.05\text{ mol}\) or showing that the mole ratio of reactant to product is \(1 : 2\) [1]
M2: Correct mass of iron as \(5.6\text{ g}\) [1]

1. Calculate the number of moles of \(\text{NaOH}\) used:
\(\text{Moles of NaOH} = \frac{20.0\text{ cm}^3}{1000} \times 0.200\text{ mol/dm}^3 = 0.0040\text{ mol}\)

2. Determine the moles of \(\text{H}_2\text{SO}_4\) that reacted:
From the balanced equation, \(2\text{ mol}\) of \(\text{NaOH}\) reacts with \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\).
\(\text{Moles of H}_2\text{SO}_4 = \frac{0.0040\text{ mol}}{2} = 0.0020\text{ mol}\)

3. Calculate the concentration of the \(\text{H}_2\text{SO}_4\):
\(\text{Concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.0020\text{ mol}}{0.0250\text{ dm}^3} = 0.080\text{ mol/dm}^3\)

M1: Correct calculation of moles of \(\text{NaOH}\) (\(0.0040\text{ mol}\)) or moles of \(\text{H}_2\text{SO}_4\) (\(0.0020\text{ mol}\)) [1]
M2: Correct concentration of \(\text{H}_2\text{SO}_4\) as \(0.080\text{ mol/dm}^3\) (accept 0.08) [1]

1. Calculate the number of moles of carbon monoxide, \(\text{CO}\):
\(\text{Moles of CO} = \frac{56.0\text{ g}}{28\text{ g/mol}} = 2.0\text{ mol}\)

2. Determine the theoretical mass of methanol, \(\text{CH}_3\text{OH}\):
Since \(1\text{ mol}\) of \(\text{CO}\) theoretically produces \(1\text{ mol}\) of \(\text{CH}_3\text{OH}\):
\(\text{Theoretical moles of CH}_3\text{OH} = 2.0\text{ mol}\)
\(\text{Theoretical mass of CH}_3\text{OH} = 2.0\text{ mol} \times 32\text{ g/mol} = 64.0\text{ g}\)

3. Calculate the percentage yield of methanol:
\(\text{Percentage yield} = \frac{51.2\text{ g}}{64.0\text{ g}} \times 100 = 80\%\)

M1: Correct calculation of theoretical mass of methanol (\(64.0\text{ g}\)) or theoretical moles of methanol (\(2.0\text{ mol}\)) [1]
M2: Correct percentage yield calculation of \(80\%\) [1]

1. Increasing the pressure shifts the equilibrium to the side with fewer gas moles (the right side has 2 moles, while the left side has 4 moles), so the yield of ammonia increases. 2. The forward reaction is exothermic. Increasing the temperature shifts the equilibrium in the endothermic direction (to the left), so the yield of ammonia decreases. 3. A catalyst increases the rate of both forward and reverse reactions equally, so it causes no change in the equilibrium yield.

M1: increases (for pressure increase) [1] M2: decreases (for temperature increase) [1] M3: no change (for catalyst) [1]

1. Increasing the temperature always increases the rate of the forward reaction because particles have more kinetic energy, leading to more frequent and successful collisions. 2. Increasing the pressure increases the rate of reaction because the concentration of gas particles increases, resulting in more frequent collisions. 3. Since the forward reaction is endothermic, decreasing the temperature shifts the equilibrium to the exothermic side (to the left), which decreases the yield of hydrogen.

M1: increases (temperature on rate) [1] M2: increases (pressure on rate) [1] M3: decreases (temperature decrease on hydrogen yield) [1]

1. Decreasing the pressure shifts the equilibrium to the side with more moles of gas (the right side has 2 moles, while the left side has 1 mole), so the yield of nitrogen dioxide increases. 2. Adding a catalyst increases the rate of both forward and reverse reactions equally, resulting in no change in the equilibrium yield. 3. Since the forward reaction is endothermic, increasing the temperature shifts the equilibrium in the endothermic direction (to the right) to absorb heat, which increases the yield of nitrogen dioxide.

