2025 Edexcel IAS-Level Chemistry (XCH11) Practice Paper with Answers

1. Calculate the mass of oxygen in the metal oxide:
\( \text{Mass of O} = 0.320\text{ g} - 0.224\text{ g} = 0.096\text{ g} \)

2. Calculate the moles of oxygen atoms:
\( n(\text{O}) = \frac{0.096\text{ g}}{16.0\text{ g mol}^{-1}} = 0.0060\text{ mol} \)

3. Determine the moles of metal \( \text{M} \) using the mole ratio in \( \text{M}_2\text{O}_3 \) (M : O = 2 : 3):
\( n(\text{M}) = n(\text{O}) \times \frac{2}{3} = 0.0060\text{ mol} \times \frac{2}{3} = 0.0040\text{ mol} \)

4. Calculate the relative atomic mass (\( A_r \)) of \( \text{M} \):
\( A_r(\text{M}) = \frac{0.224\text{ g}}{0.0040\text{ mol}} = 56.0 \)

1 mark: Correct choice of option B.

1. Look for the largest ratio jump between successive ionization energies:
- \( 1903 / 1012 = 1.88 \)
- \( 2912 / 1903 = 1.53 \)
- \( 4957 / 2912 = 1.70 \)
- \( 6274 / 4957 = 1.27 \)
- \( 21269 / 6274 = 3.39 \) (This is a very large jump!)
2. The significant jump between the 5th and 6th ionization energies shows that the 6th electron is removed from a shell closer to the nucleus.
3. This indicates that element \( \text{X} \) has 5 outer (valence) electrons and therefore belongs to Group 15.
4. Since the element is in Period 3, the Group 15 element is Phosphorus.

1 mark: Correct choice of option B.

Using VSEPR theory to determine molecular shapes:
- \( \text{SiCl}_4 \) has 4 bonding pairs and 0 lone pairs on the central Si atom, giving a tetrahedral shape.
- \( \text{H}_3\text{O}^+ \) has 3 bonding pairs and 1 lone pair on the central O atom (oxygen starts with 6 valence electrons; the +1 charge leaves 5 electrons, 3 are used for covalent single bonds to H, leaving 1 lone pair). This results in a trigonal pyramidal shape.
- \( \text{BF}_3 \) has 3 bonding pairs and 0 lone pairs on the central B atom, giving a trigonal planar shape.
- \( \text{CH}_3^+ \) has 3 bonding pairs and 0 lone pairs on the central C atom, giving a trigonal planar shape.

1 mark: Correct choice of option B.

During free-radical substitution of alkanes:
- Propagation steps must involve a radical reactant and form a radical product.
- Chlorine radicals abstract a hydrogen atom to form \( \text{HCl} \) and a butyl radical: \( \text{C}_4\text{H}_{10} + \text{Cl}^\bullet \rightarrow {}^\bullet\text{C}_4\text{H}_9 + \text{HCl} \) (Option A).
- Butyl radicals react with chlorine molecules to yield chlorobutane and regenerate a chlorine radical: \( {}^\bullet\text{C}_4\text{H}_9 + \text{Cl}_2 \rightarrow \text{C}_4\text{H}_9\text{Cl} + \text{Cl}^\bullet \) (Option B).
- Further substitution steps are also possible, such as abstracting hydrogen from a chlorobutane: \( \text{C}_4\text{H}_9\text{Cl} + \text{Cl}^\bullet \rightarrow {}^\bullet\text{C}_4\text{H}_8\text{Cl} + \text{HCl} \) (Option D).
- Option C is incorrect because highly unstable free hydrogen radicals (\( \text{H}^\bullet \)) are not formed during this mechanism.

1 mark: Correct choice of option C.

1. 2-methylbut-2-ene is \( \text{(CH}_3\text{)}_2\text{C=CHCH}_3 \).
2. Electrophilic addition begins with the addition of \( \text{H}^+ \) across the double bond.
3. If \( \text{H}^+ \) adds to C3, a tertiary carbocation is formed at C2: \( \text{(CH}_3\text{)}_2\text{C}^+-\text{CH}_2\text{CH}_3 \).
4. If \( \text{H}^+ \) adds to C2, a secondary carbocation is formed at C3: \( \text{(CH}_3\text{)}_2\text{CH}-\text{C}^+\text{HCH}_3 \).
5. Tertiary carbocations are more stable than secondary carbocations due to the greater electron-donating inductive effect of three alkyl groups compared to two.
6. The major product, 2-bromo-2-methylbutane, is formed from the more stable tertiary carbocation intermediate.

1 mark: Correct choice of option B.

1. Calculate the total mass of the reactants (or total mass of the products):
- \( M_r(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0 \)
- \( M_r(\text{NaBr}) = 23.0 + 79.9 = 102.9 \)
- \( M_r(\text{H}_2\text{SO}_4) = (2 \times 1.0) + 32.1 + (4 \times 16.0) = 98.1 \)
Total mass of reactants = \( 46.0 + 102.9 + 98.1 = 247.0 \)

2. Calculate the mass of the desired product (bromoethane):
- \( M_r(\text{C}_2\text{H}_5\text{Br}) = (2 \times 12.0) + (5 \times 1.0) + 79.9 = 108.9 \)

3. Calculate the percentage atom economy:
\( \text{Atom Economy} = \frac{108.9}{247.0} \times 100\% \approx 44.1\% \)

1 mark: Correct choice of option B.

Across Period 2, first ionization energy generally increases due to increasing nuclear charge and decreasing atomic radius with similar shielding. However, there are two distinct exceptions:
1. Boron (\( 1\text{s}^2 2\text{s}^2 2\text{p}^1 \)) has a lower first ionization energy than beryllium (\( 1\text{s}^2 2\text{s}^2 \)) because the outer electron in boron is removed from a higher-energy \( 2\text{p} \) subshell that is more shielded than the \( 2\text{s} \) subshell in beryllium. Thus, \( \text{B} < \text{Be} \).
2. Oxygen (\( 1\text{s}^2 2\text{s}^2 2\text{p}^4 \)) has a lower first ionization energy than nitrogen (\( 1\text{s}^2 2\text{s}^2 2\text{p}^3 \)) due to the spin-pair repulsion in oxygen's filled \( 2\text{p} \) orbital, which makes it easier to remove one of these paired electrons compared to nitrogen's stable, half-filled \( 2\text{p} \) subshell. Thus, \( \text{O} < \text{N} \).

Combining these facts together with the overall period trend gives the correct order:
\( \text{B} < \text{Be} < \text{O} < \text{N} \).

1 mark: Correct choice of option A.

A molecule is polar if it contains polar bonds and is asymmetrical so that the bond dipole vectors do not cancel each other out.
- \( \text{CF}_4 \) is tetrahedral and highly symmetrical; the four polar C-F bonds cancel out perfectly (non-polar).
- \( \text{CO}_2 \) is linear and symmetrical; the two polar C=O bonds cancel out perfectly (non-polar).
- \( \text{SF}_6 \) is octahedral and highly symmetrical; all six polar S-F bonds cancel out (non-polar).
- \( \text{SF}_4 \) has 4 bonding pairs and 1 lone pair on the sulfur atom, resulting in a see-saw molecular geometry. Because of the asymmetrical shape, the individual S-F dipole moments do not cancel, giving the molecule a net dipole moment.

1 mark: Correct choice of option C.

To calculate the atom economy:
$$\text{Atom Economy} = \frac{\text{Mass of desired product}}{\text{Total mass of all reactants}} \times 100$$
$$\text{Total mass of reactants} = 46.0 + 102.9 + 98.1 = 247.0\text{ g mol}^{-1}$$
$$\text{Mass of desired product (bromoethane)} = 108.9\text{ g mol}^{-1}$$
$$\text{Atom Economy} = \frac{108.9}{247.0} \times 100 \approx 44.1\%$$
Therefore, the correct option is A.

[1] Correct option is A.
- Award 1 mark for correct calculation showing 44.1%.
- Reject option B (51.4%) which might arise from incorrect reactant total.
- Reject option C (83.1%) and D (100%).

The successive ionization energies show a very large increase between the 3rd and 4th ionization energies ($2745$ to $11577\text{ kJ mol}^{-1}$). This indicates that the fourth electron is being removed from a lower, filled electron shell closer to the nucleus. Therefore, element $X$ has 3 valence electrons and forms a stable $X^{3+}$ ion. Its oxide formula will be $X_2\text{O}_3$ to balance the $O^{2-}$ charges. Thus, the correct option is B.

[1] Correct option is B.
- Award 1 mark for identifying the stable oxide formula as $X_2\text{O}_3$ based on the large jump between the 3rd and 4th ionization energies.

Let's analyze the shapes and bond angles of each species:
- $\text{NH}_4^+$ has 4 bonding pairs and 0 lone pairs. It is tetrahedral with a bond angle of $109.5^\circ$.
- $\text{H}_3\text{O}^+$ has 3 bonding pairs and 1 lone pair. The electron geometry is tetrahedral but the molecular shape is trigonal pyramidal. The lone pair repels the bonding pairs more than they repel each other, reducing the bond angle from $109.5^\circ$ to approximately $107^\circ$.
- $\text{CO}_2$ has 2 double bonds and 0 lone pairs. It is linear with a bond angle of $180^\circ$.
- $\text{BF}_3$ has 3 bonding pairs and 0 lone pairs. It is trigonal planar with a bond angle of $120^\circ$.
Therefore, the correct option is B.

