An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 Cambridge International A Level Chemistry paper. Not affiliated with or reproduced from Cambridge.
Paper 1C (4CH1/1C)
Answer all questions. Show all steps in your calculations and state correct units. Total marks: 110.
10 Question · 110 marks
Question 1 · structured
11 marks
A student carries out an experiment to find the empirical formula of an oxide of titanium. (a) Describe how the student could heat a sample of titanium metal in a crucible with oxygen to ensure complete reaction. (4 marks) (b) In the experiment, 1.44 g of titanium reacted with oxygen to form 2.40 g of titanium oxide. Calculate the empirical formula of the titanium oxide. [Ar values: Ti = 48, O = 16] (4 marks) (c) Write a balanced chemical equation for this reaction. (1 mark) (d) Calculate the mass of titanium required to produce exactly 15.0 g of this titanium oxide. (2 marks)
Show answer & marking schemeHide answer & marking scheme
Worked solution
For (a), weigh the crucible and lid, add titanium, weigh again. Heat strongly using a Bunsen burner. Periodically lift the crucible lid slightly to allow oxygen/air to enter while preventing loss of titanium oxide smoke. Heat to constant mass. For (b), mass of O = 2.40 - 1.44 = 0.96 g. Moles of Ti = 1.44 / 48 = 0.030 mol. Moles of O = 0.96 / 16 = 0.060 mol. Divide by smallest: Ti = 0.030 / 0.030 = 1, O = 0.060 / 0.030 = 2. Empirical formula is \(\text{TiO}_2\). For (c), the equation is \(\text{Ti} + \text{O}_2 \rightarrow \text{TiO}_2\). For (d), Mr of \(\text{TiO}_2\) = 48 + (16 * 2) = 80. Moles of \(\text{TiO}_2\) in 15.0 g = 15.0 / 80 = 0.1875 mol. Moles of Ti needed = 0.1875 mol. Mass of Ti = 0.1875 * 48 = 9.0 g.
Marking scheme
(a) M1: Weigh crucible, lid, and titanium; M2: Heat strongly and lift lid periodically; M3: To allow oxygen to enter but prevent loss of product; M4: Heat to constant mass (4 marks). (b) M1: Calculate mass of oxygen as 0.96 g; M2: Calculate moles of Ti (0.030) and O (0.060); M3: Find simplest ratio (1:2); M4: Deduce formula as \(\text{TiO}_2\) (4 marks). (c) M1: Correct balanced equation: \(\text{Ti} + \text{O}_2 \rightarrow \text{TiO}_2\) (1 mark). (d) M1: Calculate moles of titanium oxide (0.1875 mol); M2: Calculate mass of titanium (9.0 g) (2 marks).
Question 2 · structured
11 marks
A student investigates the displacement reaction between zinc and copper(II) sulfate. (a) The student adds excess zinc powder to 50.0 cm\(^3\) of 0.200 mol/dm\(^3\) copper(II) sulfate solution in a polystyrene cup. Explain why a polystyrene cup is used instead of a glass beaker. (2 marks) (b) The temperature of the mixture increases from 18.5 °C to 24.3 °C. Calculate the heat energy change (Q) in Joules. [Assume mass of solution is 50.0 g and specific heat capacity is 4.18 J/g/°C] (3 marks) (c) Calculate the number of moles of copper(II) sulfate used in this experiment. (2 marks) (d) Use your answers to (b) and (c) to calculate the molar enthalpy change (\(\Delta H\)) for the reaction in kJ/mol. Include a sign in your final answer. (4 marks)
Show answer & marking schemeHide answer & marking scheme
Worked solution
For (a), polystyrene is a good thermal insulator which reduces heat loss to the surroundings, making the temperature measurement more accurate. For (b), Q = m * c * \(\Delta T\) = 50.0 * 4.18 * (24.3 - 18.5) = 50.0 * 4.18 * 5.8 = 1212.2 J. For (c), moles of CuSO4 = concentration * volume = 0.200 * (50.0 / 1000) = 0.0100 mol. For (d), \(\Delta H\) = -Q / (moles * 1000) = -1212.2 / (0.0100 * 1000) = -121.22 kJ/mol, which rounds to -121 kJ/mol. Since the reaction is exothermic, the sign must be negative.
Marking scheme
(a) M1: Polystyrene is a heat insulator; M2: Reduces heat loss to the surroundings (2 marks). (b) M1: Temperature change = 5.8 °C; M2: Correct substitution into Q = mc\(\Delta T\); M3: Calculation of 1212.2 J (3 marks). (c) M1: Correct conversion of volume (0.0500 dm\(^3\)); M2: Calculation of moles = 0.0100 mol (2 marks). (d) M1: Convert Q to kJ (1.2122 kJ); M2: Divide kJ by moles (121.22); M3: State negative sign; M4: Give correct final answer of -121 kJ/mol (accept answers in range -121 to -121.2 kJ/mol) (4 marks).
Question 3 · structured
11 marks
This question is about the thermal decomposition of copper(II) carbonate. (a) State the color change observed when green copper(II) carbonate solid is heated strongly. (2 marks) (b) Write a chemical equation for this thermal decomposition, including state symbols. (3 marks) (c) A student collects 72 cm\(^3\) of carbon dioxide gas at room temperature and pressure. Calculate the amount, in moles, of carbon dioxide gas collected. [Molar volume of gas at rtp = 24 dm\(^3\)] (3 marks) (d) Carbon dioxide is a greenhouse gas. Describe how greenhouse gases contribute to global warming. (3 marks)
Show answer & marking schemeHide answer & marking scheme
Worked solution
For (a), copper(II) carbonate is green and decomposes to form copper(II) oxide, which is black. So the color change is from green to black. For (b), the equation is \(\text{CuCO}_3(s) \rightarrow \text{CuO}(s) + \text{CO}_2(g)\). For (c), convert 72 cm\(^3\) to dm\(^3\): 72 / 1000 = 0.072 dm\(^3\). Moles of CO2 = volume / molar volume = 0.072 / 24 = 0.0030 mol. For (d), greenhouse gases absorb infrared radiation emitted/reflected from the Earth's surface and re-emit it in all directions, trapping heat in the atmosphere.