M1: increases (pressure decrease) [1] M2: no change (catalyst) [1] M3: increases (temperature increase) [1]

1. Add water to the mixture and stir to dissolve the sodium chloride.
2. Filter the mixture using a filter funnel and filter paper. Barium sulfate is insoluble and remains on the filter paper as the residue.
3. Wash the residue with distilled water to remove any remaining salt solution.
4. Dry the barium sulfate residue in a warm oven or between sheets of filter paper.

M1: Add water and stir to dissolve the sodium chloride [1]
M2: Filter (to obtain barium sulfate as residue) [1]
M3: Wash the residue with (distilled) water AND dry (between filter papers / in a warm oven) [1]

1. Add excess magnesium oxide powder to dilute sulfuric acid and warm/stir until no more dissolves.
2. Filter the mixture to remove the unreacted magnesium oxide residue.
3. Heat the filtrate (magnesium sulfate solution) until it becomes a saturated solution (reaches the crystallization point).
4. Leave the hot solution to cool and form crystals, then filter to collect the crystals and dry them with filter paper.

M1: Add excess magnesium oxide to dilute sulfuric acid and warm/stir [1]
M2: Filter (to remove excess/unreacted magnesium oxide) [1]
M3: Heat filtrate to crystallization point / until saturated, allow to cool/crystallize, and dry [1]

The technique is fractional distillation.

Ethanol has a lower boiling point than water, so when the mixture is heated, ethanol evaporates first. The ethanol vapour rises up the fractionating column, enters the condenser where it cools and condenses back to liquid, and is collected as the first fraction.

M1: Fractional distillation [1]
M2: Ethanol has a lower boiling point than water, so it vaporises / distils over first [1]

To calculate the \(R_{\text{f}}\) value, the student must measure:
1. The distance from the baseline (start line) to the centre of the separated dye spot.
2. The distance from the baseline (start line) to the solvent front.

M1: Distance travelled by the dye / spot from the baseline [1]
M2: Distance travelled by the solvent front from the baseline [1]
(Note: Reject measurements not taken from the baseline / start line)

1. Calculate the mass of water lost during heating:
\( \text{Mass of H}_2\text{O} = 4.76 \text{ g} - 2.60 \text{ g} = 2.16 \text{ g} \)

2. Calculate the number of moles of anhydrous \( \text{CoCl}_2 \):
\( M_r(\text{CoCl}_2) = 59 + (35.5 \times 2) = 130 \)
\( \text{Moles of CoCl}_2 = \frac{2.60 \text{ g}}{130 \text{ g/mol}} = 0.02 \text{ mol} \)

3. Calculate the number of moles of water lost:
\( M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18 \)
\( \text{Moles of H}_2\text{O} = \frac{2.16 \text{ g}}{18 \text{ g/mol}} = 0.12 \text{ mol} \)

4. Determine the ratio of \( \text{H}_2\text{O} \) to \( \text{CoCl}_2 \):
\( y = \frac{0.12 \text{ mol}}{0.02 \text{ mol}} = 6 \)

• M1: Calculate the mass of water lost = \( 2.16 \text{ g} \) (1 mark)
• M2: Calculate the relative formula mass of \( \text{CoCl}_2 = 130 \) AND the moles of anhydrous \( \text{CoCl}_2 = 0.02 \text{ mol} \) (1 mark)
• M3: Calculate the moles of water lost = \( 0.12 \text{ mol} \) (1 mark)
• M4: Find the simplest whole number molar ratio to show \( y = 6 \) (1 mark)

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

2. Calculate the number of moles of \( \text{Li}_2\text{CO}_3 \) decomposed:
\( \text{Moles of Li}_2\text{CO}_3 = \frac{5.55 \text{ g}}{74 \text{ g/mol}} = 0.075 \text{ mol} \)

3. Use the stoichiometric ratio to find the moles of \( \text{CO}_2 \):
From the balanced equation, \( 1 \text{ mol} \) of \( \text{Li}_2\text{CO}_3 \) produces \( 1 \text{ mol} \) of \( \text{CO}_2 \).
Therefore, \( \text{Moles of CO}_2 = 0.075 \text{ mol} \).