[1] Correct option is B.
- Award 1 mark for selecting the species with a trigonal pyramidal shape and $107^\circ$ bond angle.

In free-radical substitution mechanisms:
- Initiation involves the homolytic fission of a covalent bond by UV light to produce radicals: $\text{Cl}_2 \rightarrow 2\text{Cl}\cdot$ (Option B is initiation).
- Propagation steps involve a radical reacting with a stable molecule to generate a new radical and a new molecule: $\text{CH}_4 + \text{Cl}\cdot \rightarrow \text{CH}_3\cdot + \text{HCl}$ (Option A is propagation).
- Termination steps involve two radicals combining to form a stable molecule: $\text{CH}_3\cdot + \text{Cl}\cdot \rightarrow \text{CH}_3\text{Cl}$ (Option C is termination) and $\text{CH}_3\cdot + \text{CH}_3\cdot \rightarrow \text{C}_2\text{H}_6$ (Option D is termination).
Therefore, the correct option is A.

[1] Correct option is A.
- Award 1 mark for identifying the correct propagation step equation.

For an alkene to exhibit geometric ($E$/$Z$) isomerism, each carbon atom of the double bond ($\text{C}=\text{C}$) must be bonded to two different groups.
- A: In 2-methylbut-2-ene, carbon-2 is bonded to two methyl groups ($-\text{CH}_3$), so it does not exhibit geometric isomerism.
- B: In pent-2-ene, carbon-2 is bonded to $-\text{H}$ and $-\text{CH}_3$, and carbon-3 is bonded to $-\text{H}$ and $-\text{CH}_2\text{CH}_3$. Since both carbon atoms have two different groups attached, it exhibits geometric isomerism.
- C: In 1,1-dichloroprop-1-ene, carbon-1 is bonded to two chlorine atoms ($-\text{Cl}$), so it does not exhibit geometric isomerism.
- D: In 2-methylpent-1-ene, carbon-1 is bonded to two hydrogen atoms ($-\text{H}$), so it does not exhibit geometric isomerism.
Therefore, the correct option is B.

[1] Correct option is B.
- Award 1 mark for identifying pent-2-ene as the correct alkene showing $E$/$Z$ isomerism.

Using the ideal gas equation: $PV = nRT$
Convert units to SI units:
- $P = 101\text{ kPa} = 1.01 \times 10^5\text{ Pa}$
- $V = 247\text{ cm}^3 = 247 \times 10^{-6}\text{ m}^3 = 2.47 \times 10^{-4}\text{ m}^3$
- $T = 300\text{ K}$
- $R = 8.31\text{ J mol}^{-1}\text{ K}^{-1}$

Calculate the number of moles ($n$):
$$n = \frac{PV}{RT} = \frac{1.01 \times 10^5 \times 2.47 \times 10^{-4}}{8.31 \times 300} = \frac{24.947}{2493} \approx 0.0100\text{ mol}$$

Now find the molar mass ($M$):
$$M = \frac{\text{mass}}{n} = \frac{0.581\text{ g}}{0.0100\text{ mol}} = 58.1\text{ g mol}^{-1}$$
Therefore, the relative molecular mass is 58.1. The correct option is C.

[1] Correct option is C.
- Award 1 mark for the correct conversion of units and calculation leading to $M_r = 58.1$.
- Reject A (29.1) which results from a factor of 2 error.
- Reject B (44.0).
- Reject D (116).

A neutral copper atom has the electronic configuration:
$$\text{Cu}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$$
When forming a $2+$ ion, copper first loses its outer $4s$ electron and then one of its $3d$ electrons.
This gives:
$$\text{Cu}^{2+}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^9$$
Therefore, the correct option is A.

[1] Correct option is A.
- Award 1 mark for the correct electronic configuration of $\text{Cu}^{2+}$ showing $3d^9$ and no $4s$ electrons.

Bond polarity is determined by the difference in electronegativity between the two bonded atoms. Fluorine is the most electronegative element in the periodic table (electronegativity of 4.0), and carbon has an electronegativity of 2.5. The electronegativity difference for the $\text{C}-\text{F}$ bond is 1.5, which is significantly greater than that for $\text{C}-\text{H}$, $\text{C}-\text{Cl}$, or $\text{C}-\text{O}$ bonds. Therefore, the $\text{C}-\text{F}$ bond is the most polar of the choices. Thus, the correct option is C.

[1] Correct option is C.
- Award 1 mark for identifying the $\text{C}-\text{F}$ bond as the most polar bond due to the highest difference in electronegativity.

The successive ionization energies show a massive increase between the 3rd and 4th ionization energies (from \(2745\text{ kJ mol}^{-1}\) to \(11578\text{ kJ mol}^{-1}\)). This indicates that the fourth electron is being removed from an inner shell that is closer to the nucleus and less shielded, meaning the element has three electrons in its outer shell (Group 13). Since the element is in Period 3, it must be aluminium.

1 mark for the correct option (C). No partial marks.

1. Calculate the mass of water lost:
\(5.74\text{ g} - 3.22\text{ g} = 2.52\text{ g}\)

2. Calculate the amount of anhydrous \(\text{ZnSO}_4\) in moles:
\(n(\text{ZnSO}_4) = \frac{3.22\text{ g}}{161.5\text{ g mol}^{-1}} = 0.01994\text{ mol}\)

3. Calculate the amount of water in moles:
\(n(\text{H}_2\text{O}) = \frac{2.52\text{ g}}{18.0\text{ g mol}^{-1}} = 0.140\text{ mol}\)

4. Determine the molar ratio \(x\):
\(x = \frac{0.140\text{ mol}}{0.01994\text{ mol}} \approx 7\)

1 mark for the correct option (C). No partial marks.

The hydronium ion, \(\text{H}_3\text{O}^+\), has three bonding pairs and one lone pair of electrons on the central oxygen atom. According to VSEPR theory, these four areas of electron density adopt a tetrahedral arrangement to minimize repulsion. The lone pair repels the bonding pairs more strongly than the bonding pairs repel each other, reducing the bond angle from the standard tetrahedral angle of \(109.5^\circ\) to approximately \(107^\circ\), resulting in a trigonal pyramidal shape.

- \(\text{NH}_4^+\) has four bonding pairs and no lone pairs, so it has tetrahedral geometry with a bond angle of \(109.5^\circ\).
- \(\text{BF}_3\) has three bonding pairs and no lone pairs, so it has trigonal planar geometry with a bond angle of \(120^\circ\).
- \(\text{CO}_2\) is linear with a bond angle of \(180^\circ\).

1 mark for the correct option (B). No partial marks.

1. **Stereoisomerism**: But-2-ene has two different groups attached to each carbon of the double bond (a hydrogen and a methyl group), so it exhibits \(E/Z\) isomerism. But-1-ene and 2-methylpropene do not show stereoisomerism because at least one carbon of the double bond is bonded to two identical atoms/groups.

2. **Reaction with HCl**: But-2-ene is symmetrical, so addition of \(\text{HCl}\) across either side of the double bond yields only one product: 2-chlorobutane. Pent-2-ene also exhibits \(E/Z\) isomerism but is asymmetrical, so it would react with \(\text{HCl}\) to give a mixture of 2-chloropentane and 3-chloropentane.

1 mark for the correct option (B). No partial marks.

(a)(i) For ionization: The sample is vaporised and either bombarded with high-energy electrons from an electron gun (electron impact ionization) to knock off an electron: \(\text{X}(g) \to \text{X}^+(g) + e^-\), or dissolved in a volatile solvent and injected through a fine hypodermic needle connected to a high voltage supply (electrospray ionization) to gain a proton: \(\text{X}(g) + \text{H}^+ \to \text{XH}^+(g)\).
For acceleration: The resulting positive ions are accelerated by an electric field / plates with a potential difference, giving all ions the same kinetic energy.

(a)(ii) \(\text{Relative Atomic Mass (RAM)} = \frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100} = \frac{1895.76 + 250.00 + 286.26}{100} = 24.32\).
An element with RAM of 24.32 is magnesium (Mg).

(b) Magnesium has atomic number 12. \(\text{Mg}^{2+}\) has 10 electrons: \(1s^2 2s^2 2p^6\).

(c)(i) The energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions.

(ii) \(\text{Mg}^{2+}(g) \to \text{Mg}^{3+}(g) + e^-\)

(iii) The first two electrons in magnesium are removed from the outer 3s subshell. The third electron is removed from the 2p subshell, which is an inner shell closer to the nucleus. This electron experiences significantly less shielding from inner shells and a much stronger attraction to the nucleus, requiring a much larger amount of energy to be removed.

