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In silicon dioxide (\(\text{SiO}_2\)), a large amount of energy is required to break the strong covalent bonds throughout the giant lattice. In carbon dioxide (\(\text{CO}_2\)), only weak intermolecular forces between molecules need to be overcome to melt the substance, which requires much less energy. The strong covalent bonds inside individual carbon dioxide molecules are not broken during melting.

1 mark for the correct answer B. Reject other options.

Ionic compounds have high melting points due to strong electrostatic attractions between oppositely charged ions in their giant lattice. They do not conduct electricity when solid because the ions are locked in fixed positions. However, they do conduct electricity when liquid (molten) because the ions become free to move and carry charge. Material X fits all of these characteristics.

1 mark for the correct answer B. Reject other options.

Increasing the concentration of \(\text{HCl}\) increases the number of reactant particles in a given volume, which increases the frequency of collisions, but does not alter the kinetic energy of the particles. Increasing temperature increases both collision frequency and collision energy. Adding a catalyst lowers the activation energy of the reaction. Using larger lumps decreases the surface area, which decreases the frequency of collisions.

1 mark for the correct answer A. Reject other options.

The rate of a chemical reaction is represented by the gradient (slope) of the volume-time graph. The gradient is steepest at the very start of the reaction (time = 0) because the concentrations of the reactants are highest, leading to the highest frequency of successful collisions. As the reaction continues, reactants are consumed, and the rate slows down.

1 mark for the correct answer A. Reject other options.

Smaller hydrocarbons have smaller molecules with weaker intermolecular forces between them, meaning they have lower boiling points. During fractional distillation, they rise further up the fractionating column where temperatures are lower and condense near the top. Larger hydrocarbons have stronger intermolecular forces, higher boiling points, and condense lower down where it is hotter.

1 mark for the correct answer B. Reject other options.

Cracking breaks down larger alkanes into shorter-chain alkanes (used for fuels) and alkenes (used to make polymers). In option A, decane (\(\text{C}_{10}\text{H}_{22}\)) is cracked to produce octane (\(\text{C}_8\text{H}_{18}\)) and ethene (\(\text{C}_2\text{H}_4\)). Both the carbon count (\(10 = 8 + 2\)) and hydrogen count (\(22 = 18 + 4\)) are balanced and represent realistic products.

1 mark for the correct answer A. Reject other options.

1. Calculate moles of \(\text{Na}_2\text{CO}_3\): \(\text{moles} = \frac{\text{mass}}{\text{RFM}} = \frac{5.3\text{ g}}{106\text{ g/mol}} = 0.05\text{ mol}\).
2. Convert volume from \(\text{cm}^3\) to \(\text{dm}^3\): \(\text{volume} = \frac{250}{1000} = 0.25\text{ dm}^3\).
3. Calculate concentration: \(\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.05\text{ mol}}{0.25\text{ dm}^3} = 0.20\text{ mol/dm}^3\).

1 mark for the correct answer B. Reject other options.

1. Moles of \(\text{H}_2\text{SO}_4\) used: \(\text{moles} = \text{concentration} \times \text{volume} = 0.100\text{ mol/dm}^3 \times 0.0200\text{ dm}^3 = 0.00200\text{ mol}\).
2. From the equation, 2 moles of \(\text{NaOH}\) react with 1 mole of \(\text{H}_2\text{SO}_4\). Thus, \(\text{moles of NaOH} = 2 \times 0.00200\text{ mol} = 0.00400\text{ mol}\).
3. Concentration of \(\text{NaOH}\): \(\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.00400\text{ mol}}{0.0250\text{ dm}^3} = 0.160\text{ mol/dm}^3\).

1 mark for the correct answer C. Reject other options.

Silicon dioxide contains many strong covalent bonds in a giant lattice structure. Breaking these bonds requires a very large amount of thermal energy, giving it a high melting point. Since all the valence electrons are held tightly in covalent bonds and there are no delocalised electrons or ions free to move, it does not conduct electricity.

[1 mark] C is the correct answer. Give 1 mark for C. Reject any other options.

