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).