(a)(i) Max 4 marks:
* M1: Vaporised sample is bombarded with high energy electrons / electron gun (for electron impact) OR dissolved in a solvent and forced through a high voltage needle (for electrospray) [1]
* M2: Equation: \(\text{X}(g) \to \text{X}^+(g) + e^-\)
OR \(\text{X}(g) + \text{H}^+ \to \text{XH}^+(g)\) [1]
* M3: Ions accelerated by an electric field / high voltage plates [1]
* M4: All ions accelerated to have the same kinetic energy [1]

(a)(ii) Max 3 marks:
* M1: Correct setup for RAM calculation: \(\frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100}\) [1]
* M2: RAM value calculated to 2 decimal places: 24.32 [1]
* M3: Identification of element: Magnesium / Mg (consequent on RAM value near 24.3) [1]

(b) Max 1 mark:
* M1: \(1s^2 2s^2 2p^6\) [1]

(c)(i) Max 3 marks:
* M1: Energy required to remove 1 mole of electrons from 1 mole of atoms [1]
* M2: Both species in the gaseous state [1]
* M3: To form 1 mole of gaseous 1+ ions [1]

(c)(ii) Max 1 mark:
* M1: \(\text{X}^{2+}(g) \to \text{X}^{3+}(g) + e^-\)\ (accept Mg) with correct state symbols [1]

(c)(iii) Max 3 marks:
* M1: The 3rd electron is removed from a shell closer to the nucleus / a new shell / 2p shell (whereas the first two are from 3s) [1]
* M2: The 3rd electron experiences less shielding [1]
* M3: Hence there is a stronger nuclear attraction on the 3rd electron [1]

(a)(i) First, convert units to SI:
\(p = 101\text{ kPa} = 101 \times 10^3\text{ Pa}\)
\(V = 412\text{ cm}^3 = 412 \times 10^{-6}\text{ m}^3\)
\(T = 298\text{ K}\)
Using \(pV = nRT\):
\(n = \frac{pV}{RT} = \frac{101000 \times 412 \times 10^{-6}}{8.31 \times 298} = \frac{41.612}{2476.38} = 0.0168036\text{ mol}\)
To 4 significant figures, \(n = 0.01680\text{ mol}\).

(a)(ii) Since the ratio of \(\text{MCO}_3\) to \(\text{CO}_2\) is 1:1, moles of \(\text{MCO}_3\) decomposed = \(0.0168036\text{ mol}\).
\(\text{Molar mass of } \text{MCO}_3 = \frac{\text{mass}}{\text{moles}} = \frac{2.48\text{ g}}{0.0168036\text{ mol}} = 147.59\text{ g mol}^{-1}\) (or \(147.6\text{ g mol}^{-1}\) using rounded figures).

(a)(iii) \(M_r(\text{MCO}_3) = A_r(\text{M}) + 12.0 + 3(16.0) = 147.6\)
\(A_r(\text{M}) + 60.0 = 147.6\)
\(A_r(\text{M}) = 87.6\)
Consulting the Periodic Table, the Group 2 metal with an atomic mass of approximately 87.6 is Strontium (\(\text{Sr}\)).

(a)(iv) A likely experimental error is the loss of carbon dioxide gas between adding the metal carbonate to the test tube and inserting the rubber stopper. This leads to a lower measured volume of \(\text{CO}_2\), resulting in a lower calculated value of moles, which in turn leads to a higher calculated molar mass of \(\text{MCO}_3\).

(b)(i) \(\text{Moles of } \text{HCl} = C \times V = 0.100\text{ mol dm}^{-3} \times \frac{21.40}{1000}\text{ dm}^3 = 2.14 \times 10^{-3}\text{ mol}\) (or \(0.00214\text{ mol}\)).

(b)(ii) According to the equation, 1 mole of \(\text{X}_2\text{CO}_3\) reacts with 2 moles of \(\text{HCl}\).
Moles of \(\text{X}_2\text{CO}_3\) in \(25.0\text{ cm}^3 = \frac{2.14 \times 10^{-3}}{2} = 1.07 \times 10^{-3}\text{ mol}\).
Moles of \(\text{X}_2\text{CO}_3\) in the original \(250\text{ cm}^3\) volumetric flask = \(1.07 \times 10^{-3} \times 10 = 1.07 \times 10^{-2}\text{ mol}\) (or \(0.0107\text{ mol}\)).

(b)(iii) \(\text{Molar mass of } \text{X}_2\text{CO}_3 = \frac{\text{mass}}{\text{moles}} = \frac{1.48\text{ g}}{0.0107\text{ mol}} = 138.32\text{ g mol}^{-1}\).
\(2 \times A_r(\text{X}) + 12.0 + 3(16.0) = 138.32\)
\(2 \times A_r(\text{X}) + 60.0 = 138.32\)
\(2 \times A_r(\text{X}) = 78.32\)
\(A_r(\text{X}) = 39.16\)
Consulting the Periodic Table, the metal X is Potassium (\(\text{K}\)), which has \(A_r = 39.1\).

(a)(i) Max 3 marks:
* M1: Correctly converting volume into \(\text{m}^3\) (\(412 \times 10^{-6}\)) and pressure to \(\text{Pa}\) (\(101 \times 10^3\)) [1]
* M2: Rearrangement of Ideal Gas Equation: \(n = \frac{pV}{RT}\) [1]
* M3: Evaluated to 0.01680 (must be 4 sig figs, accept 0.0168) [1]

(a)(ii) Max 2 marks:
* M1: Moles of carbonate = moles of gas (1:1 ratio used) [1]
* M2: \(M_r = 2.48 / \text{moles of gas} = 147.6\text{ g mol}^{-1}\) (allow consequential marking on M1 value) [1]

(a)(iii) Max 2 marks:
* M1: Calculation of atomic weight of M: \(A_r = M_r - 60.0 = 87.6\) (consequential on part ii) [1]
* M2: Correctly identifies Strontium / Sr [1]

(a)(iv) Max 2 marks:
* M1: Gas escapes before the bung is inserted / gas syringe sticks [1]
* M2: Volume of gas decreases, so calculated moles decrease, causing calculated molar mass to increase [1]

(b)(i) Max 1 mark:
* M1: \(0.100 \times 0.02140 = 2.14 \times 10^{-3}\text{ mol}\) [1]

(b)(ii) Max 1 mark:
* M1: \(\text{Moles of } \text{X}_2\text{CO}_3 = \frac{2.14 \times 10^{-3}}{2} \times 10 = 1.07 \times 10^{-2}\text{ mol}\) [1]

(b)(iii) Max 4 marks:
* M1: Calculates molar mass of \(\text{X}_2\text{CO}_3\) as \(\frac{1.48}{1.07 \times 10^{-2}} = 138.3\text{ g mol}^{-1}\) [1]
* M2: Sets up equation for X: \(2X + 60.0 = 138.3\) [1]
* M3: Calculates \(A_r(\text{X}) = 39.15\) [1]
* M4: Identifies X as Potassium / K [1]

(a)(i) In the dot-and-cross diagram of \(\text{PCl}_3\):
* The phosphorus atom (Group 5) has 5 outer shell electrons. It shares 1 electron with each of the three chlorine atoms, leaving one lone pair on the phosphorus.
* Each chlorine atom (Group 7) shares 1 electron with phosphorus and has 3 lone pairs (6 non-bonding electrons) in its outer shell.

(a)(ii) Shape: Trigonal pyramidal.
Bond angle: \(107^\circ\) (accept range \(100^\circ - 108^\circ\)).
Explanation:
* There are 4 regions of electron density / 4 electron pairs around the central phosphorus atom (3 bonding pairs and 1 lone pair).
* To minimise repulsion, these electron pairs arrange themselves in a tetrahedral arrangement.
* Repulsion between lone pairs and bonding pairs is stronger than the repulsion between bonding pairs. This pushes the three P-Cl bonds closer together, reducing the bond angle from the tetrahedral angle of \(109.5^\circ\) to \(107^\circ\).

(a)(iii) Shape: Trigonal bipyramidal.
Bond angles: \(90^\circ\) and \(120^\circ\) (and \(180^\circ\)).
Explanation of bonding: Phosphorus is in Period 3 and has access to empty, low-energy 3d orbitals, allowing it to expand its octet to accommodate 10 outer-shell electrons.

(b) Chlorine is more electronegative than phosphorus, so each P-Cl bond is polar, with the dipole directed towards the chlorine (\(\text{P}^{\delta+} - \text{Cl}^{\delta-}\)).
* In \(\text{PCl}_3\), the molecule has an asymmetric trigonal pyramidal shape with a lone pair on the phosphorus. The individual P-Cl bond dipoles do not cancel out, resulting in a net molecular dipole.
* In \(\text{PCl}_5\), the molecule has a highly symmetrical trigonal bipyramidal shape. The individual P-Cl dipoles cancel each other out completely, resulting in a non-polar molecule with no net molecular dipole.

(c) Both \(\text{PCl}_3\) and \(\text{PF}_3\) possess both London (dispersion) forces and permanent dipole-dipole forces.
* Chlorine atoms have significantly more electrons than fluorine atoms, meaning \(\text{PCl}_3\) has 56 electrons compared to the 42 electrons in \(\text{PF}_3\).
* Therefore, \(\text{PCl}_3\) has much stronger London forces which require more energy to break, overriding the stronger dipole-dipole forces in the more polar \(\text{PF}_3\).

