2019 DHS Prelim H2 Chem P3 ANS

© DHS 2019 Name: Answers Index Number: Class: DUNMAN HIGH SCHOOL Preliminary Examination 2019 Year 6 H2 CHEMISTRY 9729/03 Paper 3 Free Response Additional Materials: Data Booklet Answer Booklet INSTRUCTIONS TO CANDIDATES 1 Write your name, index number and class on this cover page. 2 Write your answers on the separate answer booklet provided. 3 Write in dark blue or black pen. 4 You may use an HB pencil for any diagrams or graphs. 5 Start each question on a fresh sheet of paper. *[Marks will be deducted if you fail to do so.] 6 Do not use staples, paper clips, glue or correction fluid. Section A 7 Answer all questions Section B 8 Answer one question. The number of marks is given in brackets [ ] at the end of each question or part question. You are advised to show all workings in calculations. You are reminded of the need for good English and clear presentation in your answers.

2 © DHS 2019 9729/03 Section A 1 To facilitate better plant growth, there is widespread use of fertilisers and soil additives in the agricultural sector. (a) Agricultural lime is a soil additive used to increase soil pH so as to facilitate uptake of plant nutrients such as nitrogen and phosphorous. It is usually made up of a combination of calcium carbonate and magnesium carbonate. A manufacturer claims that a 2.00 g sample of agricultural lime contains 92 % calcium carbonate (Mr = 100.1) and 8 % magnesium carbonate (Mr = 84.3) by mass. To verify the manufacturer’s claim, the following steps were carried out: Step 1 The solid carbonate mixture was dissolved in 250 cm 3 of 0.25 mol dm–3 hydrochloric acid. Step 2 A 25 cm 3 aliquot of this resultant solution was titrated with 0.14 mol dm–3 potassium hydroxide. It was found that 12.50 cm 3 of potassium hydroxide was required for complete neutralisation. (i) Based on the manufacturer’s claim of 92% calcium carbonate and 8% magnesium carbonate by mass, calculate the amount, in moles, of each carbonate present in the sample of agricultural lime. [1] Mass of CaCO3 present = 92 % × 2.00 g = 1.84 g Mass of MgCO3 present = 8 % × 2.00 g = 0.16 g Moles of CaCO3 present = 1.84 g ÷ 100.1 g mol–1 = 0.018382 = 0.0184 mol Moles of MgCO3 present = 0.16 g ÷ 84.3 g mol–1 = 0.0018980 = 0.00190 mol (ii) Use the information above to calculate the amount of HC l that reacted with the carbonate mixture in Step 1. [2] Moles of KOH reacted with excess HCl = 12.50 1000 × 0.14 = 0.00175 mol Moles of HCl present in 25 cm3 aliquot = 0.00175 mol Moles of excess HCl present in 250 cm3 solution = 0.00175 × 250 25 = 0.0175 mol Total moles of HCl used = 250 1000 × 0.25 = 0.0625 mol Moles of HCl reacted with carbonates = 0.0625 – 0.0175 = 0.045 mol (iii) The manufacturer’s claim is considere