ACSI 2022 Year 5 Chemistry HL final Exam suggested solutions

1 YEAR 5 Final Examination 2022 INTERNATIONAL BACCALAUREATE DIPLOMA PROGRAMME CHEMISTRY HIGHER LEVEL PAPER 2 Monday 12th September 2022 2 hours 15 minutes INSTRUCTIONS TO CANDIDATES • Do not open this examination paper until instructed to do so. • Write your candidate session number in the box above. • A calculator is required for this paper. • A copy of the Chemistry Data Booklet is required for this paper. • Write your answers in the boxes provided. • If you use additional sheets of paper for your answer, attach them to the booklet. Indicate the question number clearly on these sheets. • All drawings must be in ink. For examiner’s use Qn 1 /10 Qn 2 /6 Qn 3 /11 Qn 4 /5 Qn 5 /13 Qn 6 /8 Qn 7 /17 Qn 8 /11 Qn 9 /9 Wrong s.f. /units Total /90 ________________________________________________________________________ This question paper consists of 26 printed pages including this cover page.

2 Answer all questions. Write your answers in the boxes provided. n(H2SO4) = 25/1000 x 1.00 = 0.0250 mol n(H2SO4) = 16.2/1000 x 2.00 = 0.0324 mol 1. Chile saltpetre is a mineral found in Chile and Peru, which mainly consists of sodium nitrate, NaNO3. The mineral is purified to concentrate the NaNO3 which is used as a fertilizer and in some fireworks. To determine the purity of a sample of NaNO3, a (1.64 ± 0.02) g impure sample was heated in NaOH (aq) with Devarda’s alloy which contains aluminium. This reduces the NaNO3 to ammonia which is boiled off and then dissolved in acid as shown in the following equation. 3NaNO3 (aq) + 8Al (s) + 5NaOH (aq) + 18H2O (l) → 3NH3 (g) + 8NaAl(OH)4 (aq) The NH3 gas produced is dissolved in (25.00 ± 0.06) cm3 of 1.00 mol dm–3 H2SO4. 2NH3 + H2SO4 → (NH4)2SO4 The resulting solution was titrated with 2.00 mol dm–3 NaOH to determine the amount of unreacted H2SO4. (16.20 ± 0.10) cm3 of NaOH were required for complete neutralization. H2SO4 + 2NaOH → Na2SO4 + 2H2O (a) (i) Calculate the amount of H2SO4 present in 25.0 cm3 of 1.00 mol dm–3 H2SO4. [1] (ii) Calculate the amount of NaOH present in 16.2 cm3 of 2.00 mol dm-3 NaOH. [1] Marker’s comments: Well answered. 1m was deducted for incorrect unit (eg mols). Marker’s comments: Well answered.

3 n(H2SO4)reacted = 0.0250 - ½ (0.0324) = 0.00880 mol n(NaNO3) = n(NH3) = = 2 x 0.00880 = 0.0176 mol (b) (i) Use your answers in (a)(i) and (a)(ii), determine the amount of NaNO3 that is present in the sample. [2] (If you do not have answers for (a)(i) and (a)(ii), use n(H2SO4) = 0.0500 mol and n(NaOH) = 0.0250 mol, but these are not the correct answers). (ii) Calculate the percentage uncertainty for the amount NaNO3 determined in (b)(i). [2] (If you do not have answers for (a)(iii), use n(NaNO3) = 0.0100 mol but this is not the correct answer). 𝟎.𝟎𝟔 𝟐𝟓 (𝟎. 𝟎𝟐𝟓) + 𝟎.𝟏𝟎 𝟏𝟔.𝟐𝟎 ( 𝟏 𝟐 × 𝟎. 𝟎𝟑𝟐𝟒) = 0.00016 % unc for n(NaNO3) = 𝟎.𝟎𝟎𝟎𝟏𝟔 𝟎.𝟎𝟎𝟖𝟖𝟎 x 100 = 1.818% = ±1.8% (2sf; <2%) Marker’s comments: Generally well answered. Some common mistakes include: 1. calculating no. mol of H2SO4 with incorrect value from part (a). 2. oversight on taking into account of back titration. 3. missing out the mole ratio in the chemical equations involved. Marker’s comments: Poorly answered. Many candidates did not realise that the sum of absolute uncertainties is required as the corresponding calculation step in (b)(i) was subtraction. Several candidates used the incorrect values for calculation and were not awarded full marks though a correct answer was obtained. Candidates also overlooked the correct significant figure for the final calculated percentage uncertainty, which is 2 sf if it is <2%.