(a)(i) Max 2 marks:
* M1: Three shared pairs of electrons between P and each Cl [1]
* M2: Correct number of non-bonding electrons shown (one lone pair on P, three lone pairs on each Cl) [1]

(a)(ii) Max 4 marks:
* M1: Shape: Trigonal pyramidal AND Bond angle: \(107^\circ\) (accept \(100^\circ - 108^\circ\)) [1]
* M2: 4 pairs of electrons (3 bonding pairs, 1 lone pair) around phosphorus [1]
* M3: Electron pairs repel to get as far apart as possible / minimise repulsion [1]
* M4: Lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, compressing the bond angle [1]

(a)(iii) Max 3 marks:
* M1: Shape: Trigonal bipyramidal [1]
* M2: Angles: \(90^\circ\) and \(120^\circ\) (both required) [1]
* M3: Octet expansion is possible because phosphorus is in Period 3 / can access d-orbitals [1]

(b) Max 4 marks:
* M1: Chlorine is more electronegative than phosphorus, making the P-Cl bond polar [1]
* M2: \(\text{PCl}_3\) is asymmetric / has a lone pair, so the polar bonds do not cancel out [1]
* M3: Leaving a net molecular dipole (permanent dipole) [1]
* M4: \(\text{PCl}_5\) is highly symmetrical, so the bond dipoles cancel, resulting in no net dipole [1]

(c) Max 2 marks:
* M1: Both have London forces AND permanent dipole-dipole forces [1]
* M2: \(\text{PCl}_3\) has more electrons, leading to stronger London forces that override the stronger dipole-dipole forces in \(\text{PF}_3\) [1]

(a)(i) Saturated: contains only single carbon-carbon bonds.
Hydrocarbon: a compound containing hydrogen and carbon only.

(a)(ii) Initiation step: \(\text{Cl}_2 \xrightarrow{UV} 2\text{Cl}^\bullet\)
Type of bond fission: Homolytic fission.

(a)(iii) Propagation step 1: \(\text{CH}_3\text{CH}_2\text{CH}_3 + \text{Cl}^\bullet \to \text{CH}_3\dot{\text{C}}\text{HCH}_3 + \text{HCl}\)
Propagation step 2: \(\text{CH}_3\dot{\text{C}}\text{HCH}_3 + \text{Cl}_2 \to \text{CH}_3\text{CHClCH}_3 + \text{Cl}^\bullet\)

(a)(iv) During the reaction, some propyl radicals (\(\text{CH}_3\text{CH}_2\text{CH}_2^\bullet\)) are formed. In a termination step, two propyl radicals can collide and combine to form a molecule of hexane:
\(2\text{CH}_3\text{CH}_2\text{CH}_2^\bullet \to \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3\) (or \(2\text{C}_3\text{H}_7^\bullet \to \text{C}_6\text{H}_{14}\)).

(b)(i) Geometric (E/Z) isomerism requires restricted rotation about the C=C double bond, and that both carbon atoms in the double bond are attached to two different groups.
* In but-2-ene, both carbon atoms have two different groups attached (\(\text{-H}\) and \(\text{-CH}_3\)).
* In propene, one of the carbon atoms in the double bond is attached to two identical groups (two \(\text{-H}\) atoms), meaning stereoisomers cannot exist.

(b)(ii) Name of major product: 2-bromopropane.

Mechanism:
1. Electrophilic addition: Propene reacts with polar \(\text{H}^{\delta+}\text{-Br}^{\delta-}\).
2. A curly arrow is drawn starting from the C=C double bond to the \(\text{H}\) atom of \(\text{H-Br}\).
3. A curly arrow is drawn starting from the bond in \(\text{H-Br}\) to the \(\text{Br}\) atom.
4. This forms a secondary carbocation intermediate: \(\text{CH}_3\text{C}^+\text{HCH}_3\) and a bromide ion (\(\text{Br}^-\)).
5. A curly arrow is drawn from a lone pair on the \(\text{Br}^-\)[-] to the positively charged carbon atom of the carbocation, giving 2-bromopropane.

Explanation of stability:
The reaction proceeds via a secondary carbocation intermediate (\(\text{CH}_3\text{C}^+\text{HCH}_3\)) which is more stable than the alternative primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2^+\)). This is because the two electron-releasing alkyl (methyl) groups reduce the positive charge density on the carbocation carbon (positive inductive effect), lowering the activation energy for its formation.

(a)(i) Max 2 marks:
* M1: Saturated: contains only single C-C bonds (no C=C double bonds) [1]
* M2: Hydrocarbon: compound containing carbon and hydrogen only [1]

(a)(ii) Max 2 marks:
* M1: \(\text{Cl}_2 \to 2\text{Cl}^\bullet\) (with UV over arrow or in text) [1]
* M2: Homolytic fission [1]

(a)(iii) Max 2 marks:
* M1: \(\text{CH}_3\text{CH}_2\text{CH}_3 + \text{Cl}^\bullet \to \text{CH}_3\dot{\text{C}}\text{HCH}_3 + \text{HCl}\) [1]
* M2: \(\text{CH}_3\dot{\text{C}}\text{HCH}_3 + \text{Cl}_2 \to \text{CH}_3\text{CHClCH}_3 + \text{Cl}^\bullet\) [1]
(Accept structural or skeletal representations showing radical on carbon 2)

(a)(iv) Max 2 marks:
* M1: Propyl radicals colliding/combining (termination step) [1]
* M2: \(2\text{C}_3\text{H}_7^\bullet \to \text{C}_6\text{H}_{14}\) (or structural equivalent) [1]

(b)(i) Max 2 marks:
* M1: C=C double bond restricts rotation [1]
* M2: Propene has one C atom bonded to two identical groups (H atoms) whereas but-2-ene has each C bonded to different groups (H and methyl) [1]

(b)(ii) Max 5 marks:
* M1: Correct name: 2-bromopropane [1]
* M2: Curly arrow from C=C bond to \(\text{H}\) of \(\text{H-Br}\) AND dipole on \(\text{H-Br}\) (\(\delta+\) on H, \(\delta-\) on Br) AND curly arrow from H-Br bond to Br [1]
* M3: Structure of secondary carbocation (\(\text{CH}_3\text{C}^+\text{HCH}_3\)) AND \(\text{Br}^-\)[1]
* M4: Curly arrow from lone pair of \(\text{Br}^-\) to the carbocation carbon [1]
* M5: Explanation: Secondary carbocation is more stable than primary because of the inductive effect of two electron-releasing alkyl groups [1]

Boiling temperature depends on the strength of intermolecular forces. All four isomers are polar and have London forces and permanent dipole-dipole forces. 1-chlorobutane is a straight-chain molecule, meaning it has the largest surface area of contact between molecules compared to its branched isomers (2-chlorobutane, 1-chloro-2-methylpropane, and 2-chloro-2-methylpropane). This results in stronger London forces, requiring more energy to overcome, giving it the highest boiling temperature.

1 mark: Correct option (A) identified. No partial marks.

The equation for the formation of propane is: \(3\text{C}(s) + 4\text{H}_2(g) \to \text{C}_3\text{H}_8(g)\). Using a Hess cycle based on combustion data: \(\Delta H^\theta_f = \sum \Delta H^\theta_c(\text{reactants}) - \sum \Delta H^\theta_c(\text{products})\). Therefore, \(\Delta H^\theta_f = 3(-394) + 4(-286) - (-2220) = -1182 - 1144 + 2220 = -106\text{ kJ mol}^{-1}\).

1 mark: Correct option (A) identified. No partial marks.

As you go down Group 2, the cationic radius increases. Because the ionic charge remains the same (+2), the charge density of the cation decreases. Consequently, the larger cation has less polarising power, causing less distortion of the electron cloud of the carbonate ion. The C-O bond in the carbonate ion is therefore weakened to a lesser extent, making the compound more thermally stable.

1 mark: Correct option (A) identified. No partial marks.

The rate of hydrolysis of halogenoalkanes is determined by the bond enthalpy of the carbon-halogen (C-X) bond, not by its polarity. Bond strength decreases in the order C-Cl > C-Br > C-I due to the increasing atomic size of the halogen, which leads to longer and weaker bonds. The weaker the bond, the more easily it is broken during nucleophilic substitution. Therefore, 1-chlorobutane reacts the slowest and 1-iodobutane reacts the fastest.

1 mark: Correct option (B) identified. No partial marks.

An increase in temperature means that the average kinetic energy of the particles increases. In the Maxwell-Boltzmann distribution curve, this shifts the peak of the curve to the right (higher energy) and downwards (lower peak height, since the total area under the curve must remain constant, representing the constant total number of particles).

1 mark: Correct option (D) identified. No partial marks.

Potassium iodide is a strong reducing agent and reduces concentrated sulfuric acid through several stages. Sulfur dioxide (SO2) is a colourless gas; hydrogen sulfide (H2S) is a colourless toxic gas with a rotten egg smell; iodine (I2) is a grey-black solid or purple vapour; elemental sulfur (S) is formed as a yellow solid.

1 mark: Correct option (C) identified. No partial marks.