Marking scheme
(a) M1: Green; M2: turns to Black (2 marks). (b) M1: Correct formulas (\(\text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2\)); M2: All state symbols correct; M3: Equation balanced (3 marks). (c) M1: Convert cm\(^3\) to dm\(^3\) (0.072); M2: Divide by 24; M3: Correct evaluation to 0.0030 mol (3 marks). (d) M1: Greenhouse gases absorb infrared radiation; M2: That is re-radiated from the Earth's surface; M3: Re-emit this radiation back towards Earth, trapping heat (3 marks).
Question 4 · structured
11 marks
Crude oil is a complex mixture of hydrocarbons that must be processed to produce useful fuels. (a) Explain how fractional distillation separates crude oil into useful fractions. (4 marks) (b) A long-chain alkane, hexadecane (\(\text{C}_{16}\text{H}_{34}\)), is cracked to form octane (\(\text{C}_8\text{H}_{18}\)), butene (\(\text{C}_4\text{H}_8\)), and one other alkene. Write a balanced chemical equation for this reaction. (2 marks) (c) In an industrial cracking reactor, a student cracks 500 g of hexadecane. The theoretical yield of octane is 252 g. If the actual yield of octane obtained is 189 g, calculate the percentage yield of octane. (3 marks) (d) State the catalyst and temperature range used in industrial catalytic cracking. (2 marks)
Show answer & marking schemeHide answer & marking scheme
Worked solution
For (a), crude oil is vaporized and enters a fractionating column which is hot at the bottom and cool at the top. Hydrocarbons rise and condense at their respective boiling points. Long-chain hydrocarbons have high boiling points and condense near the bottom, while short-chain hydrocarbons have low boiling points and condense near the top. For (b), hexadecane cracking equation: \(\text{C}_{16}\text{H}_{34} \rightarrow \text{C}_8\text{H}_{18} + 2\text{C}_4\text{H}_8\). The missing alkene is butene, \(\text{C}_4\text{H}_8\). For (c), percentage yield = (actual yield / theoretical yield) * 100 = (189 / 252) * 100 = 75.0%. For (d), the catalyst is silica or alumina, and the temperature range is 600-700 °C.
Marking scheme
(a) M1: Crude oil is heated/vaporized; M2: Temperature gradient in the column (hotter at bottom, cooler at top); M3: Fractions condense at different heights/levels; M4: Due to having different boiling points (4 marks). (b) M1: Correct formula of missing alkene (\(\text{C}_4\text{H}_8\)); M2: Balanced equation: \(\text{C}_{16}\text{H}_{34} \rightarrow \text{C}_8\text{H}_{18} + 2\text{C}_4\text{H}_8\) (2 marks). (c) M1: State percentage yield formula; M2: Correct substitution (189 / 252); M3: Final value of 75% or 75.0% (3 marks). (d) M1: Silica / Alumina / Alumino-silicate catalyst; M2: Temperature in range 600-700 °C (2 marks).
Question 5 · structured
11 marks
An unknown ionic solid, compound X, is analyzed using chemical tests. (a) Describe how to perform a flame test on a solid sample. (3 marks) (b) Compound X does not produce a flame color. When sodium hydroxide solution is added to an aqueous solution of X, a brown precipitate forms. Identify the cation present in X and write an ionic equation for this reaction. (3 marks) (c) To another portion of the solution of X, dilute hydrochloric acid is added followed by barium chloride solution. A white precipitate forms. Identify the anion present in X and write an ionic equation for this reaction. (3 marks) (d) A student dissolves 1.60 g of anhydrous compound X in water to make 100 cm\(^3\) of solution. Calculate the concentration of the cation in this solution in mol/dm\(^3\). [Mr of compound X = 400] (2 marks)
Show answer & marking schemeHide answer & marking scheme
Worked solution
For (a), dip a clean platinum or nichrome wire into concentrated hydrochloric acid and then into the solid sample. Place the wire in the blue/roaring flame of a Bunsen burner and observe the color. For (b), a brown precipitate with NaOH indicates the iron(III) ion, \(\text{Fe}^{3+}\). The ionic equation is \(\text{Fe}^{3+}(aq) + 3\text{OH}^{-}(aq) \rightarrow \text{Fe(OH)}_3(s)\). For (c), a white precipitate with acidified barium chloride indicates the sulfate ion, \(\text{SO}_4^{2-}\). The ionic equation is \(\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)\). For (d), the formula of X is \(\text{Fe}_2(\text{SO}_4)_3\). Moles of \(\text{Fe}_2(\text{SO}_4)_3\) = 1.60 / 400 = 0.0040 mol. Since each mole of compound X contains 2 moles of \(\text{Fe}^{3+}\) ions, moles of \(\text{Fe}^{3+}\) = 0.0040 * 2 = 0.0080 mol. Concentration of \(\text{Fe}^{3+}\) = moles / volume in dm\(^3\) = 0.0080 / 0.100 = 0.080 mol/dm\(^3\).
Marking scheme
(a) M1: Use clean platinum or nichrome wire; M2: Dip wire in concentrated hydrochloric acid (to clean it); M3: Hold in roaring/blue Bunsen flame (3 marks). (b) M1: Identify iron(III) / \(\text{Fe}^{3+}\); M2: Correct formulae of reactants and product in ionic equation; M3: Correct balancing and state symbols: \(\text{Fe}^{3+}(aq) + 3\text{OH}^{-}(aq) \rightarrow \text{Fe(OH)}_3(s)\) (3 marks). (c) M1: Identify sulfate / \(\text{SO}_4^{2-}\); M2: Correct formulae in ionic equation; M3: Correct state symbols: \(\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)\) (3 marks). (d) M1: Calculate moles of X (0.0040 mol); M2: Scale up for 2 moles of cation and divide by 0.100 dm\(^3\) to get 0.080 mol/dm\(^3\) (2 marks).
