WMS12 CHEMISTRY

Part of*WMS12*

Higher Level: page 3Ordinary Level: page 19

CHEMISTRYHIGHER & ORDINARY LEVEL

Pre-Leaving Certifi cate Examination 2019MARKING SCHEME

HIGHER LEVEL CHEMISTRY | Pre-Leaving Certifi cate, 2019PAGE 2 OF 29

HIGHER LEVEL

Introduction

In considering the marking scheme, the following should be noted.

1. In many cases only key phrases are given which contain the information and ideas that must appear in the candidate’s answer in order to merit the assigned marks.

2. The descriptions, methods and defi nitions in the scheme are not exhaustive and alternative valid answers are acceptable.

3. The detail required in any answer is determined by the context and the manner in which the question is asked, and by the number of marks assigned to the answer in the examination paper and, in any instance, therefore, may vary from year to year.

4. The bold text indicates the essential points required in the candidate’s answer. A double solidus (//) separates points for which separate marks are allocated in a part of the question. Words, expressions or statements separated by a solidus (/) are alternatives which are equally acceptable for a particular point. A word or phrase in bold, given in brackets, is an acceptable alternative to the preceding word or phrase. Note, however, that words, expressions or phrases must be correctly used in context and not contradicted, and, where there is incorrect use of terminology or contradiction, the marks may not be awarded. Cancellation may apply when a candidate gives a list of correct and incorrect answers.

5. In general, names and formulas of elements and compounds are equally acceptable except in cases where either the name or the formula is specifi cally asked for in the question. However, in some cases where the name is asked for, the formula may be accepted as an alternative.

6. There is a deduction of one mark for each arithmetical slip made by a candidate in a calculation. This deduction applies to incorrect Mr values but only if a candidate shows the addition of all the correct atomic masses and the error is clearly an addition error. If the addition of atomic masses is not shown, the candidate loses 3 marks for an incorrect Mr.

HIGHER LEVEL CHEMISTRY | Pre-Leaving Certifi cate, 2019 PAGE 3 OF 29

HIGHER LEVEL

HIGHER LEVELSection A

Eight questions. These must include at least two questions from Section A.All questions carry equal marks (50 marks).

Note: (i) In many cases, most especially defi nitions, candidates’ answers must contain key phrases in order to get the assigned marks. In calculations, 3 marks are deducted for a mathematical error, 1 mark for a slip or missing units, but no further deduction is applied if the calculation, otherwise correct, is completed. (ii) Alternative valid answers are acceptable.

QUESTION 1

(a) Explain: Primary Standard:can be dissolved (used) to make up a solution of exact (known) concentration / no need to standardise by titration (can be made up directly) [any one] // pure / stable / anhydrous (not hydrated) / no water loss (no effl orescence) / not deliquescent (not hygroscopic ) / does not sublime / high formula (molecular, molar) mass (Mr). [any one]

3 + 3(b) Describe: wash (rinse) into beaker of deionised (distilled, pure) water, stir to dissolve

// pour through funnel (down glass rod) into volumetric fl ask adding rinsings to beaker // add last few drops of deionised water drop by drop (using dropper) to bring bottom of meniscus level with (up to, on, at) mark reading at eye level [all text in bold must be given in each point to

receive the mark] 6 + 3 + 3

(c) Name: Methyl Orange / Methyl Red. 3 Colour: Orange/yellow Peach/ red 3

(d) Value: Decrease 3 Explain: Less salt (sodium carbonate) present in amount used. 3

(e) Calculate: Na2CO3 = 106g [Addition must be shown for error to be treated as slip (– 1)] 1.06g in 200 cm3 = 0.01 moles (1.06/106) in 200 cm3 3 = 0.00125 moles in 25 cm3 3

Na2CO3 : 2HCl 1 : 2 0.00125 : 0.0025 moles in 20.65 cm3 3 = 0.0001210654 moles/cm3 [0.0025/20.65] 3 = 0.012 moles/l [ (0.0025/20.65) × 1000] 3 [allow 0.012 to 0.0121]

(f) What: Sodium carbonate (Na2CO3) titrated with a strong acid [hydrochloric acid 3 (HCl), sulfuric acid (H2SO4), nitric acid (HNO3)] // acid titrated with sodium hydroxide (NaOH) solution. 2 [allow 3 if given in reverse order]