The strong absorption at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C}=\text{O}\)). The absence of an absorption band in the \(3200\text{--}3600\text{ cm}^{-1}\) range indicates the absence of an alcohol group (\(\text{O}-\text{H}\)), eliminating butan-1-ol and but-3-en-1-ol. The absence of a band in the \(2500\text{--}3300\text{ cm}^{-1}\) range indicates the absence of a carboxylic acid group, eliminating butanoic acid. Therefore, the compound is the ketone, butanone.

1 mark: Correct option (C) identified. No partial marks.

The heat released is \(q = m c \Delta T\). Total mass of solution \(m = 50.0 + 50.0 = 100.0\text{ g}\). So, \(q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 6.4\text{ }^\circ\text{C} = 2675.2\text{ J} = 2.6752\text{ kJ}\). The number of moles of water formed is \(n = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). The enthalpy of neutralisation is \(\Delta H_{\text{neut}} = -\frac{q}{n} = -\frac{2.6752\text{ kJ}}{0.0500\text{ mol}} = -53.5\text{ kJ mol}^{-1}\).

1 mark: Correct option (B) identified. No partial marks.

Using Hess's Law, the enthalpy change of formation of ethanol is calculated using standard enthalpies of combustion:
\(\Delta H_f^{\ominus} = \sum \Delta H_c^{\ominus}(\text{reactants}) - \sum \Delta H_c^{\ominus}(\text{products})\)

For the reaction:
\(2\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH(l)}\)

\(\Delta H_f^{\ominus} = 2 \times \Delta H_c^{\ominus}(\text{C}) + 3 \times \Delta H_c^{\ominus}(\text{H}_2) - \Delta H_c^{\ominus}(\text{C}_2\text{H}_5\text{OH})\)
\(\Delta H_f^{\ominus} = 2(-393.5) + 3(-285.8) - (-1367.3)\)
\(\Delta H_f^{\ominus} = -787.0 - 857.4 + 1367.3 = -277.1\text{ kJ mol}^{-1}\)

1 mark: Correctly calculates \(-277.1\text{ kJ mol}^{-1}\) using the Hess cycle.
- Award [1] for option A.
- Incorrect sign, stoichiometric multipliers, or wrong cycle direction lead to other options.

Butan-1-ol has the highest boiling temperature because it can form hydrogen bonds between its molecules. Comparing the two alcohols, butan-1-ol is straight-chained, giving it a larger surface contact area and stronger London forces than the highly branched 2-methylpropan-2-ol. Butanal and ethoxyethane do not contain an \(\text{O-H}\) bond, so they cannot form intermolecular hydrogen bonds.

1 mark: Correctly identifies butan-1-ol as having the highest boiling point.
- Award [1] for option A.
- Reject other options because of either absence of hydrogen bonding or branching effects.

The reaction of chlorine with hot, concentrated sodium hydroxide is:
\(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\)
In \(\text{NaCl}\), the oxidation state of chlorine is \(-1\).
In \(\text{NaClO}_3\), the oxidation state of chlorine is calculated as: \(+1 + x + 3(-2) = 0 \Rightarrow x = +5\).
Therefore, the oxidation states of chlorine in the products are \(-1\) and \(+5\).

1 mark: Correctly identifies the two oxidation states of chlorine in the products as \(-1\) and \(+5\).
- Award [1] for option B.
- Option A corresponds to reaction with cold, dilute NaOH (which yields \(-1\) and \(+1\)).

The molecular formula \(\text{C}_3\text{H}_6\text{O}\) indicates one degree of unsaturation (a ring or double bond). The strong, sharp peak at \(1715\text{ cm}^{-1}\) is characteristic of a carbonyl group, \(\text{C=O}\). The absence of a broad absorption in the \(3200-3600\text{ cm}^{-1}\) range indicates that there is no alcohol \(\text{O-H}\) group. Therefore, the compound must be a carbonyl compound with formula \(\text{C}_3\text{H}_6\text{O}\), which matches propanone.

1 mark: Correctly identifies the compound as propanone.
- Award [1] for option C.
- Reject option A and B (wrong molecular formulas, contain \(\text{O-H}\)).
- Reject option D (wrong molecular formula).

A catalyst provides an alternative pathway with a lower activation energy, which effectively shifts the activation energy barrier, \(E_a\), to the left (decreases it). The catalyst does not alter the energy distribution of the reacting molecules, so the Maxwell-Boltzmann distribution curve itself remains completely unchanged.

1 mark: Correctly identifies that the curve is unchanged and the activation energy shifts to the left.
- Award [1] for option B.

Down Group 2 from magnesium to barium:
1. The solubility of Group 2 hydroxides increases (magnesium hydroxide is sparingly soluble, while barium hydroxide is highly soluble).
2. The thermal stability of Group 2 carbonates increases because the ionic radius of the cation increases down the group, decreasing its charge density and polarizing power, making it less able to polarize and weaken the carbonate ion's C-O bond.

1 mark: Identifies that hydroxides become more soluble and carbonates become more thermally stable down the group.
- Award [1] for option A.

1. Calculate mass of solution:
\(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ g}\)

2. Calculate heat energy released (\(q\)):
\(q = mc\Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 5.8\text{ K} = 2424.4\text{ J} = 2.4244\text{ kJ}\)

3. Calculate moles of water formed:
\(n(\text{H}_2\text{O}) = n(\text{H}^+) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)

4. Calculate \(\Delta H_{neut}\):
\(\Delta H_{neut} = -\frac{2.4244\text{ kJ}}{0.0500\text{ mol}} = -48.488\text{ kJ mol}^{-1} \approx -48.5\text{ kJ mol}^{-1}\)

1 mark: Correctly calculates \(-48.5\text{ kJ mol}^{-1}\).
- Award [1] for option B.
- Option A is obtained by dividing by total volume or incorrect mole calculation.
- Option C is obtained by using only 50 g of solution.
- Option D has the incorrect sign.

1. The rate of hydrolysis of halogenoalkanes is determined by the strength of the carbon-halogen bond. The C-I bond is weaker than the C-Cl bond, meaning iodoalkanes react faster than chloroalkanes.
2. Tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react much faster than primary halogenoalkanes (such as 1-iodobutane) because tertiary halogenoalkanes react via the \(S_N1\) mechanism, which involves a stable tertiary carbocation intermediate, whereas primary halogenoalkanes react via the slower \(S_N2\) mechanism.
Therefore, 2-iodo-2-methylpropane reacts fastest, via the \(S_N1\) mechanism.

1 mark: Identifies 2-iodo-2-methylpropane and the \(S_N1\) mechanism.
- Award [1] for option D.

To calculate the standard enthalpy change of reaction from enthalpy changes of combustion, we use the Hess's Law relationship:

\(\Delta_r H^\ominus = \sum \Delta_c H^\ominus(\text{Reactants}) - \sum \Delta_c H^\ominus(\text{Products})\)

Substituting the given values:

\(\Delta_r H^\ominus = [\Delta_c H^\ominus(C_3H_6(g)) + \Delta_c H^\ominus(H_2(g))] - [\Delta_c H^\ominus(C_3H_8(g))]\)

\(\Delta_r H^\ominus = [(-2058) + (-286)] - [-2220]\)

\(\Delta_r H^\ominus = -2344 + 2220 = -124 \text{ kJ mol}^{-1}\)

Award 1 mark for selecting correct option A.
- Correct calculation of sum of reactants combustion: \(-2344 \text{ kJ mol}^{-1}\)
- Correct application of Hess cycle (Reactants - Products)
- Reject: B (incorrect sign), C or D (incorrect application of Hess cycle)

Both propan-1-ol and propan-2-ol can form hydrogen bonds because they contain polar \(O-H\) groups. However, propan-1-ol is a straight-chain isomer with a larger surface area of contact than the branched propan-2-ol. This allows propan-1-ol to form stronger London forces of attraction between its molecules. Propanal possesses permanent dipole-dipole forces but not hydrogen bonds, and propane has only weak London forces. Thus, propan-1-ol has the highest boiling temperature.

Award 1 mark for selecting correct option A.
- Distinguishing that alcohols have higher boiling points than aldehydes and alkanes of similar molar masses due to hydrogen bonding.
- Identifying that less branched/straight-chain isomers have stronger London forces and higher boiling points.

Potassium bromide reacts with concentrated sulfuric acid to initially produce hydrogen bromide, \(HBr\) (an acid-base reaction). Because bromide ions are reasonably strong reducing agents, they reduce the concentrated sulfuric acid to sulfur dioxide, \(SO_2(g)\), while being oxidized to bromine, \(Br_2(g)\). Bromide is not strong enough to reduce sulfuric acid further to elemental sulfur or hydrogen sulfide, \(H_2S\) (only iodide can do this). Thus, \(SO_2\) is the correct reduction product of sulfuric acid.