Using larger marble chips decreases the total surface area of the reactant. This reduces the frequency of collisions between reactant particles, thereby decreasing the initial rate of reaction. All other options would increase the rate of reaction.

[1 mark] C is the correct answer. Give 1 mark for C. Reject any other options.

The temperature is lowest at the top of the fractionating column. Fractions collected here consist of smaller hydrocarbon molecules with weaker intermolecular forces, which means they have the lowest boiling points.

[1 mark] A is the correct answer. Give 1 mark for A. Reject any other options.

First, convert the volume from \(\text{cm}^3\) to \(\text{dm}^3\): \(250\text{ cm}^3 = 0.250\text{ dm}^3\). Then calculate the concentration using: \(\text{Concentration} = \frac{\text{mass}}{\text{volume}} = \frac{10.0\text{ g}}{0.250\text{ dm}^3} = 40.0\text{ g/dm}^3\).

[1 mark] C is the correct answer. Give 1 mark for C. Reject any other options.

As you go down Group 7, the size of the halogen molecules increases, leading to stronger intermolecular forces and therefore an increased boiling point. Additionally, the outer shell is further away from the nucleus, meaning electrons are less strongly attracted and reactivity decreases.

[1 mark] B is the correct answer. Give 1 mark for B. Reject any other options.

Metallic structures consist of a lattice of positive metal ions surrounded by a sea of delocalised electrons. These delocalised electrons are free to move throughout the giant structure and can carry electrical charge under an applied potential difference.

[1 mark] C is the correct answer. Give 1 mark for C. Reject any other options.

The stage directly concerned with disposal and waste management of a product is the end-of-life displacement or waste management stage, which analyzes the environmental impacts of recycling, landfilling, or incineration.

[1 mark] D is the correct answer. Give 1 mark for D. Reject any other options.

Diamond is a giant covalent lattice where each carbon atom is joined to four others by strong covalent bonds. These require a large amount of energy to break. In contrast, carbon dioxide exists as simple molecules. Although the covalent bonds within the molecules are strong, the intermolecular forces between the molecules are very weak and require very little energy to overcome, making it a gas at room temperature.

1 mark: Statement that diamond has a giant covalent structure with strong covalent bonds that need a lot of energy to break. 1 mark: Statement that carbon dioxide has simple molecules with weak intermolecular forces that require little energy to overcome.

In solid sodium chloride, the sodium and chloride ions are held tightly in a giant ionic lattice and are not free to move. When melted (molten), the strong electrostatic forces are overcome, allowing the ions to move freely and carry an electric current.

1 mark: States that in solid sodium chloride, the ions are fixed in position and cannot move. 1 mark: States that in molten sodium chloride, the ions are free to move and carry charge. (Do not accept 'electrons' moving).

The rate of reaction in Experiment B (powder) is faster than in Experiment A (chips). This is because the powder has a larger surface area to volume ratio, exposing more particles to the acid. This results in a higher frequency of successful collisions (more collisions per second).

1 mark: States that the rate of reaction in Experiment B is faster due to a larger surface area. 1 mark: Explains that this leads to more frequent collisions / higher frequency of successful collisions (do not accept 'more collisions' without a time reference).

Catalysts increase the rate of reaction by providing an alternative reaction pathway that has a lower activation energy. The catalyst is not chemically consumed in the reaction, so its mass remains completely unchanged at the end of the reaction.

1 mark: For stating that a catalyst provides an alternative reaction pathway with a lower activation energy. 1 mark: For stating that the mass of the catalyst remains unchanged.

Crude oil is heated to vaporise it and passed into a fractionating column which is hot at the bottom and cooler at the top (a temperature gradient). The hydrocarbons rise up the column and condense when they reach the level that matches their boiling point, allowing them to be collected as separate fractions.

1 mark: States that crude oil is heated/vaporised and enters a fractionating column with a temperature gradient (hotter at the bottom, cooler at the top). 1 mark: States that hydrocarbons condense at different heights/temperatures based on their boiling points.