Question 6 · structured
11 marks
A student performs a titration to find the concentration of a sodium hydroxide solution. They titrate 25.0 cm\(^3\) of sodium hydroxide solution with 0.150 mol/dm\(^3\) sulfuric acid. The mean volume of sulfuric acid needed for neutralization is 18.40 cm\(^3\). (a) Write a balanced chemical equation for the reaction. (2 marks) (b) Calculate the amount, in moles, of sulfuric acid used. (2 marks) (c) Calculate the concentration of the sodium hydroxide solution in mol/dm\(^3\). (4 marks) (d) Calculate the concentration of the sodium hydroxide solution in g/dm\(^3\). [Ar values: Na = 23, O = 16, H = 1] (3 marks)
Show answer & marking schemeHide answer & marking scheme
Worked solution
For (a), the equation is \(\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\). For (b), moles of \(\text{H}_2\text{SO}_4\) = concentration * volume in dm\(^3\) = 0.150 * (18.40 / 1000) = 0.00276 mol. For (c), from the chemical equation, 1 mole of acid reacts with 2 moles of alkali. Moles of NaOH = 2 * 0.00276 = 0.00552 mol. Concentration of NaOH = moles / volume in dm\(^3\) = 0.00552 / (25.0 / 1000) = 0.2208 mol/dm\(^3\). For (d), Mr of NaOH = 23 + 16 + 1 = 40. Concentration in g/dm\(^3\) = concentration in mol/dm\(^3\) * Mr = 0.2208 * 40 = 8.832 g/dm\(^3\).
Marking scheme
(a) M1: Correct reactants and products (\(\text{H}_2\text{SO}_4\) and \(\text{NaOH}\)); M2: Correctly balanced (\(\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\)) (2 marks). (b) M1: Convert volume to dm\(^3\) (0.0184); M2: Calculate moles (0.00276 mol) (2 marks). (c) M1: Identify 1:2 reaction ratio; M2: Moles of NaOH = 0.00552 mol; M3: Divide by 0.0250 dm\(^3\); M4: Calculate concentration as 0.221 or 0.2208 mol/dm\(^3\) (4 marks). (d) M1: Calculate Mr of NaOH = 40; M2: Multiply concentration by Mr; M3: Correct evaluation to 8.83 g/dm\(^3\) (accept range 8.8 to 8.84 g/dm\(^3\)) (3 marks).
Question 7 · structured
11 marks
An experiment is set up to determine the percentage by volume of oxygen in the air. A student places wet iron wool in the bottom of a 100 cm\(^3\) measuring cylinder and inverts it in a trough of water. (a) State the percentage by volume of the two most abundant gases in clean, dry air. (2 marks) (b) Describe how the student should use this setup over the course of a week to find the volume of oxygen that has reacted. (3 marks) (c) At the start of the experiment, the volume of air inside the cylinder is 84 cm\(^3\). After one week, the water level stops rising and the remaining gas volume is 67 cm\(^3\). Calculate the percentage of oxygen in this sample of air. (3 marks) (d) State why this reaction occurs slowly, and suggest how the student could accelerate the rusting process without changing the temperature or main reactants. (3 marks)
Show answer & marking schemeHide answer & marking scheme
Worked solution
For (a), nitrogen is approximately 78% and oxygen is approximately 21%. For (b), the student should record the initial volume of air in the measuring cylinder. Leave the apparatus for a week until the water level stops rising (constant volume is reached). Record the final volume of gas. Subtract the final volume from the initial volume to find the volume of oxygen reacted. For (c), volume of oxygen reacted = 84 - 67 = 17 cm\(^3\). Percentage of oxygen = (volume of oxygen / initial volume of air) * 100 = (17 / 84) * 100 = 20.24%, which rounds to 20.2%. For (d), the reaction is slow because rusting of iron has a high activation energy at room temperature. To speed it up, the student could use a catalyst (such as adding sodium chloride solution to the iron wool) or use more finely divided iron wool to increase surface area.
Marking scheme
(a) M1: Nitrogen 78%; M2: Oxygen 21% (accept 78-79% and 20-21%) (2 marks). (b) M1: Measure and record the initial air volume; M2: Leave the apparatus until the water level stops rising; M3: Measure the final volume of gas (3 marks). (c) M1: Calculate volume change = 17 cm\(^3\); M2: Show fraction (17 / 84); M3: Correct final percentage calculation of 20.2% (accept range 20% to 20.3%) (3 marks). (d) M1: Slow because activation energy is high at room temperature; M2: Speed up by adding an ionic solution/salt to the iron wool; M3: Use finer iron wool to increase surface area (3 marks).
Question 8 · structured
11 marks
This question is about energetics and bond calculations. (a) Explain how you would identify an exothermic reaction from a reaction profile (energy level diagram), stating what the labels \(\Delta H\) and \(E_a\) represent on such a diagram. (4 marks) (b) Methane burns in oxygen according to the equation: \(\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)\). Use the following bond energies to calculate the enthalpy change (\(\Delta H\)) for this reaction in kJ/mol: C-H = 413 kJ/mol, O=O = 498 kJ/mol, C=O = 805 kJ/mol, O-H = 464 kJ/mol. (5 marks) (c) State whether the reaction is endothermic or exothermic, and explain this in terms of the energy associated with bond breaking and bond making. (2 marks)
Show answer & marking schemeHide answer & marking scheme
Worked solution
For (a), in an exothermic reaction, the products have less energy than the reactants, so the energy level curve goes down. \(\Delta H\) represents the overall enthalpy change (difference in energy between reactants and products) and is negative. \(E_a\) represents the activation energy (minimum energy required to start the reaction), represented by the arrow from the reactant level to the peak of the curve. For (b), Bonds broken (reactants): 4 * (C-H) + 2 * (O=O) = 4 * 413 + 2 * 498 = 1652 + 996 = 2648 kJ/mol. Bonds made (products): 2 * (C=O) + 4 * (O-H) = 2 * 805 + 4 * 464 = 1610 + 1856 = 3466 kJ/mol. \(\Delta H\) = Bonds broken - Bonds made = 2648 - 3466 = -818 kJ/mol. For (c), the reaction is exothermic because the energy released when forming new bonds (3466 kJ/mol) is greater than the energy taken in to break the existing bonds (2648 kJ/mol).
Marking scheme
(a) M1: Product level is lower than reactant level; M2: \(\Delta H\) is the overall enthalpy change; M3: \(E_a\) is activation energy; M4: Peak represents the transition state, with arrow from reactant level to peak representing activation energy (4 marks). (b) M1: Calculate energy needed to break reactant bonds (1652 + 996 = 2648); M2: Correct calculation of C-H total (1652) and O=O total (996); M3: Calculate energy released making product bonds (1610 + 1856 = 3466); M4: Correct calculation of C=O total (1610) and O-H total (1856); M5: Correct subtraction resulting in -818 kJ/mol (5 marks). (c) M1: Exothermic; M2: Because more energy is released when forming bonds than is absorbed when breaking bonds (2 marks).