HIGHER LEVEL

QUESTION 2

(a) Why: to heat (provide energy) to reaction mixture / to keep solvent (mixture) boiling (hot, heated) / to keep reaction (mixture) at constant temperature above room temperature / to help reaction

go to(wards) completion (equilibrium) / to maximise yield / to speed up reaction / to reduce time for reaction / to help reaction reach

(exceed, overcome) activation energy [any one] // without loss of solvent (vapour, ethanol, volatile material*) / without solvent (vapour, ethanol, volatile material*) boiling away (off )

[any one] [Allow ‘evaporation’ or ‘boil dry’ instead of ‘boiling away (off )’.] [*Allow ‘reactants’.] 2 + 2

(b) Type: base (alkaline)-catalysed hydrolysis / saponifi cation 4 [Accept ‘substitution’.]

(c) Why: It acts as a solvent. 3 Why remove: To increase the yield / so soap is not kept in solution / allow soap to precipitate 3

(d) Name: Brine (saturated salt solution) 3 Function: to precipitate (isolate, separate) the soap / soap insoluble in brine 3

(e) Draw: 6

Name: / CH2OHCHOHCH2 OH or propane-1,2,3-triol / glycerol / glycerine [In the case of single H atoms bonded to a C atom in the expanded formula, the H symbol may be omitted.] [Incorrect name cancels correct formula and vice versa.]

State: in the brine / fi ltrate 3


HIGHER LEVEL

(f) Draw: 6

At least one correct label. [no labels – 3]

(g) Calculate: Glyceryl tripalmitate = 806g [Addition must be shown for error to be treated as slip (– 1)] 3 4.03g = 0.005 moles 3 0.005 × 3 = 0.015 moles of soap formed 3 Soap =278 [Addition must be shown for error to be treated as slip (– 1)] 0.015 × 278 = 4.17g (100%) 3 2.92g = 70% [allow 2.919g – 2.92g] 3 [1 mark to be deducted for incorrect rounding off resulting in candidate’s

answers lying outside given ranges but deduction to be made once only in (g).]

Soap

Buchner Funnel

To vacuum pump

Filtrate

Buchner Flask


HIGHER LEVEL

QUESTION 3

(a) Explain: Volatile: A liquid which is easily changed to gas / easily vaporised / low boiling point 3

Relative molecular mass: average mass of a molecule of the substance 3/ average of mass numbers of isotopes of the elements in the molecule taking abundances into account (as they occur naturally) [any one] // compared to (based on) 1/12th carbon–12 isotope 2

(b) Describe:

Apparatus A diagram: fl ask, sealed (covered) with foil with small hole (pinhole)* , immersed so that at least half is under water. 3[* Accept if hole mentioned in account of experiment. Label required: anyone correct label.] mass: get mass of fl ask and foil 3[add liquid and arrange as in diagram]heat until liquid gone / heat until fl ask appears empty / vaporised 3cool (dry) and reweigh 3get mass of sample by subtraction (Get diff erence) 3volume: fi ll fl ask and empty into graduated (measuring) cylinder 3temperature: use thermometer (probe, sensor) to read temperature 3of water (or got from diagram).

OR

Apparatus B diagram: gas syringe with self-sealing cap (septum cap, can be shown sealed), surrounded by heating device (oven, steam jacket, beaker of water). 3[Label required: any one correct label.]mass: get mass of hypodermic (syringe) containing liquid 3inject some liquid into gas syringe 3reweigh hypodermic (syringe) 3get mass by subtraction (Get diff erence) 3volume: read from gas syringe. 3temperature: read from thermometer (probe, sensor) in heating device (or got from diagram). 3

pinhole

rubber band

aluminium foil

vapour

boiling water

thermometer

hotplate

plunger

thermometer gas syringe oven vapour

rubber seal

small syringe


HIGHER LEVEL

(c) How: Using a barometer / bourdon guage / barograph (barothermograph) / pressure sensor (not probe) 3

(d) Calculate: PV = nRT n = PV 3 RT n = 1 × 105 × 77 × 10– 6

8.3 × 370 n = 0.0025 moles 3

OR

P1V1 = P2V2 => 1 × 105 × 77 = 1.013 x105 × V2 6 T1 T2 370 273 V2 = 56.08 / 56.1 cm3 3 Moles 56.08 / 56.1 = 0.0025 3 22400

(e) Calculate: 0.0025 moles = 0.21g 3 1.0 mole = 84g = molecular mass 3


HIGHER LEVEL

Section B QUESTION 4

Eight items to be answered. Six marks to be allocated to each item and one additional mark to be added to each of the fi rst two items for which the highest marks are awarded.