To distinguish between an alkane and an alkene, add bromine water to each sample. The alkene is unsaturated and will react with the bromine, decolourising the bromine water from orange-brown to colourless. The alkane is saturated and does not react, so the mixture remains orange-brown.

1 mark: For identifying the reagent as bromine water. 1 mark: For stating that the alkene decolourises the bromine water (turns orange-brown to colourless) while the alkane remains orange-brown (no change).

First, convert the volume of the solution from cubic centimetres to cubic decimetres: \(250\text{ cm}^3 / 1000 = 0.25\text{ dm}^3\). Then, calculate the concentration by dividing the mass of the solute by the volume of the solution: \(\text{Concentration} = 10.0\text{ g} / 0.25\text{ dm}^3 = 40\text{ g/dm}^3\).

1 mark: For converting the volume to \(0.25\text{ dm}^3\) (or showing the division of 250 by 1000). 1 mark: For calculating the final concentration as \(40\text{ (g/dm}^3)\). Award 2 marks for a correct final answer of 40.

First, find the number of moles of \(\text{HCl}\) in the solution: \(\text{Moles} = \text{concentration} \times \text{volume} = 0.100\text{ mol/dm}^3 \times 2.00\text{ dm}^3 = 0.200\text{ mol}\). Next, calculate the mass of \(\text{HCl}\) using the formula: \(\text{Mass} = \text{moles} \times M_r = 0.200\text{ mol} \times 36.5\text{ g/mol} = 7.3\text{ g}\).

1 mark: For calculating the number of moles as \(0.200\text{ mol}\). 1 mark: For calculating the mass of \(\text{HCl}\) as \(7.3\text{ g}\) (or \(7.30\text{ g}\)). Award 2 marks for a correct final answer of 7.3.

In graphite, each carbon atom forms three covalent bonds in layered hexagonal rings. This leaves one outer electron per carbon atom delocalised. These delocalised electrons are free to move throughout the structure and carry electrical charge. In contrast, in diamond, each carbon atom is covalently bonded to four other carbon atoms in a rigid, giant tetrahedral structure. All outer-shell electrons are localized in these strong covalent bonds, meaning there are no free or delocalised electrons to conduct electricity.

1 mark: State that graphite has delocalised / free electrons that can move (and carry charge). 1 mark: State that diamond has no free / delocalised electrons because all outer electrons are shared in covalent bonds.

Sodium chloride has a giant ionic lattice structure. There are strong electrostatic forces of attraction (ionic bonds) between the oppositely charged sodium and chloride ions, which require a large amount of heat energy to overcome. Water is a simple covalent substance. While the covalent bonds holding the hydrogen and oxygen atoms together within each molecule are strong, the intermolecular forces between the water molecules are very weak. Consequently, very little heat energy is required to overcome these weak intermolecular forces and melt the ice.

1 mark: Identifies that sodium chloride has strong electrostatic forces of attraction / ionic bonds between oppositely charged ions which require a large amount of energy to break. 1 mark: Identifies that water has weak intermolecular forces between molecules which require very little energy to overcome (do not credit references to breaking covalent bonds in water).

When temperature increases, the reacting particles gain kinetic energy and move faster. This has two key effects: firstly, they collide more frequently (higher collision frequency). Secondly, and more significantly, a much larger proportion of the colliding particles possess energy equal to or greater than the activation energy. This results in a higher rate of successful collisions, thereby increasing the rate of reaction.

1 mark: Mentions that particles collide more frequently (higher collision frequency). 1 mark: Mentions that a greater proportion of collisions have energy equal to or greater than the activation energy (leading to more successful collisions per second).

A catalyst increases the rate of reaction by providing an alternative reaction pathway that has a lower activation energy. This allows a greater fraction of collisions to be successful. Because a catalyst is not chemically consumed or permanently changed during the reaction, its total mass remains exactly the same at the end of the reaction as it was at the start.

1 mark: Identifies that a catalyst provides an alternative reaction pathway with a lower activation energy. 1 mark: States that the mass of the catalyst remains unchanged / constant at the end of the reaction.

