Christchurch 4E Prelim Chemistry 6092 P3 MS draft 3 2025

Christ Church Secondary School Secondary Four Express 2025 Prelim Examination Chemistry 6092 1 Marking Scheme Paper 3 6092/03 1ai PDO Results table: Records initial burette readings, final burette readings and volume added with correct headings and units in a titration table All burette readings for all accurate titres in titration table are recorded to nearest 0.05 cm3 T 1 D 1 [5] MMO Titration results Accuracy For the average titre (of consistent readings) within 0.20 cm3 of supervisor’s average value score 2 marks For the average titre (of consistent readings) within 0.30 cm3 of supervisor’s average value score 1 marks Concordance At least two titre values are within 0.20 cm3 (using uncorrected titres) (teacher’s result) Titration number 1 2 3 Final burette reading / cm3 Initial burette reading / cm3 Volume of Q used / cm3 Best titration results (✓) R 2 C 1 1aii MMO Appropriate average volume of Q in 2d.p. from closest titre values (titres should be identified either in the table by a tick, or in a calculation). Teacher’s result: 19.5 – 20.5 cm3 1 [1] 1bi ACE No of moles of NaOH = 25/1000 x 0.0984 = 0.00246 mol 1 [1] 1bii ACE Concentration, in mol/dm3, of H3PO3 in Q = 5.04 / 82 = 0.061463 mol/dm3 = 0.0615 mol/dm3 1 1 [2] 1biii ACE No of moles of H3PO3 in 20.1 cm3 of Q = [1aii] / 1000 x 0.061463 = x mol Since no of moles of NaOH that reacted = 0.00246 mol, No of moles of sodium hydroxide that react with 1 mole of H3PO3 = 0.00246 / x = 2.00 mol (Accept: 1.95 – 2.05) 1 1 [2] 1biv ACE 2 NaOH + H3PO3 → Na2HPO3 + 2 H2O 1 [1] 1c MMO Repeat the titration without the indicator, using 25.0cm3 of P and the volume of Q found in (a)(ii). Transfer the salt solution produced to an evaporating dish. 1 1 [3]

Christ Church Secondary School Secondary Four Express 2025 Prelim Examination Chemistry 6092 2 Heat the salt solution until saturated solution is formed. Cool the solution to allow crystallisation. Filter to obtain crystals, wash with cool distilled water and pat dry crystals with filter paper. 1 1d ACE There would be no change to the titration results. Mr of H2SO3 is also 82 / the mole concentration of the acid is the same as H3PO3 Accept calculations to show that the mole concentration is the same. Number of moles of hydroxide that react with 1 mole of H2SO3 is the same as that of H3PO3 / the number of moles of H2SO3 required for neutralisation is the same as that of H3PO3. 1 1 [2] [Total: 17] 2ai MMO PDO Any TWO from below • effervescence, gas forms a white precipitate when bubbled into limewater • solid dissolves to form a colourless solution • solution becomes warm 2 [2] 2aii MMO PDO White precipitate forms Dissolves in excess to form a colourless solution 1 1 [2] 2aiii MMO PDO Any TWO from below • solid turns yellow • condensation/ water droplets at top of tube • turns white again (when cooled) 2 [2] 2aiv ACE Carbon dioxide 1 [1] 2av ACE Zinc and / or aluminium (ions) give the same result 1 [1] 2avi ACE Add aqueous ammonia 1 [1] 2bi MMO PDO White precipitate forms 1 [1] 2bii MMO PDO White precipitate forms 1 [1] 2biii MMO PDO White precipitate forms Insoluble in excess 1 1 [2] 2biv ACE Calcium chloride / CaCl2 From (b)(ii), white precipitate formed with acidified silver nitrate shows presence of chloride ions. From (b)(iii), white precipitate that is insoluble in excess sodium hydroxide shows the presence of calcium ions. (Evidence must specify the test itself) 1 1 1 [3] [Total: 16]

Christ Church Secondary School Secondary Four Express 2025 Prelim Examination Chemistry 6092 3 3a P Labelled diagram Procedure (accept lesser details if they can be complemented with the labelled diagram): 1. Measure the mass of the rusty nail using an electronic balance, M g. 2. Add the rusty nail into a conical flask and add excess dilute acid to the conical flask. 3. Stopper the conical flask immediately with a delivery tube connected to a calibrated gas syringe. 4. Measure the maximum volume of gas collected in the syringe, V cm3. 5. Calculate the mass of rust using the following method: Fe + 2H+ → Fe2+ + H2 No. of moles of hydrogen = (V/24000) mol No. of moles of iron = (V/24000) mol Mass of iron = (V/24000) x 56 = (7V/3000) g Mass of rust = M – (7V/3000) g 2 2 [4] 3b P Correct scale Correct points plotted Correct line of best fit with all points 1 1 1 [3] rusty nail gas syringe retort stand conical flask dilute acid

Christ Church Secondary School Secondary Four Express 2025 Prelim Examination Chemistry 6092 4 [Total: 7]