(a) A neutron changes into a proton / 10 n1

1 p // and an electron / + e– that is emitted 3 [‘Proton (atomic) number increases by one’ is allowable for (3) if no other mark awarded. 3 No marks for beta particle emitted as given in the question]

(b) half internuclear distance (half distance between the centres of the atoms) in a single homonuclear bond (of singly-bonded atoms of the same element) 6

(c) (i) +4 3 (ii) +7 3

(d) Ca(HCO3)2 CaCO3 + CO2 + H2O 6 or Mg(HCO3)2 MgCO3 + CO2 + H2O [Formulas (3) correctly balanced (3)]

(e) 6.0% w/v = 6g in 100 cm3 // 3 = 12g in 200 cm3 3(f) Kerosene 6

(g) Correct structural formula // 3 Correct functional group 3 [Hs not necessary on 1st carbon, but is necessary in functional group]

(h) reactant(s) (substrate) and catalyst in diff erent phase(s) // 6 boundary between reactants and catalyst

(i) [marks only awarded for two units] 6

(j) melts sharply / over a narrow range // 3 close to correct value /close to value from tables of melting points 3

(k) A prevention of the escape of infrared radiation (accept heat) by gases in the atmosphere // 3 H2O / CO2 / CH4 / CFC 3 B have variable valencies / form coloured compounds / 3 have catalytic activity. [any 2] 3

C

Cl

H

C

H

H

C

Cl

H

C

H

H

H H

H

H C C

O


HIGHER LEVEL

QUESTION 5 (a) Atomic no.: The number of protons in the nucleus of an atom 4

Atomic Orb: space (volume, region) around nucleus of an atom // where there is a 2 relatively high probability (possibility) of fi nding an electron / where an 2 electron is likely to be found [‘Area’ around nucleus not acceptable.] or approximate solution // to (of) a Schrödinger wave equation.

(b) Explain: in the ground state the hydrogen electron occupies the lowest available 3 energy level // the electron can jump (move, become excited) to a higher energy level (state) if it receives (absorbs) a certain amount of energy 3 (light, heat, electricity, a photon) // excited (higher energy) state unstable (temporary) // electron falls back to a lower level [any 3] 3

energy emitted (given out) as photon (light of defi nite frequency, light 3of defi nite wavelength, hf, hυ) thus giving rise to a spectrum / energy emitted (hf, hυ) corresponds to (=) diff erence between the two energy levels (E2 – E1) thus giving rise to a line on the spectrum / E2 – E1 = hf (hυ)[Marks not awarded wherever ‘atom’ is incorrectly used instead of ‘electron’.] [Some marks, maximum (6), available from a good labelled diagram.]

(c) Write: 1s2 2s2 2p6 3s2 3p4 // [Ne] 3s2 3p4 3

Energy levels: 3 energy levels 3

Orbitals: 9 orbitals 3

(d) Name: Gold 3 Alpha particles 3

Observations: fi rst observation most of particles went straight through, (undefl ected). 3 second observation: some alpha particles were defl ected as they passed 3 through the gold foil. third observation: a small few alpha particles were refl ected 3 (defl ected straight back)

(e) Distinguish: sigma (σ) bond is formed by head (end) on overlap of atomic orbitals // 3 pi (π) bond is formed by sideways overlap of atomic orbitals 3 [Orbitals need not be mentioned twice.] [Allow ‘collision’ of orbitals.] or sigma (σ) bond symmetrical with respect to rotation about bond axis // pi (π) bond asymmetrical (unsymmetrical) with respect to rotation about bond axis or sigma (σ) bond allows rotation about the bond axis // pi (π) bond does not allow rotation about the bond axis or sigma (σ) bond has one region of overlap // pi (π) bond has two regions of overlap or in a multiple (double, triple) covalent bond // sigma (σ) bond is strongest bond / sigma (σ) bond is formed fi rst / pi (π) bond is weaker bond / pi (π) bond is formed only after a sigma (σ) bond has already been formed [Marks may be obtained from good clear diagrams with orbitals labelled if present.]