Crude oil is heated and vaporised before entering the bottom of a fractionating column. The column is hotter at the bottom and cooler at the top (a temperature gradient). As the vaporised hydrocarbons rise up the column, they cool. They condense back into liquid fractions at different levels when they reach temperatures equal to their boiling points: longer hydrocarbons with high boiling points condense near the bottom, while shorter hydrocarbons with lower boiling points condense near the top.

1 mark: Mentions that the fractionating column has a temperature gradient (hotter at the bottom, cooler at the top). 1 mark: Mentions that different fractions condense at different heights/levels depending on their boiling points.

Fractional distillation of crude oil produces an excess of long-chain hydrocarbons, which are thick liquids that do not burn easily and are in low demand. In contrast, there is a very high demand for short-chain hydrocarbons, which make excellent fuels. Cracking is used to break down these less useful long-chain alkanes into more valuable short-chain alkanes and highly reactive alkenes, which are used as raw materials to make plastics and polymers.

1 mark: Explanation that cracking converts low-demand / less useful long-chain hydrocarbons into high-demand / more useful short-chain hydrocarbons. 1 mark: Correctly names the two products of cracking as alkanes and alkenes.

Step 1: Calculate the number of moles of sodium carbonate: \(\text{moles} = \frac{\text{mass}}{M_r} = \frac{10.6\text{ g}}{106\text{ g/mol}} = 0.10\text{ mol}\). Step 2: Convert the volume from \(\text{cm}^3\) to \(\text{dm}^3\): \(\text{volume} = \frac{250\text{ cm}^3}{1000} = 0.25\text{ dm}^3\). Step 3: Calculate the concentration in \(\text{mol/dm}^3\): \(\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.10\text{ mol}}{0.25\text{ dm}^3} = 0.40\text{ mol/dm}^3\).

1 mark: Correctly calculates moles of sodium carbonate as \(0.10\text{ mol}\) OR converts volume to \(0.25\text{ dm}^3\). 1 mark: Correctly calculates the final concentration of \(0.40\text{ mol/dm}^3\) (allow \(0.4\text{ mol/dm}^3\)).

Step 1: Calculate the amount of hydrochloric acid (\(\text{HCl}\)) used: \(\text{moles of HCl} = \text{concentration} \times \text{volume} = 0.100\text{ mol/dm}^3 \times 0.0200\text{ dm}^3 = 0.0020\text{ mol}\). Step 2: Use the stoichiometry of the reaction. Since the ratio of \(\text{NaOH}\) to \(\text{HCl}\) is \(1:1\), the moles of \(\text{NaOH}\) neutralised must be \(0.0020\text{ mol}\). Step 3: Calculate the concentration of the \(\text{NaOH}\) solution: \(\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.0020\text{ mol}}{0.0250\text{ dm}^3} = 0.080\text{ mol/dm}^3\).

1 mark: Correctly calculates moles of hydrochloric acid as \(0.0020\text{ mol}\) (or \(2.0 \times 10^{-3}\text{ mol}\)). 1 mark: Correctly calculates the concentration of the sodium hydroxide solution as \(0.080\text{ mol/dm}^3\) (allow \(0.08\text{ mol/dm}^3\)).

In graphite, each carbon atom is covalently bonded to three others in layers. This leaves one outer shell electron per carbon atom delocalised. These delocalised electrons are free to move throughout the layers and carry electrical charge. In contrast, in diamond, each carbon atom is covalently bonded to four others in a giant tetrahedral arrangement. All valence electrons are held tightly in localized covalent bonds, leaving no free or delocalised electrons to conduct electricity.

1 mark: For stating that graphite contains delocalised electrons that are free to move (and carry charge/current) because each carbon atom forms three bonds.
1 mark: For stating that diamond does not conduct because it has no delocalised/free electrons as each carbon atom forms four bonds.

When temperature increases, reactant particles gain kinetic energy and move faster. This results in more frequent collisions (more collisions per unit time). Additionally, because particles have more kinetic energy, a larger proportion of these collisions have energy equal to or greater than the activation energy, making them successful. Both factors contribute to an increased rate of reaction.