Question 9 · Structured
11 marks
A student performs an experiment to determine the formula of hydrated cobalt(II) chloride, \(\text{CoCl}_2 \cdot x\text{H}_2\text{O}\), by heating a sample to remove the water of crystallisation. (a) State why the crucible is heated, cooled and reweighed multiple times during this experiment. [2 marks] (b) Describe the colour change when hydrated cobalt(II) chloride is heated. [2 marks] (c) A student heats a sample of hydrated cobalt(II) chloride in a crucible. The following results are obtained: mass of empty crucible = 22.40 g; mass of crucible + hydrated cobalt(II) chloride = 27.16 g; mass of crucible + anhydrous cobalt(II) chloride = 25.00 g. Calculate the value of \(x\) in \(\text{CoCl}_2 \cdot x\text{H}_2\text{O}\). Show your working. (\(A_r\): \(\text{Co} = 59\), \(\text{Cl} = 35.5\), \(\text{H} = 1\), \(\text{O} = 16\)) [5 marks] (d) The student noticed some condensation forming on the upper inside walls of the crucible which disappeared upon stronger heating. Explain how the calculated value of \(x\) would be affected if this condensation did not completely evaporate before weighing. [2 marks]
Show answer & marking schemeHide answer & marking scheme
Worked solution
For (a), the crucible is heated, cooled, and reweighed multiple times to ensure all the water of crystallisation is completely driven off, which is known as heating to a constant mass. For (b), the colour change is from pink to blue. For (c), we first calculate the mass of anhydrous \(\text{CoCl}_2\): \(25.00\text{ g} - 22.40\text{ g} = 2.60\text{ g}\). The mass of water lost: \(27.16\text{ g} - 25.00\text{ g} = 2.16\text{ g}\). Next, calculate the number of moles of anhydrous \(\text{CoCl}_2\): \(M_r\) of \(\text{CoCl}_2 = 59 + 2(35.5) = 130\). Moles of \(\text{CoCl}_2 = \frac{2.60\text{ g}}{130\text{ g/mol}} = 0.020\text{ mol}\). Calculate the number of moles of water: \(M_r\) of \(\text{H}_2\text{O} = 18\). Moles of \(\text{H}_2\text{O} = \frac{2.16\text{ g}}{18\text{ g/mol}} = 0.120\text{ mol}\). Determine the simplest molar ratio: \(\text{H}_2\text{O}\) to \(\text{CoCl}_2 = \frac{0.120}{0.020} = 6\). Therefore, \(x = 6\). For (d), if the condensation did not completely evaporate, the measured mass of the anhydrous salt would be higher than it should be. This would make the calculated mass of water lost lower, which would result in a lower calculated value of \(x\).
Marking scheme
(a) M1: to ensure all water of crystallisation has been removed / driven off (1) M2: to heat to a constant mass (1). (b) M1: pink (1) M2: to blue (1). (c) M1: calculation of mass of anhydrous \(\text{CoCl}_2\) (2.60 g) and mass of water lost (2.16 g) (1) M2: calculation of \(M_r\) of \(\text{CoCl}_2 = 130\) (1) M3: calculation of moles of \(\text{CoCl}_2 = 0.020\text{ mol}\) and moles of \(\text{H}_2\text{O} = 0.120\text{ mol}\) (1) M4: division of moles by the smallest value to get ratio 1 : 6 (1) M5: final answer \(x = 6\) (1). (d) M1: the calculated mass of water lost would be smaller / the measured mass of anhydrous salt would be larger (1) M2: therefore, the calculated value of \(x\) would be smaller (1).
Question 10 · Structured
11 marks
A student uses a polystyrene cup calorimeter to determine the enthalpy change (\(\Delta H\)) for the displacement reaction: \(\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}\). (a) Explain why a polystyrene cup is used instead of a copper beaker in this experiment. [2 marks] (b) In an experiment, a student adds excess zinc powder to \(50.0\text{ cm}^3\) of \(0.400\text{ mol/dm}^3\) copper(II) sulfate solution. The initial temperature of the solution is \(19.5\text{ }^\circ\text{C}\) and the maximum temperature reached is \(30.7\text{ }^\circ\text{C}\). Calculate the heat energy change (\(Q\)) in joules for this reaction. Assume the density of the solution is \(1.00\text{ g/cm}^3\) and its specific heat capacity is \(4.18\text{ J/g/}^\circ\text{C}\). [4 marks] (c) Use your answer to part (b) to calculate the enthalpy change (\(\Delta H\)) for this reaction in kJ/mol. Include a sign in your final answer. If you could not calculate \(Q\), use a value of \(2500\text{ J}\). [4 marks] (d) Suggest one source of error, other than heat loss, that could lead to an inaccurate value of \(\Delta H\). [1 mark]
Show answer & marking schemeHide answer & marking scheme
Worked solution
For (a), polystyrene is a good thermal insulator which reduces heat loss to the surroundings, making the temperature measurement more accurate. A copper beaker is a good thermal conductor and would lose heat quickly. For (b), use \(Q = m c \Delta T\). Mass (\(m\)) = \(50.0\text{ cm}^3 \times 1.00\text{ g/cm}^3 = 50.0\text{ g}\). Temperature change (\(\Delta T\)) = \(30.7\text{ }^\circ\text{C} - 19.5\text{ }^\circ\text{C} = 11.2\text{ }^\circ\text{C}\). \(Q = 50.0\text{ g} \times 4.18\text{ J/g/}^\circ\text{C} \times 11.2\text{ }^\circ\text{C} = 2340.8\text{ J}\). For (c), calculate the moles of \(\text{CuSO}_4\): \(\text{moles} = \text{volume (dm}^3\text{)} \times \text{concentration} = 0.0500\text{ dm}^3 \times 0.400\text{ mol/dm}^3 = 0.0200\text{ mol}\). Since zinc is in excess, \(\text{CuSO}_4\) is the limiting reactant. Convert \(Q\) to kJ: \(\frac{2340.8\text{ J}}{1000} = 2.3408\text{ kJ}\). Enthalpy change \(\Delta H = -\frac{Q}{\text{moles}} = -\frac{2.3408\text{ kJ}}{0.0200\text{ mol}} = -117.04\text{ kJ/mol}\). Rounded to 3 significant figures, this is \(-117\text{ kJ/mol}\). If using the dummy value of \(2500\text{ J}\): moles = \(0.0200\text{ mol}\), \(Q = 2.50\text{ kJ}\), \(\Delta H = -\frac{2.50}{0.0200} = -125\text{ kJ/mol}\). For (d), other sources of error include: incomplete reaction, slow transfer of zinc causing heat loss before maximum temperature is reached, or the specific heat capacity of the copper(II) sulfate solution is not exactly the same as pure water.