HIGHER LEVEL

QUESTION 6

(a) What: measure of tendency (likelihood) to auto-ignite (knock, pink, pre-ignite, 5 ignite early, ignite before spark) / number representing ability (tendency,

measure) to resist autoigniting (knocking, etc.) / number based on a scale where 2,2,4-trimethylpentane (iso-octane) is assigned a rating of 100 and heptane (n-heptane) a value of 0 / percentage by volume of 2,2,4-trimethylpentane (iso-octane) in a blend (mix) with heptane (n-heptane) that matches the behaviour of the fuel

[any one] Name: heptane / n-heptane 3 2,2,4-trimethylpentane (isooctane) 3

State: Short chain / highly branched / cyclic(aromatic) [any 2] 3 + 3

(b) What: general formula / diff er by CH2 / same functional group / similar 3 chemical properties / gradation in physical properties / similar 3 method of preparation [Accept “uniform chemical type” for “similar chemical properties”.] [any 2]

Draw: methyl butane (2 methyl butane) // 3 CH3 CH(CH3) CH2 CH3 3

dimethyl propane (2,2 dimethyl propane) // 3 CH3 C(CH3)2CH3 3

Which: pentane 3

(c) Calculate: 5C + 5O2 5CO2 ΔH = – 1970 kJ 3 6H2 + 3O2 6H2O ΔH = – 1716 kJ 3 C5H12 5C + 6H2 ΔH = 174 kJ 3 ____________________________________________________ C5H12 + 8O2 5CO2 + 6H2O ΔH = – 3512 kJ mol–1 3

Or ΣΔHc = ΣΔHf(products) – ΣΔHf(reactants) ΣΔHc = [ 5(–394) + 6(–286)] – (– 174)] ΔHc(pentane) = 1970 // – 1716 // + 174 3 + 3 + 3 = – 3512 kJ mol–1 3 [ +3512 by this method is not a slip and is worth maximum (9).] [+3512 with no work shown is worth (3).]

H

C

H

H

H

C

C

H

H

C

H

H

H

C

H

HH

HHH

H H

HHH

H

H HHC

C

C

CC


HIGHER LEVEL

QUESTION 7

(a) Le Chatelier’s Principle: systems in (at) equilibrium // react to oppose (minimise, relieve) 3 + 2 applied stress(es) (disturbance(s)) [Instead of ‘stress(es){disturbances}’ accept ‘change in temperature, pressure

or number of moles (concentrations)’ if all three {temperature, pressure and moles (concentrations)} are given.]

Equilibrium: forward reaction and reverse reaction continue at equal rates / rr = rf / 3 rb = rf or state reached at which concentrations of reactants and products

are constant

Why: reaction has not stopped (is continuing) / forward & reverse reaction(s) 3 still occurring

(b) (i) Explain: no 3 both forward and reverse reactions continue at same rate / 3 reactants changing to products and products changing to reactants continuously

(ii) State: blue to pink / becomes pink 3 Explain: cooling always shifts in the exothermic (heat producing) direction 3 (iii) Reverse: add conc. hydrochloric acid (HCl) / add chloride ions (Cl¯) / 6 e.g. NaCl / remove water [any 1]

(c) Write: Kc = [CH3COOC2H5][H2O] [CH3COOH][C2H5OH] [Square brackets essential.] 6

Find the Mass: CH3COOH + C2H5OH CH3COOC2H5 + H2O 0.2 mol* 0.2 mol* 0 mol 0 mol 3 [* addition must be shown for error to be treated as slip]

0.2 – x 0.2 – x x x 3 x2 = 4 3 (0.2 – x)2

x = 2 (0.2 – x) x = 0.13• [allow 0.13 / 0.14] 3 Mass = 0.13• × 88 = 11.44g [11.4 – 12.3] 3 Or CH3COOH + C2H5OH CH3COOC2H5 + H2O 0.2 mol* 0.2 mol* 0 mol 0 mol [* addition must be shown for error to be treated as slip] x x 0.2 – x 0.2 – x (0.2 – x)2 = 4 x2