4. Calculate the volume of \( \text{CO}_2 \) at r.t.p.:
\( \text{Volume of CO}_2 = 0.075 \text{ mol} \times 24 \text{ dm}^3\text{/mol} = 1.8 \text{ dm}^3 \)

• M1: Calculate \( M_r(\text{Li}_2\text{CO}_3) = 74 \) (1 mark)
• M2: Calculate the moles of \( \text{Li}_2\text{CO}_3 = 0.075 \text{ mol} \) (1 mark)
• M3: Deduce that the moles of \( \text{CO}_2 = 0.075 \text{ mol} \) (1 mark)
• M4: Calculate the volume of \( \text{CO}_2 = 1.8 \text{ dm}^3 \) (1 mark)

1. Express the percentages as mass in grams (per \( 100 \text{ g} \) of compound X):
\( \text{Mass of C} = 37.21 \text{ g} \)
\( \text{Mass of H} = 7.75 \text{ g} \)
\( \text{Mass of Cl} = 55.04 \text{ g} \)

2. Calculate the number of moles of each element:
\( \text{Moles of C} = \frac{37.21}{12} = 3.101 \text{ mol} \)
\( \text{Moles of H} = \frac{7.75}{1} = 7.750 \text{ mol} \)
\( \text{Moles of Cl} = \frac{55.04}{35.5} = 1.550 \text{ mol} \)

3. Divide each value by the smallest number of moles (\( 1.550 \)):
\( \text{C} = \frac{3.101}{1.550} \approx 2 \)
\( \text{H} = \frac{7.750}{1.550} = 5 \)
\( \text{Cl} = \frac{1.550}{1.550} = 1 \)

4. Write down the empirical formula:
\( \text{C}_2\text{H}_5\text{Cl} \)

• M1: Calculate the moles of Carbon \( = 3.101 \text{ mol} \) AND Hydrogen \( = 7.750 \text{ mol} \) (1 mark)
• M2: Calculate the moles of Chlorine \( = 1.550 \text{ mol} \) (1 mark)
• M3: Divide all moles by the smallest value (\( 1.550 \)) to get the relative ratio of \( 2 : 5 : 1 \) (1 mark)
• M4: Deduce the final empirical formula as \( \text{C}_2\text{H}_5\text{Cl} \) (1 mark)

The main acidic impurity in the blast furnace is silicon(IV) oxide (silica, \(\text{SiO}_2\)). This reacts with the basic calcium oxide (\(\text{CaO}\)) in a neutralisation reaction to form a liquid slag, which is calcium silicate (\(\text{CaSiO}_3\)).

Equation:
\(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\)

Product name:
Calcium silicate

M1: Correctly balanced chemical equation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\) [1]
M2: State the name 'calcium silicate' (accept 'slag') [1]

In the extraction of zinc, zinc oxide is reduced by carbon:
\(\text{ZnO} + \text{C} \rightarrow \text{Zn} + \text{CO}\)
Here, carbon gains oxygen and acts as a reducing agent.
Aluminium is higher in the reactivity series than carbon. Therefore, carbon is not a strong enough reducing agent to displace oxygen from aluminium oxide.

M1: Carbon acts as a reducing agent / reduces zinc oxide / removes oxygen from zinc oxide [1]
M2: Aluminium is more reactive than carbon (or has a stronger affinity for oxygen than carbon) [1]

In pure copper, the atoms are all of the same size and are arranged in regular layers which can easily slide over each other when a force is applied.

In brass, zinc atoms of a different size are introduced. This disrupts the regular arrangement of the layers of copper atoms, making it much more difficult for the layers to slide over one another, resulting in a harder and stronger alloy.

M1: Zinc atoms have a different size to copper atoms [1]
M2: This disrupts the regular structure/layers of atoms, preventing them from sliding over each other easily [1]

Sodium is a very reactive metal (Group I), so its carbonate is thermally stable and does not decompose upon heating (no visible change).