Award 1 mark for selecting correct option C.
- Reject: A (HBr is not a reduction product; sulfur is not reduced)
- Reject: B (bromine is the oxidation product of bromide)
- Reject: D (H2S is only produced when stronger reducing agents like iodide are used)

The oxidation of \(C_4H_{10}O\) (an alcohol) to \(C_4H_8O\) indicates the formation of either a ketone or an aldehyde. The peak at \(1715 \text{ cm}^{-1}\) confirms the presence of a carbonyl group (\(C=O\)). The absence of an absorption in the range \(2700 - 2900 \text{ cm}^{-1}\) (the aldehyde \(C-H\) stretch) and the lack of an \(O-H\) stretch show that the product is a ketone (butanone) rather than an aldehyde or unreacted alcohol. Ketones are produced by the oxidation of secondary alcohols. Butan-2-ol is the only secondary alcohol among the options. Butan-1-ol and 2-methylpropan-1-ol are primary alcohols (which would oxidize to aldehydes), and 2-methylpropan-2-ol is a tertiary alcohol (which does not oxidize under these conditions).

Award 1 mark for selecting correct option B.
- Recognising that the lack of aldehyde C-H and alcohol O-H bands signifies a ketone product.
- Deduencing that ketones are formed only from secondary alcohols (butan-2-ol).

(a)(i) Reagents: Sodium bromide (or potassium bromide) and concentrated sulfuric acid (or 50% sulfuric acid). Alternatively, phosphorus(III) bromide, \(PBr_3\).
Conditions: Heat under reflux.

(ii) \(C_4H_{10}O + HBr \rightarrow C_4H_9Br + H_2O\) (or \(C_4H_9OH + HBr \rightarrow C_4H_9Br + H_2O\))

(b) Mechanism of nucleophilic substitution (either \(S_N1\) or \(S_N2\) is acceptable):
For \(S_N2\):
- Curly arrow from the lone pair of \(OH^-\) to the carbon atom bonded to bromine.
- Dipoles shown on the C-Br bond (\(C^{\delta+} - Br^{\delta-}\)) and a curly arrow from the C-Br bond to the bromine atom.
- Correctly drawn transition state with partial bonds to OH and Br, and an overall negative charge, OR a single-step concerted process showing the products directly.
- Correct structure of the organic product butan-2-ol and the leaving group \(Br^-\).

For \(S_N1\):
- Curly arrow from the C-Br bond to the bromine atom with dipoles shown on C-Br.
- Correct carbocation intermediate (\(CH_3CH^+CH_2CH_3\)) and \(Br^-\) ion.
- Curly arrow from the lone pair of \(OH^-\) to the positive carbon of the carbocation.
- Correct structure of the organic product butan-2-ol.

(c)
- Trend: Rate of hydrolysis increases down Group 7 (1-chlorobutane < 1-bromobutane < 1-iodobutane) / 1-iodobutane reacts the fastest and 1-chlorobutane reacts the slowest.
- The reaction involves breaking the carbon-halogen bond.
- Bond enthalpy decreases down the group / the C-I bond is the weakest and C-Cl bond is the strongest.
- Less energy is required to break the C-I bond, so it breaks more easily, making bond enthalpy the dominant factor over bond polarity.

(d)
- The IR spectrum of the product will show a broad absorption band for the O-H (alcohol) group at 3200-3750 \(cm^{-1}\).
- The absorption band for the C-Br bond at 500-600 \(cm^{-1}\) will have disappeared (or will not be present in the spectrum of the product).

(a)(i)
- M1: NaBr / KBr and concentrated \(H_2SO_4\) (accept 50% \(H_2SO_4\) or phosphoric acid) [1]
- M2: Heat under reflux [1]
(Alternative for 2 marks: \(PBr_3\) [1] and heat / reflux [1])

(a)(ii)
- \(C_4H_{9}OH + HBr \rightarrow C_4H_9Br + H_2O\) (accept molecular formulas: \(C_4H_{10}O + HBr \rightarrow C_4H_9Br + H_2O\)) [1]

(b)
- M1: Curly arrow from lone pair on \(OH^-\) (or \(O\) of \(OH^-\)) to the carbon bonded to Br [1]
- M2: Dipoles on C-Br bond (\(C^{\delta+}\) and \(Br^{\delta-}\)) and curly arrow from C-Br bond to Br [1]
- M3: Correct transition state structure showing partial bonds and negative charge (or if \(S_N1\) pathway chosen: correct carbocation intermediate structure) [1]
- M4: Correct structures of butan-2-ol and bromide ion products [1]

(c)
- M1: Rate increases from 1-chlorobutane to 1-iodobutane (or equivalent statement) [1]
- M2: Hydrolysis rate depends on breaking the C-Halogen bond [1]
- M3: Bond enthalpy decreases down the group (C-Cl > C-Br > C-I) / C-I bond is the weakest [1]
- M4: Bond enthalpy is the deciding factor / bond polarity would predict the opposite trend [1]

(d)
- M1: Presence of broad O-H absorption band at 3200-3750 \(cm^{-1}\) [1]
- M2: Disappearance of C-Br absorption band at 500-600 \(cm^{-1}\) [1]

(a) The enthalpy change when one mole of a substance is burned completely in oxygen under standard conditions (298 K and 1 bar / 100 kPa), with all reactants and products in their standard states.

(b)(i) \(q = m c \Delta T = 150.0 \times 4.18 \times (45.8 - 20.2) = 150.0 \times 4.18 \times 25.6 = 16051.2\ J = 16.1\ kJ\) (or 16.05 kJ)

(ii) \(n = \frac{\text{mass}}{\text{Molar Mass}} = \frac{0.960}{32.0} = 0.0300\ mol\)

(iii) \(\Delta H_c = -\frac{q}{n} = -\frac{16.0512}{0.0300} = -535\ kJ\ mol^{-1}\) (to 3 sig figs)

(iv) Any two of:
- Heat lost to the surrounding air / calorimeter.
- Incomplete combustion of methanol (producing carbon monoxide/soot).
- Evaporation of methanol from the wick before/after weighing.

(c)(i) \(3C(s) + 3H_2(g) + O_2(g) \rightarrow C_2H_5COOH(l)\) (State symbols graphite/s, g, g, l must be present and correct)

(ii) Hess's Law Cycle:
Using combustion data: \(\Delta H_f = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})\)
\(\sum \Delta H_c(\text{reactants}) = 3 \times \Delta H_c(C) + 3 \times \Delta H_c(H_2)\)
\(\sum \Delta H_c(\text{reactants}) = 3(-394) + 3(-286) = -1182 - 858 = -2040\ kJ\ mol^{-1}\)
\(\Delta H_f = -2040 - (-1527) = -513\ kJ\ mol^{-1}\)

(a)
- M1: Enthalpy change when one mole of a substance is burned completely in excess oxygen [1]
- M2: Under standard conditions (298 K, 1 atm/100 kPa) with reactants and products in their standard states [1]

(b)(i)
- \(q = 150.0 \times 4.18 \times 25.6 = 16.1\ (kJ)\) (accept 16.05 / 16.0512 kJ) [1]

(b)(ii)
- \(n = 0.0300\ (mol)\) [1]

(b)(iii)
- M1: Divide energy by moles: \(\frac{16.0512}{0.0300} = 535\) [1]
- M2: Correct negative sign and unit: \(-535\ kJ\ mol^{-1}\) [1]
(Allow ECF from part (i) and (ii))

(b)(iv)
- M1: Heat loss to the surroundings / beaker [1]
- M2: Incomplete combustion of methanol / evaporation of methanol [1]

(c)(i)
- M1: Correct species: \(3C + 3H_2 + O_2 \rightarrow C_2H_5COOH\) (or \(C_3H_6O_2\)) [1]
- M2: Correct state symbols: \(C(s)\) (or \(C(s, graphite)\)), \(H_2(g)\), \(O_2(g)\), \(C_2H_5COOH(l)\) [1]

(c)(ii)
- M1: Correct Hess cycle drawing or algebraic expression: \(\Delta H_f = 3\Delta H_c(C) + 3\Delta H_c(H_2) - \Delta H_c(\text{propanoic acid})\) [1]
- M2: Calculation of \(\sum \Delta H_c(\text{reactants}) = 3(-394) + 3(-286) = -2040\ (kJ\ mol^{-1})\) [1]
- M3: Final answer: \(-513\ (kJ\ mol^{-1})\) [1]

(a)(i) \(MgCO_3(s) \rightarrow MgO(s) + CO_2(g)\)

(ii)
- Thermal stability increases down the group (from magnesium carbonate to barium carbonate).
- Cation size / ionic radius increases down the group while the overall charge remains \(2+\).
- Therefore, the charge density of the cation decreases.
- The larger cation has less polarizing power / distorts the carbonate ion (C-O bond) less, requiring more heat energy to break the bond and decompose the carbonate.

(b)(i)
- X: \(NaCl\) (sodium chloride) [accept \(NaF\)]
- Y: \(NaI\) (sodium iodide)
- Z: \(NaBr\) (sodium bromide)

(ii)
- Purple vapour: Iodine (\(I_2\))
- Yellow solid: Sulfur (\(S\))

(iii) \(NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl\) (or \(2NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2HCl\))

(iv)
- Iodide ions (\(I^-\)) are stronger reducing agents than chloride ions (\(Cl^-\)).
- This is because iodide ions are larger and have more electron shielding, meaning the outermost electrons are less strongly attracted to the nucleus and are lost more easily to reduce sulfuric acid.