0.2 – x = 2 x x = 0.13• [allow 0.13 / 0.14] Mass = 0.13• × 88 = 11.44g [11.4 – 12.3]


HIGHER LEVEL

QUESTION 8

(a) Identify: A / C2H5OH / Ethanol 5

(b) Name: A = Ethanol 3 B = Ethanal 3 C = Ethanoic acid 3 D = Methyl ethanoate 3

(c) Test: B: test for oxidation Add Fehling’s (Benedict’s) solution (reagent) and 3 + 3 warm / changes from blue to red // add ammoniacal silver nitrate (Tollens’ reagent) / silver mirror [accept acidifi ed potassium permanganate/pink to colourless] [must link

positive result to B] Or C: Test for an acid any indicator with correct acid colour /pH paper or

sensor showing a value less than 7. [must link positive result to C]

(d) Classify: X = Oxidation 3 Y = Oxidation 3 Z = substitution 3

(e) Reagent: sodium dichromate (Na2Cr2O7) // 3 sulphuric acid (H2SO4) 3

(f) Name: Propanoic Acid 3

Draw: 3

[OH is acceptable for O – H]

(g) Write 2CH3COOH + Na2CO3 2CH3COONa + CO2 + H2O 3 + 3 [Formulas (3) correctly balanced (3)]

H

H

H C C C

O

O

HH

H


HIGHER LEVEL

QUESTION 9

(a) Defi ne: (i) An acid and a base that diff er by a single proton (H+) 4 (ii) poor proton donor / slightly (poorly, weakly) dissociated 4 [Allow ‘Partly (partially, not fully) dissociated’ and ‘does not readily donate protons’ for weak acid description.]

Identify: base HSO3 – // H2O 3 Conjugate acid H2SO3 // H3O+ 3 [base and conjugate acid must match]

(b) Defi ne: – log10[H+] / – log10[H3O+ ] (3) negative log to base ten of hydrogen 6 (hydronium) ion concentration in moles per litre

What: reliable (accurate, suitable) only for dilute solutions / only valid 3 + 3 (useful) in 0 - 14 range / applies to aqueous solutions only / unreliable (inaccurate, unsuitable) in very concentrated solutions / unreliable for negative pH values. [any 2] [‘Temperature dependent / (25 °C )’ acceptable.] [‘Valid range 1 – 14’ unacceptable but does not cancel.]

(c) State: (i) HCl – Colourless 3 (ii) NaOH – Pink 3

Calculate: [OH-] = √ 0.003 × (6.65 × 10 -6) 3 = √ 1.995 × 10 –8

= 1.412 × 10 –4 [accept 1.4 – 1.412] 3 pOH = – log10(1.412 × 10-4 ) = 3.85 pH = 14 – 3.85 = 10.15 [ Accept 10.15 – 10.2] 3

(d) Why: danger to health (toxic, poisonous) / may cause foetal abnormalities / 3 minamata disease

Name: atomic absorption spectroscopy (atomic absorption spectrometry) 3 [Allow AAS]

Explain: precipitation / coagulation / complexation / adsorption / absorption / 3 reverse osmosis / ion exchange / deionising


HIGHER LEVEL

QUESTION 10

(a) Defi ne: change in concentration per unit time of one reactant 4 (product)/ rate of change of concentration of one reactant (product) /

change in concentration of one reactant (product) time

Factors: Particle size/nature of reactants/presence of a catalyst [any 2] 3 + 3

Explain: minimum energy required for colliding particles (molecules) to react 3 / minimum energy required for eff ective collisions between particles

(molecules) [Accept ‘energy needed for colliding particles to initiate reaction’./ or ‘energy

required for reaction to take place’] A collision which reaches (exceeds) activation energy / results in 3 (brings about, causes) reaction between colliding particles (molecules) /

results in product formation

Eff ect: (i) small (slight, tiny, < 5 %) increase 3 (ii) large (big, substantial, > 50 %) increase in eff ective collisions 3 (iii) no eff ect / none 3

(b) Explain: production of chemical change (reaction, decomposition) when 4 electricity {electrical energy (current)} passes through an electrolyte (ionic melt, ionic solution) [Do not accept “chemical change produced by electrical energy (current)”. ]

Explain: Loss of electrons 3

Write: Anode I¯ ½I2 + e– // 2I¯ I2 + 2e– 6 [Formulas (3) correctly balanced (3)] [Electrons may be shown subtracted on the

left. Neg. charge on electron need not be shown.]