1 mark: Mentioning that particles move faster/have more kinetic energy leading to more frequent collisions (or more collisions per second/unit time). [Do not accept 'more collisions' alone without reference to time/frequency]
1 mark: Mentioning that a higher proportion/fraction of collisions are successful because more particles have energy greater than or equal to the activation energy.

The crude oil is heated and vaporised before entering the fractionating column. The column is hot at the bottom and becomes progressively cooler towards the top (creating a temperature gradient). As the vaporised hydrocarbons rise up the column, they cool down. Each hydrocarbon condenses back into a liquid and is piped off at the specific height where the temperature of the column falls below its boiling point.

1 mark: For stating there is a temperature gradient in the column (hotter at the bottom, cooler at the top).
1 mark: For stating that hydrocarbons condense and are collected at different heights/levels based on their different boiling points.

To find the concentration in \(\text{g/dm}^3\), we use the formula:

\(\text{Concentration } (\text{g/dm}^3) = \frac{\text{Mass of solute } (\text{g})}{\text{Volume of solution } (\text{dm}^3)}\)

First, convert the volume from \(\text{cm}^3\) to \(\text{dm}^3\):
\(\text{Volume} = \frac{250}{1000} = 0.25\text{ dm}^3\)

Next, calculate the concentration:
\(\text{Concentration} = \frac{5.3\text{ g}}{0.25\text{ dm}^3} = 21.2\text{ g/dm}^3\)

1 mark: For converting the volume to \(\text{dm}^3\) (\(0.25\text{ dm}^3\)) OR for showing a correct division method, e.g., \(\frac{5.3}{250} \times 1000\).
1 mark: For the correct final answer of 21.2.

Step 1: Calculate the amount in moles of sodium carbonate.
\(\text{Moles} = \frac{\text{mass}}{\text{relative formula mass}} = \frac{10.6\text{ g}}{106} = 0.10\text{ mol}\)

Step 2: Convert the volume from \(\text{cm}^3\) to \(\text{dm}^3\).
\(\text{Volume in dm}^3 = \frac{250\text{ cm}^3}{1000} = 0.25\text{ dm}^3\)

Step 3: Calculate the concentration in \(\text{mol/dm}^3\).
\(\text{Concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.10\text{ mol}}{0.25\text{ dm}^3} = 0.40\text{ mol/dm}^3\)

• 1 mark for calculating moles of \(\text{Na}_2\text{CO}_3\): \(0.1\text{ mol}\)
• 1 mark for converting volume to \(\text{dm}^3\): \(0.25\text{ dm}^3\) (or dividing by 0.25)
• 1 mark for the correct final answer: \(0.4\) or \(0.40\) (\(\text{mol/dm}^3\))

Step 1: Calculate the reacting masses from the balanced equation.
1 mole of \(\text{Fe}_2\text{O}_3\) produces 2 moles of \(\text{Fe}\).
Mass of \(\text{Fe}_2\text{O}_3 = 160\text{ g}\)
Mass of \(2\text{Fe} = 2 \times 56 = 112\text{ g}\)

Step 2: Use ratios to find the mass produced from \(240\text{ kg}\).
\(160\text{ kg}\) of \(\text{Fe}_2\text{O}_3\) produces \(112\text{ kg}\) of \(\text{Fe}\).
Ratio = \(\frac{240}{160} = 1.5\)

Step 3: Calculate the mass of iron.
\(\text{Mass of iron} = 1.5 \times 112 = 168\text{ kg}\)

• 1 mark for determining the relative mass of the product: \(2 \times 56 = 112\)
• 1 mark for setting up the correct calculation ratio: \(\frac{240}{160} \times 112\) (or equivalent working)
• 1 mark for the correct final mass: \(168\) (kg)

Step 1: Calculate the relative formula mass (\(M_r\)) of \(\text{MgCl}_2\).
\(M_r = 24 + (2 \times 35.5) = 95\)