(a)(i)
- \(MgCO_3(s) \rightarrow MgO(s) + CO_2(g)\) (All species and state symbols must be correct for 1 mark) [1]

(a)(ii)
- M1: Thermal stability increases down the group [1]
- M2: Cation size / ionic radius increases (charge remains 2+) [1]
- M3: Charge density decreases / polarizing power of the cation decreases [1]
- M4: Polarisation / distortion of the carbonate ion decreases (so C-O bond is harder to break) [1]

(b)(i)
- X: \(NaCl\) / \(NaF\) [1]
- Y: \(NaI\) [1]
- Z: \(NaBr\) [1]

(b)(ii)
- Purple vapour: Iodine / \(I_2\) [1]
- Yellow solid: Sulfur / \(S\) (Reject: sulfur dioxide, hydrogen sulfide) [1]

(b)(iii)
- \(NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl\) (or \(2NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2HCl\)) [1]

(b)(iv)
- M1: Iodide (\(I^-\)) is a stronger reducing agent than chloride (\(Cl^-\)) / \(I^-\) reduces sulfuric acid while \(Cl^-\) cannot [1]
- M2: Iodide is larger / has more shielding, so it loses electrons more easily [1]

**(a)(i)**
- Sulfur dioxide (\(\text{SO}_2\))
- Bromine (\(\text{Br}_2\))
*(Accept: Hydrogen sulfide / \(\text{H}_2\text{S}\), Sulfur / \(\text{S}\))*

**(a)(ii)**
- Chloride ions are weaker reducing agents than bromide ions (as chlorine is more electronegative/retains its electrons more strongly), so no redox reaction occurs. The oxidation state of chlorine remains \(-1\) and sulfur remains \(+6\).

**(b)**
- Moles of butan-1-ol = \(\frac{11.1}{74.0} = 0.150\text{ mol}\)
- Theoretical mass of 1-bromobutane = \(0.150\text{ mol} \times 137.0\text{ g mol}^{-1} = 20.55\text{ g}\)
- Percentage yield = \(\frac{13.7}{20.55} \times 100\% = 66.666...\% \approx 66.7\%\)
*(Alternative method using moles: Moles of 1-bromobutane = \(\frac{13.7}{137.0} = 0.100\text{ mol}\). Percentage yield = \(\frac{0.100}{0.150} \times 100\% = 66.7\%\))*

**(c)(i)**
- Add the crude 1-bromobutane and aqueous sodium hydrogencarbonate to the separating funnel, insert the stopper, and shake/invert the funnel.
- Periodically release pressure by opening the tap with the funnel inverted (to release \(\text{CO}_2\) gas generated during neutralization).
- Allow the mixture to settle into two distinct layers. The lower layer is the organic layer (1-bromobutane) because its density (\(1.27\text{ g cm}^{-3}\)) is greater than that of water (\(1.00\text{ g cm}^{-3}\)).
- Remove the stopper and run off the lower organic layer into a clean flask.

**(c)(ii)**
- Drying agent: Anhydrous calcium chloride (\(\text{CaCl}_2\)) or anhydrous sodium sulfate (\(\text{Na}_2\text{SO}_4\)) or anhydrous magnesium sulfate (\(\text{MgSO}_4\)).
- Appearance: The organic liquid changes from cloudy to clear/transparent.

**(d)(i)**
- Butan-1-ol has a broad \(\text{O-H}\) alcohol absorption at \(3200\text{--}3750\text{ cm}^{-1}\) which is absent in pure 1-bromobutane.
- Butan-1-ol has a \(\text{C-O}\) absorption at \(1000\text{--}1300\text{ cm}^{-1}\) which is absent in pure 1-bromobutane.
- 1-bromobutane has a \(\text{C-Br}\) absorption at \(500\text{--}600\text{ cm}^{-1}\) which is absent in butan-1-ol.
*(Award marks for any two clear differences, with correct ranges and bond assignments)*

**(d)(ii)**
- The IR spectrum of the product will show absolutely no absorption peak in the \(3200\text{--}3750\text{ cm}^{-1}\) region (no \(\text{O-H}\) stretch present).

**(e)(i)**
- Ethanol acts as a mutual solvent. Halogenoalkanes are insoluble in water but soluble in ethanol, allowing the reactants (halogenoalkane and aqueous silver nitrate) to mix into a single phase to react.

**(e)(ii)**
- Cream precipitate.
- Ionic equation: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\)

**(e)(iii)**
- The rate of hydrolysis increases in the order: 1-chlorobutane < 1-bromobutane < 1-iodobutane (i.e., 1-iodobutane reacts fastest, 1-chlorobutane reacts slowest).
- Down the group, bond length increases, and bond enthalpy decreases (the \(\text{C-I}\) bond is much weaker than \(\text{C-Br}\), which is weaker than \(\text{C-Cl}\)). Therefore, less energy is required to break the carbon-halogen bond, leading to a faster reaction.

**Part (a)** [Total: 3 marks]
- **(i)** 1 mark for sulfur dioxide (\(\text{SO}_2\)). 1 mark for bromine (\(\text{Br}_2\)). (Accept sulfur or hydrogen sulfide for one of these marks)
- **(ii)** 1 mark for stating that chloride ions cannot reduce concentrated sulfuric acid because they are weaker reducing agents / chlorine remains in the \(-1\) oxidation state.

**Part (b)** [Total: 3 marks]
- **1 mark** for calculating moles of butan-1-ol: \(0.150\text{ mol}\).
- **1 mark** for calculating theoretical mass of 1-bromobutane: \(20.55\text{ g}\) (or moles of 1-bromobutane obtained: \(0.100\text{ mol}\)).
- **1 mark** for correct percentage yield to 3 significant figures: \(66.7\%\). (Allow TE for calculation error but must be to 3 sig figs for final accuracy mark).

**Part (c)** [Total: 6 marks]
- **(i)** [4 marks total]:
- **1 mark** for shaking/inverting the separating funnel with the mixture.
- **1 mark** for venting/releasing pressure to prevent pressure build-up from carbon dioxide.
- **1 mark** for identifying the lower layer as 1-bromobutane due to its higher density.
- **1 mark** for running off the lower organic layer (or discarding the upper aqueous layer).
- **(ii)** [2 marks total]:
- **1 mark** for naming a suitable anhydrous drying agent (e.g., anhydrous \(\text{CaCl}_2\), \(\text{MgSO}_4\), or \(\text{Na}_2\text{SO}_4\)). *Reject: anhydrous cobalt chloride, anhydrous copper sulfate.*
- **1 mark** for stating the liquid changes from cloudy to clear/transparent.

**Part (d)** [Total: 4 marks]
- **(i)** [3 marks total]:
- **1 mark** for identifying the broad \(\text{O-H}\) stretch at \(3200\text{--}3750\text{ cm}^{-1}\) in the alcohol (absent in the halogenoalkane).
- **1 mark** for identifying the \(\text{C-Br}\) stretch at \(500\text{--}600\text{ cm}^{-1}\) in the halogenoalkane (absent in the alcohol).
- **1 mark** for identifying the \(\text{C-O}\) stretch at \(1000\text{--}1300\text{ cm}^{-1}\) in the alcohol (absent in the halogenoalkane).
- **(ii)** [1 mark]:
- **1 mark** for stating there is no peak/absorption at \(3200\text{--}3750\text{ cm}^{-1}\).

**Part (e)** [Total: 5 marks]
- **(i)** [1 mark] for explaining that ethanol acts as a mutual solvent/cosolvent to allow the halogenoalkane and aqueous silver nitrate to mix.
- **(ii)** [2 marks total]:
- **1 mark** for "cream precipitate" (reject other colors).
- **1 mark** for correct ionic equation with state symbols: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\).
- **(iii)** [2 marks total]:
- **1 mark** for stating the correct trend (rate increases: \(\text{Cl} < \text{Br} < \text{I}\)).
- **1 mark** for explaining that bond enthalpy decreases down the group (\(\text{C-I}\) is the weakest bond), so it is easiest to break.

(a) Pipette / volumetric pipette.

(b) \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\)

(c) Yellow to orange (or peach).

(d)
(i) \(\text{Moles of HCl} = 0.100 \times \frac{25.00}{1000} = 0.00250\text{ mol}\)
(ii) \(\text{Moles of Na}_2\text{CO}_3\text{ in 25.0 cm}^3 = \frac{0.00250}{2} = 0.00125\text{ mol}\)
(iii) \(\text{Moles of Na}_2\text{CO}_3\text{ in 250.0 cm}^3 = 0.00125 \times 10 = 0.0125\text{ mol}\)

(e) \(\text{Molar mass of hydrated salt} = \frac{3.58}{0.0125} = 286.4\text{ g mol}^{-1}\)
\(\text{Molar mass of anhydrous Na}_2\text{CO}_3 = 2(23.0) + 12.0 + 3(16.0) = 106.0\text{ g mol}^{-1}\)
\(\text{Molar mass of } x\text{H}_2\text{O} = 286.4 - 106.0 = 180.4\text{ g mol}^{-1}\)
\(x = \frac{180.4}{18.0} = 10.02\)
Therefore, the integer value is \(x = 10\).