Cathode 2H2O + 2e– H2 + 2OH¯ // H2O + e– ½H2 + OH¯ 6 [Formulas (3) correctly balanced (3)] [Electrons may be shown subtracted on the right. Neg. charge on the electron need

not be shown.]

Describe: Colour at the anode red/orange/yellow/brown is due to the iodine 3 Colour at the cathode is pink as phenolphthalein is pink in base (OH–) 3


HIGHER LEVEL

(c) Defi ne: relative (measure of force of, number expressing the) attraction that 6 an atom has // for shared pair of electrons / for electrons in a covalent bond

Describe: [Correct shape not reqd. Accept all dots or all crosses. For bonds accept × – .]

6

What: three bonding and one non-bonding (lone) pair / four electron (valence) 3 pairs* // giving bond arrangement (shape of molecule) to be pyramidal 3

Which: 107o 3

Explain greater repelling power (repulsion) of lone pair // lone pair pushes bonds closer together [Allow “l.p.: b.p. > b.p.: b.p.” or “l.p.: l.p. > l.p.: b.p. > b.p.: .p.”]

4

H H H H

H

XX X

XX

X NH

N


HIGHER LEVEL

QUESTION 11

(a) Describe: add freshly-prepared iron(II) sulfate (FeSO4) solution // 4 + 3 + 3 trickle H2SO4 (concentrated sulfuric acid) dropwise (carefully, slowly) down side of (into slanting) test-tube // brown ring forms at junction of the two liquids [All marks available from a good labelled diagram.].

Describe: Method 1 clean a platinum (nichrome) wire (rod, probe) in 3×3 concentrated hydrochloric acid (HCl) // dip rod in salt and hold salt in hot (blue) part of Bunsen fl ame// Observe the colour of the fl ame produced.

or Method 2 soak wood (splint, stick) overnight in water / use damp (wet) 3×3 wood (splint, stick) //dip splint (stick) in salt and hold salt in hot (blue) part of Bunsen fl ame // Observe the colour of the fl ame produced

or Method 3 prepare a solution of the given salt in water and ethanol 3×3 (propanol) //spray solution onto (into) hot (blue) part of Bunsen fl ame // Observe the colour of the fl ame produced

What: Copper is Green fl ame(blue green) 3 Potassium is lilac fl ame 3

(b) Distinguish: Aliphatic: These are hydrocarbons consisting of straight / branched 2 chains of carbon atoms /rings of carbon other than benzene rings Aromatic: hydrocarbons which contain benzene ring 2

Draw: Eugenol 6

What: double (triple, multiple) carbon-to-carbon bond present / 6 undergoes addition

Name: bromine water / bromine soln / Br2 or 3 acidifi ed potassium manganate(VII) soln /acidifi ed KMnO4 soln / H+ + MnO4

– soln / KMnO4 + H2SO4 soln. [any 1]

(bromine) brown solution // decolorises / changes to colourless 3 (not ‘clear’) 3

or (manganate) pink solution decolorises // changes to colourless (not ‘clear’) [The two (3)s are linked i.e. if bromine is given as reagent then colour change must match]

OCH3

CH CH2H2C

HO


HIGHER LEVEL

(c) A

Name: coke (charcoal, C, carbon) // 3 limestone (CaCO3) 3 Write: Fe2O3 + 3CO 2Fe + 3CO2 / Fe2O3 + 3C 2Fe + 3CO / 2Fe2O3 + 3C 4Fe + 3CO2 [Formulas (3) correctly balanced (3)] 6

Name: Impurity calcium oxide. 3 CaO + SiO2 CaSiO3 / 6 CaO + Al2O3 CaAl2O4 {Ca(AlO2)2} [Formulas (3) correctly balanced (3)]

Why: Pig iron has limited uses and is brittle with a high carbon content, 4 rusts / steel is much more useful and is harder stronger and can be corrosion resistant (doesn’t rust) [any bad property of iron or good property of steel]

B

What: Dust / moisture / carbon dioxide [any 2] 3 + 3

Describe: Compressed (liquid) air warms up in column 3 // liquid oxygen (highest boiling point) collects 3 at base of column and is removed // gaseous nitrogen (lowest boiling point) 3 comes off at top of column [good labelled diagram – full marks relevant diagram with one correct label – 3 marks]

Explain: Continuous // 2 Air goes in at one end and product comes out at the other 2 in a continuous manner [Air is cooled and fractionated continuously [Give the two (2)s for this.]