4 ``` ` `` (iii) Use your answer in (b)(i) to determine the percentage composition of NaNO3 present in the impure sample. [2] (If you do not have answers for (c), use n(NaNO3) = 0.0100 mol but this is not the correct answer). % composition (NaNO3) = [0.0176 mol x Mr(NaNO3) / 1.64] x 100 = 91.21951 ≈ 91.2% (3 sf) (iv) Calculate the absolute uncertainty of the percentage composition of NaNO3. [2] [𝟎. 𝟗𝟏% + 𝟎.𝟎𝟐 𝟏.𝟔𝟒 × 𝟏𝟎𝟎%)] x 91.21951 = 1.942 ≈ ±2% (1sf) Deduct 1m for incorrect sf (overall) Marker’s comments: Fairly well answered. Some candidates were unable to obtain full marks as they were supposed to determine the absolute uncertainty by multiplying the percentage uncertainty with the percentage composition value. Many also did not leave the final absolute uncertainty in 1 sf. Marker’s comments: Generally well answered. However a few candidates were not familiar with the formula and performed incorrect calculation. Some candidates also calculated the percentage composition in moles instead of in mass, which should be discouraged.

5 ClF5 BrF3 Lewis structure Bond angle of F-X-F 81 o to 89 o 81o to 89 o Accept 110 o to 119 o or slightly >180 o Molecular Geometry Square pyramidal T-shape 2. (a) Fluorine reacts with other elements in Group 17 to form interhalogen compounds. The formulae of two such compounds are as shown. (i) Draw the Lewis structure. [1] (ii) State the correct bond angle. [1] (iii) State the molecular geometry for each molecule. [2] Marker’s comments: Poorly attempted. Common mistakes include: 1. missing lone pairs on the central and terminal atoms 2. not stating a specific bond angle (the term <90o is not allowed for question command term “state”) 3. unfamiliar with the assigned shape given the no. bond pairs and lone pairs.

6 Both molecules are polar and thus there is presence of permanent dipole-dipole forces between the molecules. However, the permanent dipole-dipole force between BrF3 molecules is stronger than that in ClF5 due to greater dipole as a result of larger electronegativity difference between Br-F than Cl-F. Hence more energy is required to overcome the dipole-dipole forces between BrF3 molecules. Copper forms Cu2+ ions while Zn forms Zn2+ ions. For Cu2+, it has a partially filled 3d orbitals while for Zn2+, the 3d orbitals are fully filled. (b) The boiling point of the molecules are given in the table. compound ClF5 BrF3 boiling point / oC -13.1 125.7 [2] Suggest why the boiling points of the molecules are significantly different. 3. (a) (i) Explain why copper is considered a transition metal while zinc is not. [2] Marker’s comments: Poorly attempted. Most candidates were unable to identify the correct reasons behind the variation in boiling point by stating irrelevant justifications including polar vs non polar molecules, or varying in London dispersion force due to different Mr. Marker’s comments: Very poorly attempted. Most of the students did not know the definition of transition elements and were unable to answer this question.

7 For Zn2+ complexes, the 3d orbitals are fully filled and no d-d transition can take place. For Cu2+ complexes, the presence of ligands split the 3d orbitals into 2 sets of different energies✔. Cu2+ has partially filled d-orbitals. The 3d electron from the lower energy d orbital can absorb light of a particular wavelength and move up to a higher energy d orbital. ✔ Energy gap corresponds to light in the visible region of the spectrum. ✔ The colour seen is t