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Chemistry Solutions (Practical)
a. i) Any three from: A method of weighing by difference / wash the solid from its weighing container into the beaker Wash the (wet) rod into the flask / beaker after use Wash the (wet) beaker into the flask after transfer Wash the filter funnel (after transfer) into the flask Use a teat pipette to make up to the mark on the volumetric flask Ensure the bottom of the (liquid) meniscus is on the graduation mark Mix / shake the final solution in the flask / invert flask [1 mark for each] ii) Do (a) further titration(s) [1 mark] to obtain concordant results [1 mark] b. i) Space will fill during titration / titres or volumes added are too high [1 mark] ii) Less chance of losing liquid on swirling / liquid doesn’t splash on swirling [1 mark] - Do not accept ‘easier to swirl’ on its own. iii) Returns reagent on the sides of the flask to the reaction mixture (to ensure that all of the acid / alkali reacts) [1 mark] This does not change the number of moles of reagents / water is not a reagent / water is one of the products. [1 mark] iv) (idea that) a single titration could be flawed / anomalous. or To obtain concordant results. [1 mark]
c. Pipette = 0.05 × 100 / 25.0 = 0.2% [1 mark] Burette = 0.15 × 100 / 24.25 cm^3 = 0.6% [1 mark]
a. Measured volume would be greater [1 mark] Level in burette falls as tap is filled before any liquid is delivered [1 mark] b. Drop sizes vary or percentage error for amount of oil will be large as the amount used is so small [1 mark] c. Use a larger single volume of oil [1 mark] Dissolve this oil in the organic solvent [1 mark] Transfer to a conical flask and make up to 250 cm^3 with the solvent [1 mark] Titrate (25 cm^3 ) samples from flask [1 mark] d. Mass of oil = 0.92 * (0.05 * 5) = 0.23 g [1 mark] Moles of oil = 0.23 / 885 = 2.6 x 10-4^ mol [1 mark] Moles of bromine = 0.02 * 0.0394 = 7.9 x 10-4^ mol [1 mark] So reacting ratio is 7.9 / 2.6 = 3.03 : 1 → 3 : 1 [1 mark] So there are 3 C=C bonds [1 mark]
e. i) or using structural formula: C 6 H 5 NHCOCH 3 + NO 2 +^ → C 6 H 4 (NHCOCH 3 )NO 2 + H+^ [1 mark] not formation equation: (HNO 3 + 2 H 2 SO 4 → NO 2 +^ + 2 HSO 4 -^ + H 3 O+) ii) Electrophilic substitution [1 mark] f. Hydrolysis [1 mark] g. (Sn or Fe) / HCl or H 2 / Ni [1 mark for either pair] not LiAlH 4
a. 2 NaBr + 2 H 2 SO 4 → Na 2 SO 4 + Br 2 + SO 2 + 2 H 2 O or ionic equation: 2 Br -^ + 2 H 2 SO 4 → Br 2 + SO 2 + SO 4 2-^ + 2 H 2 O [1 mark] Br -^ ions are larger (have greater ionic radius) than Cl-^ ions [1 mark] Therefore Br -^ ions are more easily oxidised / lose an electron more easily (than Cl-). [1 mark]
b. Stage 1: formation of precipitates ● Add silver nitrate ● To form precipitates of AgCl and AgBr ● AgNO 3 + NaCl → AgCl + NaNO 3 ● AgNO 3 + NaBr → AgBr + NaNO 3 Stage 2: selective dissolving of AgCl ● Add excess of dilute ammonia to the mixture of precipitates ● The AgCl precipitate dissolves ● AgCl + 2 NH 3 → Ag(NH 3 ) 2 +^ + Cl- Stage 3: separation and purification of AgBr ● Filter off the remaining AgBr precipitate ● Wash to remove soluble compounds ● Dry to remove water [8-10 points, including all equations → 5-6 marks] [5-7 (from all three stages) points → 3-4 marks] [3-4 points → 1-2 marks] [Must show clear, concise, logical order to gain full marks] c. Cl 2 + 2 OH-^ → ClO-^ + Cl-^ + H 2 O [1 mark] ClO-^ is +1, Cl-^ is -1 [1 mark for both] [Accept HO-^ for hydroxide and OCl-^ for chlorate(I)]
a. Compound 1 [1 mark] No visible change with H 2 SO 4 , [1 mark] gives white precipitate with NaOH [1 mark] b. BaCO 3 [1 mark; allow name ‘barium carbonate’] The carbonate ion released CO 2 [1 mark] But the BaSO 4 formed is highly insoluble. [1 mark] c. Compound 4 [1 mark] Sr(OH) 2 + H 2 SO 4 → SrSO 4 + 2 H 2 O [1 mark; allow ionic eqn, ignore states]
d. Any one reason from: Small amount of tea used / consumed Concentration of acid in tea is low High temperature decomposes the acid Calcium ions in milk form a precipitate with the acid [1 mark] not tea is not consumed often e. Mass of acid = 180 and mass of reagents = 450 [1 mark] Atom economy = 180 / 450 * 100 = 40%. [1 mark]
a. Mr = 164.0 [1 mark] % N = 28 / 164 * 100 = 17.1 % [1 mark] b. i) Absorption depends on (is proportional to) path length / distance travelled through solution [1 mark] ii) To select the colour / frequency / wavelength that is (most strongly) absorbed (by the solution) or filter is chosen to complement the colour of the solution [1 mark] iii) Quicker to analyse extracted samples than by titration or uses smaller volumes of solution [1 mark]
Step 1: (ester → alcohol) Name: (acid/base) hydrolysis Reagents: water and dilute HCl or NaOH (aq) then HCl [1 mark for both name and reagents] Apparatus: reflux condenser, round bottomed flask, bunsen burner [2 marks] Step 2 : (decarboxylation) Apparatus: distillation apparatus (condenser, flask, thermometer) [1 mark] Step 3: (benzene → nitro) Name: nitration Reagents: conc H 2 SO 4 and conc HNO 3 [1 mark for name and reagents] Apparatus: ice bath, thermometer [1 mark] Step 4: (nitro → amine) Name: reduction Reagents: Sn / HCl or Fe / HCl or H 2 / Ni or NaBH 4 [1 mark for both name and reagents] Step 5: (amine → amide) Name: acylation Reagents: ethanoyl chloride or ethanoic anhydride [1 mark for both name and reagents] Apparatus: Buchner flask, Buchner funnel, vacuum/suction apparatus [2 marks]