Marking scheme
(a) M1: polystyrene is a good thermal insulator / poor conductor of heat (1) M2: this reduces heat loss to the surroundings (1). (b) M1: state formula \(Q = mc\Delta T\) (1) M2: calculate \(\Delta T = 11.2\text{ }^\circ\text{C}\) (1) M3: calculate mass of solution = 50.0 g (1) M4: calculate \(Q = 2340.8\text{ J}\) (accept 2340 J or 2341 J) (1). (c) M1: calculate moles of \(\text{CuSO}_4 = 0.0500 \times 0.400 = 0.0200\text{ mol}\) (1) M2: convert \(Q\) to kJ = 2.3408 kJ (1) M3: divide kJ by moles to get 117 (1) M4: include correct negative sign and units (accept -117 kJ/mol, allow ECF from part b) (1). [If dummy value used: \(\frac{2.50}{0.0200} = -125\text{ kJ/mol}\) scores max 4 marks]. (d) M1: any one from: incomplete reaction / some zinc did not react (1) OR copper(II) sulfate solution concentration was not exact (1) OR heat absorbed by the polystyrene cup / thermometer itself (1).
Paper 2C (4CH1/2C)
Answer all questions. Show all calculation steps. Total marks: 70.
7 Question · 70 marks
Question 1 · Advanced Structured Question
10 marks
A student designs an experiment to determine the enthalpy change of displacement for the reaction between zinc and copper(II) sulfate solution.\ \ \( \text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)} \)\ \ The student uses a polystyrene cup as a calorimeter. They place \( 50.0\text{ cm}^3 \) of \( 0.200\text{ mol/dm}^3 \) copper(II) sulfate solution in the cup and record its initial temperature. They then add an excess of zinc powder and stir the mixture, recording the maximum temperature reached. The temperature of the solution increases by \( 10.5\text{ }^\circ\text{C} \).\ \ (a) Calculate the heat energy released, in joules, during this reaction. (Assume the specific heat capacity of the solution is \( 4.18\text{ J/g/}^\circ\text{C} \) and its density is \( 1.00\text{ g/cm}^3 \)). [2 marks]\ \ (b) Calculate the number of moles of copper(II) sulfate used in this experiment. [1 mark]\ \ (c) Calculate the molar enthalpy change (\( \Delta H \)) for this displacement reaction in \( \text{kJ/mol} \). State the correct sign and express your final answer to 3 significant figures. [3 marks]\ \ (d) Explain why a polystyrene cup is preferred over a glass beaker for this experiment, and why it is important to use an excess of zinc. [2 marks]\ \ (e) The student's experimental value is less exothermic than the accepted data book value. Suggest two changes to the apparatus or procedure (excluding using a polystyrene cup) that would improve the accuracy of the results. [2 marks]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Heat energy released: \( q = m \times c \times \Delta T \)\ \( m = 50.0\text{ cm}^3 \times 1.00\text{ g/cm}^3 = 50.0\text{ g} \)\ \( q = 50.0 \times 4.18 \times 10.5 = 2194.5\text{ J} \) (or \( 2190\text{ J} \) to 3 s.f.)\ \ (b) Moles of \( \text{CuSO}_4 = \text{concentration} \times \text{volume} = 0.200\text{ mol/dm}^3 \times 0.0500\text{ dm}^3 = 0.0100\text{ mol} \)\ \ (c) Molar enthalpy change: \( \Delta H = -\frac{q}{\text{moles}} \)\ \( \Delta H = -\frac{2194.5\text{ J}}{0.0100\text{ mol}} = -219450\text{ J/mol} = -219.45\text{ kJ/mol} \)\ Rounding to 3 s.f. gives \( -219\text{ kJ/mol} \).\ \ (d) A polystyrene cup is a good thermal insulator, reducing heat loss to the surroundings. An excess of zinc ensures that all the copper(II) sulfate reacts completely. \ (e) Put a lid on the polystyrene cup to reduce heat loss by convection/evaporation. Use a wind shield or wrap the cup in cotton wool/insulating material.
Marking scheme
M1: Heat energy calculation: \( 50.0 \times 4.18 \times 10.5 = 2194.5\text{ J} \) (1)\ M2: Correct units or 3 s.f. conversion (1)\ M3: Moles calculation: \( 0.0500 \times 0.200 = 0.0100\text{ mol} \) (1)\ M4: Division of energy by moles: \( 2194.5 / 0.0100 = 219.45\text{ kJ/mol} \) (1)\ M5: Correct negative sign (1)\ M6: Expressed to 3 significant figures: \( -219\text{ kJ/mol} \) (1)\ M7: Polystyrene cup reduces heat loss (1)\ M8: Excess zinc ensures complete reaction of copper(II) sulfate (1)\ M9: Suggestion 1: Use a lid on the cup (1)\ M10: Suggestion 2: Extra insulation around the cup (e.g. cotton wool, beaker) or use a digital thermometer/data logger with higher resolution (1)
Question 2 · Advanced Structured Question
10 marks
A student uses heated copper to determine the percentage by volume of oxygen in a sample of air.\ \ (a) Describe how the student can use two gas syringes and heated copper wire to determine the percentage by volume of oxygen in a sample of air. Your description should include the observations made during the experiment. [4 marks]\ \ (b) In an experiment, the initial volume of air in one syringe is \( 80.0\text{ cm}^3 \). After heating the copper and passing the air over it until the volume remains constant, the apparatus is allowed to cool. The final volume of gas is \( 63.6\text{ cm}^3 \).\ \ (i) Explain why the gas must be cooled to room temperature before measuring the final volume. [2 marks]\ \ (ii) Calculate the percentage by volume of oxygen in this sample of air. [2 marks]\ \ (iii) Name the copper compound formed on the surface of the copper wire and write a balanced chemical equation for its formation. [2 marks]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Place copper wire in a silica tube connected between two gas syringes containing a known volume of air (e.g. \( 80.0\text{ cm}^3 \)). Heat the copper strongly using a Bunsen burner. Push the plunger of one syringe to pass the air over the heated copper into the other syringe, and repeat this process continuously. The pink/brown copper wire turns black. Continue heating until there is no further decrease in the volume of the gas. Allow to cool and measure the final volume.\ \ (b)(i) Gases expand when heated / occupy a larger volume at higher temperatures. Cooling to room temperature ensures that the initial and final volumes are measured under the same conditions for a fair and accurate comparison.\ \ (ii) Volume of oxygen reacted = \( 80.0\text{ cm}^3 - 63.6\text{ cm}^3 = 16.4\text{ cm}^3 \).\ Percentage of oxygen = \( \frac{16.4}{80.0} \times 100 = 20.5\\% \).\ \ (iii) Copper(II) oxide.\ Equation: \( 2\text{Cu(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{CuO(s)} \).