Copper is a less reactive metal, so copper(II) carbonate decomposes easily upon heating to form copper(II) oxide (a black solid) and carbon dioxide gas (bubbles/gas evolved):
\(\text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2\)

M1: Sodium carbonate: no change / no reaction [1]
M2: Copper(II) carbonate: green solid turns black (accept: gas produced that turns limewater milky / thermal decomposition occurs) [1]

Pure aluminium oxide (alumina) has an extremely high melting point of over \(2000^\circ\text{C}\). Dissolving it in molten cryolite:
1. Lowers the operating temperature to about \(950^\circ\text{C}\), which saves huge amounts of thermal energy and reduces costs.
2. Increases the electrical conductivity of the electrolyte.

M1: Lowers the melting point of the electrolyte / reduces energy costs [1]
M2: Increases electrical conductivity / acts as a solvent [1]

(i) Carboxylic acids have the functional group \(\text{---COOH}\). A carboxylic acid with three carbons is propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).

(ii) Esters have the functional group \(\text{---COO---}\). For a total of three carbons, the possible esters are:
- Methyl ethanoate: \(\text{CH}_3\text{COOCH}_3\)
- Ethyl methanoate: \(\text{HCOOCH}_2\text{CH}_3\)

(i) propanoic acid [1]
(ii) methyl ethanoate [1], ethyl methanoate [1] (any order)

(i) The diol part of the polyester repeating unit is \(\text{---O---CH}_2\text{---CH}_2\text{---O---}\). Restoring the hydrogen atoms to the oxygen atoms gives the diol monomer: \(\text{HO---CH}_2\text{---CH}_2\text{---OH}\), which is ethane-1,2-diol (or ethylene glycol).

(ii) The dicarboxylic acid part is \(\text{---CO---CH}_2\text{---CH}_2\text{---CO---}\). Restoring the hydroxyl groups to the carbonyl carbons gives the dicarboxylic acid monomer: \(\text{HOOC---CH}_2\text{---CH}_2\text{---COOH}\), which has four carbon atoms and is named butanedioic acid (or butane-1,4-dioic acid).

(iii) Polyester formation is a condensation polymerisation reaction, which eliminates a molecule of water (\(\text{H}_2\text{O}\)) for each ester link formed.

(i) ethane-1,2-diol / 1,2-ethanediol [1] (accept ethylene glycol)
(ii) butanedioic acid / butane-1,4-dioic acid [1]
(iii) water / \(\text{H}_2\text{O}\) [1]

(i) But-2-ene is an alkene containing a carbon-carbon double bond (\(\text{C}=\text{C}\)) which can break open to link monomer units together. Butane is an alkane and contains only single bonds (is saturated), so it cannot undergo addition polymerisation.

(ii) The monomer is \(\text{CH}_3\text{---CH}=\text{CH---CH}_3\). During polymerisation, the double bond breaks to form single bonds extending out of the carbons. The repeating unit is therefore represented as \(\text{---CH(CH}_3\text{)---CH(CH}_3\text{)---}\).

(iii) The monomer formula is \(\text{C}_4\text{H}_8\). In addition polymerisation, no atoms are lost, so the polymer has the same empirical formula as the monomer. The simplest ratio of carbon to hydrogen in \(\text{C}_4\text{H}_8\) is 1:2, which gives \(\text{CH}_2\).

(i) but-2-ene is unsaturated / has a carbon-carbon double bond / has a \(\text{C}=\text{C}\) bond (whereas butane is saturated / contains only single bonds) [1]
(ii) \(\text{---CH(CH}_3\text{)---CH(CH}_3\text{)---}\) [1]
(iii) \(\text{CH}_2\) [1]

(i) The remaining two isomers are branched-chain alcohols with a 3-carbon parent chain (propane):
- A methyl group on carbon 2 and the alcohol group on carbon 1: 2-methylpropan-1-ol.
- A methyl group on carbon 2 and the alcohol group on carbon 2: 2-methylpropan-2-ol.

(ii) Oxidation of a primary alcohol (butan-1-ol) with acidified potassium manganate(VII) yields a carboxylic acid with the same number of carbon atoms. Therefore, butan-1-ol (4 carbons) oxidises to form butanoic acid.

(i) 2-methylpropan-1-ol [1] (accept methylpropan-1-ol), 2-methylpropan-2-ol [1] (accept methylpropan-2-ol)
(ii) butanoic acid [1]