Step 2: Calculate the theoretical yield of \(\text{MgCl}_2\).
\(\text{Moles of Mg} = \frac{6.0\text{ g}}{24} = 0.25\text{ mol}\)
Since 1 mole of \(\text{Mg}\) forms 1 mole of \(\text{MgCl}_2\), the theoretical yield of \(\text{MgCl}_2\) is:
\(0.25\text{ mol} \times 95 = 23.75\text{ g}\)

Step 3: Calculate the percentage yield.
\(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 = \frac{19.0\text{ g}}{23.75\text{ g}} \times 100 = 80\%\)

• 1 mark for calculating the theoretical yield of \(\text{MgCl}_2\): \(23.75\text{ g}\) (or showing working that achieves this)
• 1 mark for correctly setting up the percentage yield expression: \(\frac{19.0}{\text{their theoretical yield}} \times 100\)
• 1 mark for the correct final answer: \(80\) (%)

Step 1: Calculate the change in the volume of carbon dioxide gas collected between 20 seconds and 50 seconds.
\(\text{Change in volume} = 42\text{ cm}^3 - 15\text{ cm}^3 = 27\text{ cm}^3\)

Step 2: Calculate the time interval.
\(\text{Time interval} = 50\text{ s} - 20\text{ s} = 30\text{ s}\)

Step 3: Calculate the mean rate of reaction.
\(\text{Mean rate} = \frac{\text{Change in volume}}{\text{Time interval}} = \frac{27\text{ cm}^3}{30\text{ s}} = 0.9\text{ cm}^3\text{/s}\)

• 1 mark for calculating the change in volume: \(27\text{ cm}^3\)
• 1 mark for calculating the time interval: \(30\text{ s}\)
• 1 mark for the correct final value of the rate: \(0.9\) (\(\text{cm}^3\text{/s}\))

Step 1: Convert percentages to relative molar amounts by dividing each percentage by its relative atomic mass (\(A_r\)).
• For Carbon: \(\frac{82.8}{12} = 6.90\)
• For Hydrogen: \(\frac{17.2}{1} = 17.20\)

Step 2: Divide both values by the smaller of the two numbers (6.90) to find the simplest ratio.
• Carbon ratio: \(\frac{6.90}{6.90} = 1.0\)
• Hydrogen ratio: \(\frac{17.20}{6.90} \approx 2.5\)

Step 3: Convert the decimal ratio to whole numbers by multiplying both values by 2.
• Carbon: \(1.0 \times 2 = 2\)
• Hydrogen: \(2.5 \times 2 = 5\)

The empirical formula is \(\text{C}_2\text{H}_5\).

• 1 mark for calculating the correct molar ratio values (dividing by atomic mass): \(6.90\) and \(17.20\)
• 1 mark for finding the simplified ratio of elements: \(1 : 2.5\)
• 1 mark for scaling to the correct whole-number ratio empirical formula: \(\text{C}_2\text{H}_5\)

To measure the rate of reaction by gas collection:
- Place a set mass of calcium carbonate (marble chips) in a conical flask.
- Add a set volume of a specific concentration of hydrochloric acid.
- Immediately seal the flask with a stopper connected via a delivery tube to a gas syringe (or an inverted measuring cylinder filled with water) and start a timer.
- Record the volume of gas produced at regular intervals (e.g., every 10 or 20 seconds) for a set period of time.
- To investigate concentration, repeat the experiment with the same volume and mass of reactants, at the same temperature, but with a different concentration of hydrochloric acid.

Award 1 mark for each of the following points, up to a maximum of 4 marks:
- **M1**: Place a known/measured mass of calcium carbonate (marble chips) into a flask and add a known volume of acid (1 mark).
- **M2**: Connect the flask immediately to a gas syringe (or delivery tube leading to a measuring cylinder inverted over water) and start a stopwatch/timer (1 mark).
- **M3**: Measure and record the volume of gas collected at regular, specified time intervals (e.g. every 10 seconds) (1 mark).
- **M4**: State that the experiment must be repeated with different concentrations of hydrochloric acid while keeping controlled variables (such as temperature, volume of acid, or surface area of chips) constant (1 mark).