Marking scheme
M1: Heat the copper wire strongly and pass air back and forth over it using the plungers (1)\ M2: Heat until the volume of gas stops decreasing / remains constant (1)\ M3: Correct observation: copper turns from pink-brown to black (1)\ M4: Allow to cool before taking final volume reading (1)\ M5: Reference to gas expanding when hot / volume increasing with temperature (1)\ M6: Need to measure both volumes at the same temperature (1)\ M7: Calculate decrease in volume: \( 80.0 - 63.6 = 16.4\text{ cm}^3 \) (1)\ M8: Calculate percentage: \( \frac{16.4}{80.0} \times 100 = 20.5\\% \) (1)\ M9: Copper(II) oxide (1)\ M10: \( 2\text{Cu} + \text{O}_2 \rightarrow 2\text{CuO} \) (1)
Question 3 · Advanced Structured Question
10 marks
Crude oil is a complex mixture of hydrocarbons that is separated into fractions using fractional distillation. Shorter-chain alkanes and alkenes are then obtained from heavier fractions by catalytic cracking.\ \ (a) Explain how the properties of fractions change from the top to the bottom of a fractional distillation column. Refer to boiling point, viscosity, and flammability in your answer. [3 marks]\ \ (b) Catalytic cracking is used to break down larger alkanes into smaller, more useful molecules.\ \ (i) State the catalyst and temperature range used in industrial catalytic cracking. [2 marks]\ \ (ii) Explain why cracking is an economically vital process in the oil industry. [2 marks]\ \ (c) A molecule of decane, \( \text{C}_{10}\text{H}_{22} \), undergoes cracking to form one molecule of propene, one molecule of but-1-ene, and one molecule of an alkane \( X \).\ \ (i) State the molecular formula of alkane \( X \). [1 mark]\ \ (ii) Write a balanced chemical equation for this cracking reaction. [1 mark]\ \ (iii) Draw the displayed formula of but-1-ene. [1 mark]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) As you move from the top to the bottom of the column, the molecules become larger. This results in: 1. Boiling points increase (due to stronger intermolecular forces). 2. Viscosity increases (the liquids become thicker and flow less easily). 3. Flammability decreases (they become harder to ignite).\ \ (b)(i) Catalyst: Alumina (aluminium oxide) or silica (silicon dioxide). Temperature: 600–700 °C (accept any temperature in the range 500–800 °C).\ \ (ii) Fractional distillation produces a surplus of long-chain hydrocarbons and a deficit of short-chain hydrocarbons compared to market demand. Cracking converts surplus long-chain alkanes into high-demand shorter-chain alkanes (used as fuels) and alkenes (used to make polymers/plastics).\ \ (c)(i) Propene is \( \text{C}_3\text{H}_6 \). But-1-ene is \( \text{C}_4\text{H}_8 \).\ Number of carbon atoms in alkane \( X = 10 - 3 - 4 = 3 \).\ Formula of alkane \( X = \text{C}_3\text{H}_8 \) (propane).\ \ (ii) Equation: \( \text{C}_{10}\text{H}_{22} \rightarrow \text{C}_3\text{H}_6 + \text{C}_4\text{H}_8 + \text{C}_3\text{H}_8 \).\ \ (iii) Displayed formula of but-1-ene showing all atoms and bonds: H2C=CH-CH2-CH3 with each carbon having 4 bonds.
Marking scheme
M1: Boiling points increase down the column (1)\ M2: Viscosity increases / becomes thicker down the column (1)\ M3: Flammability decreases / harder to ignite down the column (1)\ M4: Silica / Alumina / Aluminosilicates (1)\ M5: Temperature 600-700 °C (accept 500-800 °C) (1)\ M6: Explains that short-chain alkanes are in higher demand as fuels (1)\ M7: Explains that alkenes produced are used for polymerisation / making plastics (1)\ M8: \( \text{C}_3\text{H}_8 \) (1)\ M9: Balanced equation: \( \text{C}_{10}\text{H}_{22} \rightarrow \text{C}_3\text{H}_6 + \text{C}_4\text{H}_8 + \text{C}_3\text{H}_8 \) (1)\ M10: Correct displayed formula of but-1-ene (showing C=C double bond, and correct number of H atoms on each C) (1)
Question 4 · Advanced Structured Question
10 marks
An unknown solid compound, \( Y \), contains one metal cation and one halide anion. A student carries out tests to identify these ions.\ \ (a) The student carries out a flame test on solid \( Y \) and observes a lilac flame. Identify the cation present. [1 mark]\ \ (b) The student dissolves solid \( Y \) in deionised water to form a solution. They split the solution into two portions.\ \ (i) To the first portion, the student adds dilute nitric acid followed by silver nitrate solution. A cream precipitate forms. Identify this precipitate and write an ionic equation, including state symbols, for its formation. [3 marks]\ \ (ii) Identify compound \( Y \). [1 mark]\ \ (iii) State why dilute nitric acid is added before the silver nitrate solution. [2 marks]\ \ (c) If compound \( Y \) were instead replaced with ammonium bromide, \( \text{NH}_4\text{Br} \):\ \ (i) Describe the tests and observations you would make to confirm the presence of the ammonium ion. [2 marks]\ \ (ii) Write a balanced chemical equation for the reaction that occurs when ammonium bromide is heated with sodium hydroxide solution. [1 mark]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Potassium ion / \( \text{K}^+ \).\ \ (b)(i) Cream precipitate: silver bromide (\( \text{AgBr} \)).\ Ionic equation: \( \text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)} \).\ \ (ii) Potassium bromide / \( \text{KBr} \).\ \ (iii) Dilute nitric acid is added to react with and remove any carbonate or sulfite ions that might be present in the solution. If not removed, these ions would form a precipitate with silver ions, giving a false positive result.\ \ (c)(i) Add sodium hydroxide solution and warm the mixture. Test the gas produced with damp red litmus paper. Observation: gas turns damp red litmus paper blue.\ \ (ii) \( \text{NH}_4\text{Br} + \text{NaOH} \rightarrow \text{NaBr} + \text{H}_2\text{O} + \text{NH}_3 \).