(a) Pipette / volumetric pipette [1 mark] (Reject: graduated pipette, measuring cylinder).
(b) \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\) (or ionic equivalent) [1 mark].
(c) Yellow [1 mark] to orange / peach [1 mark] (Accept: yellow to pink / red-orange; Reject: yellow to red).
(d)
(i) Moles of \(\text{HCl} = 0.00250\text{ mol}\) [1 mark].
(ii) Moles of \(\text{Na}_2\text{CO}_3 = 0.00125\text{ mol}\) (allow TE for ans to d(i) divided by 2) [1 mark].
(iii) Moles of \(\text{Na}_2\text{CO}_3\) in 250 cm3 \(= 0.0125\text{ mol}\) (allow TE for ans to d(ii) multiplied by 10) [1 mark].
(e)
- Molar mass of hydrated sodium carbonate \(= 286.4\text{ g mol}^{-1}\) (allow TE for \(3.58 / \text{ans to d(iii)}\)) [1 mark].
- Subtracts molar mass of anhydrous sodium carbonate (\(106.0\text{ g mol}^{-1}\)) to find mass of water \(= 180.4\text{ g mol}^{-1}\) [1 mark].
- Divides by 18.0 to obtain \(10\) [1 mark].

(a) Polystyrene is an effective thermal insulator, which minimises heat transfer/loss to the surroundings.

(b) Zinc powder has a larger surface area than granules, increasing the rate of reaction so that the maximum temperature is reached more quickly (reducing heat loss during the reaction).

(c) Extrapolation allows the student to estimate the maximum temperature that would have been reached if the reaction had occurred instantaneously, correcting for the heat lost to the surroundings between the start of the reaction and the time when the maximum temperature was reached.

(d) \(q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 11.6\text{ K} = 2424.4\text{ J}\).

(e) Moles of \(\text{CuSO}_4\) (limiting reactant): \(n = 0.200\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.0100\text{ mol}\).
\(\Delta H = - \frac{q}{1000 \times n} = - \frac{2424.4}{1000 \times 0.0100} = -242.44\text{ kJ mol}^{-1}\).
To 3 significant figures: \(-242\text{ kJ mol}^{-1}\).

(a) Polystyrene is a (good) thermal insulator / reduces heat loss to the surroundings [1 mark].
(b) Powder has a larger surface area / increases reaction rate / reaction completes faster [1 mark].
(c)
- Reaction is not instantaneous / heat is lost while the reaction is occurring [1 mark].
- Extrapolation estimates the maximum temperature rise if the reaction had been instantaneous [1 mark].
(d)
- Substitution of values: \(q = 50.0 \times 4.18 \times 11.6\) [1 mark].
- Evaluation: \(2424.4\text{ J}\) (or \(2.4244\text{ kJ}\)) [1 mark].
(e)
- Calculates moles of copper(II) sulfate \(= 0.0100\text{ mol}\) [1 mark].
- Divides energy in kJ by moles: \(2.4244 / 0.0100 = 242.44\) [1 mark].
- Negative sign [1 mark].
- Correct rounding to 3 significant figures: \(-242\text{ kJ mol}^{-1}\) [1 mark] (Allow TE from d).

(a)
(i) Dip a platinum or nichrome wire into concentrated hydrochloric acid and heat it in a Bunsen burner flame until no color is observed to clean it. Dip the clean wire into the solid sample, then place it into a non-luminous (blue) Bunsen flame.
(ii) The brick-red flame indicates the presence of calcium ions, \(\text{Ca}^{2+}\).

(b)
(i) The cream precipitate indicates the presence of bromide ions, \(\text{Br}^-\). The ionic equation is: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\).
(ii) Silver bromide dissolves in concentrated aqueous ammonia, so the cream precipitate dissolves to form a clear/colourless solution.

(c)
(i) Concentrated sulfuric acid reacts with solid calcium bromide to produce hydrogen bromide and bromine. Observations include: steamy/misty fumes (from \(\text{HBr}\)), brown fumes/vapor (from \(\text{Br}_2\)), or a choking gas (from \(\text{SO}_2\)).
(ii) Bromine (\(\text{Br}_2\)) is responsible for the brown fumes; hydrogen bromide (\(\text{HBr}\)) is responsible for the misty fumes; sulfur dioxide (\(\text{SO}_2\)) is responsible for the choking gas.

(a)(i)
- Use a nichrome or platinum wire [1 mark].
- Clean by dipping in concentrated hydrochloric acid and heating in a Bunsen flame until no colour remains [1 mark].
- Dip in sample and place in a non-luminous / blue Bunsen burner flame [1 mark].
(a)(ii) Calcium / \(\text{Ca}^{2+}\) [1 mark] (Reject: Ca / calcium metal).
(b)(i)
- Bromide / \(\text{Br}^-\)[1 mark].
- \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\) (including state symbols) [1 mark].
(b)(ii) Precipitate dissolves / disappears / clear solution forms [1 mark] (Reject: partially dissolves).
(c)(i) Any two from:
- Misty / steamy fumes
- Brown fumes / gas / vapor
- Choking / pungent gas
- Orange-brown liquid condensing on the tube walls
[2 marks].
(c)(ii) Bromine / \(\text{Br}_2\) (or hydrogen bromide / \(\text{HBr}\) or sulfur dioxide / \(\text{SO}_2\)) [1 mark].

(a) Bromine water can be used to test for cyclohexene. Reagent: Bromine water (or bromine in an organic solvent). Observation: Orange-brown solution decolourises (becomes colourless).

(b) Reagent: Acidified potassium dichromate(VI) (solution). Colour change: Orange to green (as primary alcohols are oxidised, while tertiary alcohols are resistant to oxidation).

(c)
(i) Ethanol acts as a common solvent, allowing the halogenoalkane and the aqueous silver nitrate to mix/dissolve together.
(ii) 1-bromobutane reacts faster because the \(\text{C}-\text{Br}\) bond is weaker (has lower bond enthalpy) than the \(\text{C}-\text{Cl}\) bond, so it requires less energy to break and is broken more rapidly.
(iii) Precipitates form (cream for bromide, white for chloride). The faster reaction is confirmed by the precipitate forming more quickly / in less time.

(a)
- Reagent: Bromine water / aqueous bromine / bromine in organic solvent [1 mark].
- Observation: Decolourises / goes from orange-brown to colourless [1 mark] (Reject: goes clear).
(b)
- Reagent: Acidified potassium dichromate(VI) / \(\text{K}_2\text{Cr}_2\text{O}_7\) and \(\text{H}^+\) [1 mark].
- Colour change: Orange to green [1 mark].
(c)(i) Common solvent / cosolvent / dissolves both organic and aqueous reactants [1 mark].
(c)(ii)
- 1-bromobutane (reacts faster) [1 mark].
- The \(\text{C}-\text{Br}\) bond is weaker / has lower bond enthalpy than the \(\text{C}-\text{Cl}\) bond [1 mark].
- Less energy is needed to break the \(\text{C}-\text{Br}\) bond [1 mark].
(c)(iii)
- Cream / white precipitate forms [1 mark].
- The precipitate forms faster / more rapidly with 1-bromobutane [1 mark].

(a) Concentrated sulfuric acid is a strong oxidising agent and would oxidise cyclohexanol/cyclohexene to carbon or sulfur dioxide, causing side-reactions and lowering the yield. Phosphoric acid is non-oxidising.

(b)
(i) The cyclohexene product is in the upper layer because its density is lower than that of the aqueous layer.
(ii) Place the mixture in the separating funnel, stopper it, and shake. Periodically invert the funnel and open the tap to release pressure (venting gas). Allow the layers to separate, remove the stopper, then run off the lower aqueous layer through the tap, and collect the top organic layer through the top of the funnel.

(c)
(i) Anhydrous calcium chloride / anhydrous magnesium sulfate / anhydrous sodium sulfate.
(ii) Clear / transparent (not cloudy).

(d)
- Moles of cyclohexanol \(= \frac{12.0}{100.2} = 0.11976\text{ mol}\).
- Theoretical mass of cyclohexene \(= 0.11976\text{ mol} \times 82.1\text{ g mol}^{-1} = 9.832\text{ g}\).
- Percentage yield \(= \frac{5.90}{9.832} \times 100\% = 60.0\%\).

(a) Sulfuric acid is an oxidising agent / would oxidise cyclohexanol / causes side reactions / produces carbon or sulfur dioxide [1 mark].
(b)(i) Top layer [1 mark].
(b)(ii)
- Shake the mixture in the funnel with the stopper on [1 mark].
- Invert the funnel and open tap periodically to release pressure / gas build-up [1 mark].
- Remove stopper, allow layers to separate, and run off the lower layer, keeping the top layer [1 mark].
(c)(i) Anhydrous calcium chloride / magnesium sulfate / sodium sulfate [1 mark] (Reject: anhydrous copper(II) sulfate, cobalt chloride, calcium oxide, silica gel).
(c)(ii) Clear / transparent / no longer cloudy / misty [1 mark] (Reject: colourless, unless specified as clear).
(d)
- Calculates moles of cyclohexanol \(= 0.1198\text{ mol}\) [1 mark].
- Calculates theoretical mass of cyclohexene \(= 9.83\text{ g}\) [1 mark].
- Calculates percentage yield \(= 60.0\%\) (Accept 60% or 60.01%) [1 mark].