An accurate titration requires precise equipment and technique:
- Use a volumetric pipette and pipette filler to measure exactly \(25.0\text{ cm}^3\) of the sodium hydroxide into a conical flask.
- Add an appropriate indicator (like phenolphthalein which changes from pink to colorless, or methyl orange which changes from yellow to peach/orange). Do not use universal indicator as it has a gradual color change.
- Fill a burette with hydrochloric acid and note the starting value. Run the acid into the flask slowly, swirling the flask continuously to ensure mixing.
- Near the end point, add the acid dropwise until a single drop causes a permanent color change. Record the final volume. Repeat the procedure to obtain concordant results (within \(0.1\text{ cm}^3\) of each other) to calculate a mean.

Award 1 mark for each of the following points, up to a maximum of 4 marks:
- **M1**: Use a pipette (and pipette filler) to measure and transfer the \(25.0\text{ cm}^3\) of sodium hydroxide into a conical flask (1 mark).
- **M2**: Add a few drops of a named single-color-change indicator (e.g. phenolphthalein or methyl orange; reject universal indicator) (1 mark).
- **M3**: Add hydrochloric acid from a burette slowly, swirling the flask, and add dropwise near the end-point until the indicator changes color permanently (1 mark).
- **M4**: Record the final volume (titre) and repeat the process to obtain concordant results (within \(0.1\text{ cm}^3\)) (1 mark).

To confirm the identity of the solid:
1. **For the cation (Sodium, \(\text{Na}^+\))**: Perform a flame test. Clean a nichrome or platinum wire by dipping it in concentrated hydrochloric acid and placing it in a hot Bunsen flame. Dip the wire into the sample, then hold it in the outer blue flame of the Bunsen burner. A persistent, intense yellow flame indicates the presence of sodium ions.
2. **For the anion (Chloride, \(\text{Cl}^-\))**: Perform a precipitation test. Dissolve a small amount of the sample in distilled water. Add a few drops of dilute nitric acid to remove any carbonate impurities, followed by a few drops of silver nitrate solution. The formation of a white precipitate (silver chloride) confirms the presence of chloride ions.

Award 1 mark for each of the following points, up to a maximum of 4 marks:
- **M1 (Sodium preparation)**: Describe cleaning a wire loop (nichrome/platinum) in acid and dipping it into the sample (1 mark).
- **M2 (Sodium observation)**: Place the wire in a blue/roaring Bunsen flame and observe a yellow/orange-yellow flame (1 mark).
- **M3 (Chloride preparation)**: Make a solution of the sample in distilled water and add dilute nitric acid (1 mark).
- **M4 (Chloride observation)**: Add silver nitrate solution and observe a white precipitate (1 mark). (Reject hydrochloric acid for acidification of the chloride test)

Fractional distillation is used to separate miscible liquids with different boiling points:
- The mixture of ethanol and water is heated in a distillation flask.
- The vapors rise up a fractionating column. The column has a high surface area (often containing glass beads), which allows repeated condensation and evaporation, enriching the vapor in the more volatile component (ethanol).
- A thermometer at the top of the column measures the temperature of the vapor entering the condenser. This should read approximately \(78^\circ\text{C}\) while ethanol is distilling.
- The vapor enters a Liebig condenser, where it is cooled by a flow of cold water entering the bottom jacket and exiting the top. The vapor condenses back into liquid ethanol and is collected in a receiving vessel.

Award 1 mark for each of the following points, up to a maximum of 4 marks:
- **M1**: Heat the mixture in a flask connected to a fractionating column (containing glass beads/packing to provide a temperature gradient) (1 mark).
- **M2**: Use a thermometer placed at the top of the fractionating column/entrance of the condenser to monitor the vapor temperature (1 mark).
- **M3**: Pass the vapor through a condenser cooled by a flow of cold water running in the opposite direction (water in at bottom, out at top) (1 mark).
- **M4**: Control the heating to maintain the temperature at/near \(78^\circ\text{C}\) (the boiling point of ethanol) to distill and collect the ethanol first (1 mark).