Marking scheme
M1: Potassium ion / \( \text{K}^+ \) (1)\ M2: Silver bromide (1)\ M3: Ionic equation: \( \text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr} \) (1)\ M4: State symbols: \( \text{(aq)} \) for reactants, \( \text{(s)} \) for product (1)\ M5: Potassium bromide / \( \text{KBr} \) (1)\ M6: To remove/destroy carbonate ions / impurities (1)\ M7: To prevent the formation of other silver precipitates / false positive (1)\ M8: Warm with sodium hydroxide solution (1)\ M9: Gas (ammonia) turns damp red litmus paper blue (1)\ M10: \( \text{NH}_4\text{Br} + \text{NaOH} \rightarrow \text{NaBr} + \text{H}_2\text{O} + \text{NH}_3 \) (1)
Question 5 · Advanced Structured Question
10 marks
A student determines the formula of hydrated magnesium sulfate, \( \text{MgSO}_4 \cdot x\text{H}_2\text{O} \), by heating a sample in a crucible to drive off the water of crystallisation.\ \ (a) Describe how the student should carry out this experiment to ensure that all the water of crystallisation has been removed. [3 marks]\ \ (b) The student records the following data:\ - Mass of empty crucible = \( 22.45\text{ g} \)\ - Mass of crucible + hydrated magnesium sulfate = \( 24.91\text{ g} \)\ - Mass of crucible + anhydrous magnesium sulfate = \( 23.65\text{ g} \)\ \ (i) Calculate the mass of anhydrous magnesium sulfate, \( \text{MgSO}_4 \), obtained. [1 mark]\ \ (ii) Calculate the mass of water of crystallisation lost. [1 mark]\ \ (iii) Calculate the value of \( x \) in the formula \( \text{MgSO}_4 \cdot x\text{H}_2\text{O} \). Show your working. (Relative formula masses: \( \text{MgSO}_4 = 120.4 \), \( \text{H}_2\text{O} = 18.0 \)). [4 marks]\ \ (c) Suggest why the calculated value of \( x \) would be lower than the true value if the sample were not heated to constant mass. [1 mark]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Heat the hydrated salt in the crucible gently at first, then strongly. Allow the crucible to cool and then weigh it. Repeat the heating, cooling, and weighing process until two consecutive mass readings are identical (heating to constant mass).\ \ (b)(i) Mass of anhydrous \( \text{MgSO}_4 = 23.65\text{ g} - 22.45\text{ g} = 1.20\text{ g} \).\ \ (ii) Mass of water lost = \( 24.91\text{ g} - 23.65\text{ g} = 1.26\text{ g} \).\ \ (iii) Moles of \( \text{MgSO}_4 = \frac{1.20\text{ g}}{120.4\text{ g/mol}} = 0.009967\text{ mol} \).\ Moles of \( \text{H}_2\text{O} = \frac{1.26\text{ g}}{18.0\text{ g/mol}} = 0.07000\text{ mol} \).\ Ratio of \( \text{H}_2\text{O} : \text{MgSO}_4 = \frac{0.07000}{0.009967} = 7.02 \).\ Since \( x \) must be an integer, \( x = 7 \).\ \ (c) If it is not heated to constant mass, some water of crystallisation remains in the solid. This makes the measured mass of water lost smaller than it should be, resulting in a lower ratio of water to salt and a lower value of \( x \).
Marking scheme
M1: Heat the crucible containing the hydrated salt (1)\ M2: Cool and weigh (1)\ M3: Repeat heating and weighing until mass is constant (1)\ M4: Mass of anhydrous salt = 1.20 g (1)\ M5: Mass of water lost = 1.26 g (1)\ M6: Calculate moles of anhydrous salt: \( 1.20 / 120.4 = 0.00997\text{ mol} \) (1)\ M7: Calculate moles of water: \( 1.26 / 18.0 = 0.0700\text{ mol} \) (1)\ M8: Find the simplest whole number ratio: \( 0.0700 / 0.00997 = 7 \) (1)\ M9: Deduce \( x = 7 \) (1)\ M10: Explain that some water remains in the sample, making the mass loss (and therefore calculated moles of water) too small (1)
Question 6 · Advanced Structured Question
10 marks
This question is about the thermal decomposition of calcium carbonate, \( \text{CaCO}_3 \).\ \ (a) Write a balanced chemical equation, including state symbols, for the thermal decomposition of calcium carbonate. [2 marks]\ \ (b) A student heats \( 5.00\text{ g} \) of pure calcium carbonate until it completely decomposes. \ (i) Calculate the maximum theoretical volume, in \( \text{dm}^3 \), of carbon dioxide gas produced. Assume the reaction takes place at room temperature and pressure (rtp), where 1 mole of gas occupies \( 24.0\text{ dm}^3 \). (Relative formula mass: \( \text{CaCO}_3 = 100.1 \)). [3 marks]\ \ (ii) In the experiment, the student collects \( 1.08\text{ dm}^3 \) of carbon dioxide gas. Calculate the percentage yield of carbon dioxide. [2 marks]\ \ (c) The student performs the decomposition in an open crucible on a balance.\ \ (i) State and explain the change in mass of the crucible and its contents during the reaction. [1 mark]\ \ (ii) Sketch a graph to show how the mass of the crucible and its contents changes with time from the moment heating begins until the reaction is complete. Label the axes. [2 marks]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Equation: \( \text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)} \).\ \ (b)(i) Moles of \( \text{CaCO}_3 = \frac{5.00}{100.1} = 0.04995\text{ mol} \).\ From the 1:1 stoichiometry, moles of \( \text{CO}_2 = 0.04995\text{ mol} \).\ Volume of \( \text{CO}_2 = 0.04995 \times 24.0 = 1.199\text{ dm}^3 \) (accept \( 1.20\text{ dm}^3 \)).\ \ (ii) Percentage yield = \( \frac{\text{Actual Volume}}{\text{Theoretical Volume}} \times 100 \)\ Percentage yield = \( \frac{1.08}{1.199} \times 100 = 90.07\\% \) (accept \( 90.0\\% \) or \( 90\\% \)).\ \ (c)(i) The mass decreases because carbon dioxide gas is produced and escapes from the open crucible into the atmosphere.\ \ (ii) Sketch graph: Y-axis labeled 'Mass / g' (or 'Mass'), X-axis labeled 'Time / s' (or 'Time'). The curve should start at a high value, decrease with a decreasing gradient, and then become perfectly horizontal (constant mass) when the reaction is complete.
Marking scheme
M1: Balanced chemical equation: \( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \) (1)\ M2: State symbols: \( \text{(s)} \) for \( \text{CaCO}_3 \) and \( \text{CaO} \), \( \text{(g)} \) for \( \text{CO}_2 \) (1)\ M3: Calculate moles of \( \text{CaCO}_3 = 5.00 / 100.1 = 0.04995\text{ mol} \) (1)\ M4: Equate moles of \( \text{CO}_2 \) to moles of \( \text{CaCO}_3 \) (1:1 ratio) (1)\ M5: Calculate volume of gas: \( 0.04995 \times 24.0 = 1.20\text{ dm}^3 \) (1)\ M6: Percentage yield formula / substitution: \( 1.08 / 1.20 \times 100 \) (1)\ M7: Correct final percentage yield: \( 90.0\\% \) (accept range 90.0% to 90.1%) (1)\ M8: Explains that mass decreases because CO2 gas escapes (1)\ M9: Graph axes labeled correctly (y-axis: mass, x-axis: time) (1)\ M10: Curve starts high, slopes downwards with decreasing rate, and flattens out to horizontal (1)
Question 7 · Advanced Structured Question
10 marks
A student determines the concentration of a sample of hydrochloric acid by titrating it against a standard solution of sodium hydroxide.\ \ (a) First, the student prepares a standard solution by dissolving \( 2.00\text{ g} \) of solid sodium hydroxide, \( \text{NaOH} \), in deionised water to make exactly \( 250\text{ cm}^3 \) of solution.\ Calculate the concentration, in \( \text{mol/dm}^3 \), of this sodium hydroxide solution. (Relative formula mass: \( \text{NaOH} = 40.0 \)). [3 marks]\ \ (b) The student transfers \( 25.0\text{ cm}^3 \) of the sodium hydroxide solution to a conical flask.\ \ (i) Describe the practical steps to carry out the titration to find the volume of hydrochloric acid needed to neutralise the sodium hydroxide. Name the piece of apparatus used to measure the sodium hydroxide solution, the apparatus containing the acid, and state the color change of a suitable indicator. [4 marks]\ \ (ii) The average titre of the hydrochloric acid used is \( 21.50\text{ cm}^3 \). The equation for the reaction is:\ \ \( \text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)} \)\ \ Calculate the concentration, in \( \text{mol/dm}^3 \), of the hydrochloric acid. Give your answer to 3 significant figures. [3 marks]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Moles of \( \text{NaOH} = \frac{2.00\text{ g}}{40.0\text{ g/mol}} = 0.0500\text{ mol} \).\ Volume of solution in \( \text{dm}^3 = \frac{250}{1000} = 0.250\text{ dm}^3 \).\ Concentration of \( \text{NaOH} = \frac{0.0500\text{ mol}}{0.250\text{ dm}^3} = 0.200\text{ mol/dm}^3 \).\ \ (b)(i) Use a pipette (and pipette filler) to measure and transfer exactly \( 25.0\text{ cm}^3 \) of sodium hydroxide solution into a clean conical flask. Add a few drops of phenolphthalein indicator to the flask, which turns pink. Fill a burette with hydrochloric acid and record the initial reading. Add the acid slowly from the burette to the conical flask while swirling constantly. As the end point is approached, add the acid dropwise until the pink color just disappears (turns colourless). Record the final volume and repeat to obtain concordant results.\ \ (ii) Moles of \( \text{NaOH} = \text{concentration} \times \text{volume} = 0.200\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.00500\text{ mol} \).\ From the 1:1 reaction ratio, moles of \( \text{HCl} = 0.00500\text{ mol} \).\ Concentration of \( \text{HCl} = \frac{\text{moles}}{\text{volume}} = \frac{0.00500\text{ mol}}{0.02150\text{ dm}^3} = 0.23256\text{ mol/dm}^3 \).\ Rounding to 3 s.f. gives \( 0.233\text{ mol/dm}^3 \).
Marking scheme
M1: Calculate moles of NaOH: \( 2.00 / 40.0 = 0.0500\text{ mol} \) (1)\ M2: Convert volume to dm3: \( 250 / 1000 = 0.250\text{ dm}^3 \) (1)\ M3: Concentration of NaOH: \( 0.0500 / 0.250 = 0.200\text{ mol/dm}^3 \) (1)\ M4: Use a pipette to transfer the sodium hydroxide (1)\ M5: Use a burette to add the hydrochloric acid (1)\ M6: Name indicator and correct colour change (e.g. phenolphthalein from pink to colourless, or methyl orange from yellow to orange/red) (1)\ M7: Add dropwise near the end-point and swirl (1)\ M8: Calculate moles of NaOH in 25.0 cm3: \( 0.200 \times 0.0250 = 0.00500\text{ mol} \) (1)\ M9: Relate moles of HCl to NaOH (1:1 ratio, so moles of HCl = 0.00500) (1)\ M10: Calculate concentration of HCl: \( 0.00500 / 0.02150 = 0.233\text{ mol/dm}^3 \) (1)
Wondering how well you actually know this?
Thinka is an AI practice app for DSE students — unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.