Name: Co-product Liquid nitrogen / neon / argon / krypton / xenon 3 Nitrogen: Inert atmosphere (or example) / fast freezing of fruit 3 (sperm samples) / gas chromatography / spectroscopy Neon: Inert atmosphere (or example) / lighting (lasers) / tv tubes / refrigerant Argon: Inert atmosphere (or example) / lighting (lasers / MRI / CT) Xenon: Inert atmosphere (or example) / photography / lighting (lasers) / MRI /CT (3) [Use must be matched with named co-product]

nitrogen

oxygen

compressedair(liquid air)

fractionatingcolumn

ORDINARY LEVEL CHEMISTRY | Pre-Leaving Certifi cate, 2019PAGE 18 OF 29

ORDINARY LEVEL

Introduction

In considering the marking scheme the following should be noted.

1. In many cases only key phrases are given which contain the information and ideas that must appear in the candidate’s answer in order to merit the assigned marks.

2. The descriptions, methods and defi nitions in the scheme are not exhaustive and alternative valid answers are acceptable.

3. The detail required in any answer is determined by the context and the manner in which the question is asked, and by the number of marks assigned to the answer in the examination paper, and in any instance, therefore, may vary from year to year.

4. Words, expressions or statements separated by a solidus (/) are alternatives which are equally acceptable. A word or phrase in bold, given in brackets, is an acceptable alternative to the preceding word or phrase. Note, however, that words, expressions or phrases must be correctly used in context and not contradicted, and where there is evidence of incorrect use or contradiction, the marks may not be awarded.

5. In general, names and formulas of elements and compounds are equally acceptable except in cases where either the name or the formula is specifi cally asked for in the question. However, in some cases where the name is asked for, the formula may be accepted as an alternative.

6. There is a deduction of one mark for each arithmetical slip made by a candidate in a calculation.

ORDINARY LEVEL CHEMISTRY | Pre-Leaving Certifi cate, 2019 PAGE 19 OF 29

ORDINARY LEVEL

ORDINARY LEVELSection A

Eight questions. These must include at least two questions from Section A.All questions carry equal marks (50 marks).

Note: (i) In many cases, most especially defi nitions, candidates’ answers must contain key phrases in order to get the assigned marks In calculations, 3 marks are deducted for a mathematical error, 1 mark for a slip or missing units, but no further deduction is applied if the calculation, otherwise correct, is completed

(ii) Alternative valid answers are acceptable.

QUESTION 1

(a) What: (i) Refl ux 3 (ii) Distillation 3 (b) What: Anti-bumping granules / porcelain chips (bits, pieces) / pumice / 6 any named inert material

(c) Why: It acts as a solvent / to get them into solution / it dissolves them 6

(d) Why: To maximise the yield / to minimise the loss 6

(e) Why: To redissolve the mixture / keep soap in solution / 6 make it easier to pour out of fl ask

(f) Name: Brine / salt (NaCl) solution 6

(g) Why: to remove any sodium hydroxide 6

(h) What: (i) a lather forms 4

(ii) a scum is formed / doesn’t easily form a lather / need a lot of soap 4 to form a lather / no lather


ORDINARY LEVEL

Question 2

(a) Name: Pipette 3

How: Rinse pipette with deionised water // 3 and then rinse with hydrochloric acid solution (solution that it will contain) 3

(b) Name: Conical Flask [allow 3 for titration fl ask] 6

State: Swirl (shake) after each addition // 3 wash down the side of the fl ask with deionised water. 3

(c) What: A solution whose exact concentration is known 5

(d) Name: methyl orange / methyl red / phenolphthalein 3

What: methyl orange: yellow (orange) // red (pink, peach) 3 or methyl red: yellow // red 3 or phenolphthalein: purple (pink, violet) // colourless [6 marks may be given for correct colours of an incorrect indicator. Allow (3)

for correct colours reversed (and matching a named indicator). (9) marks only available for correct indicator and correct colours in correct order.]

(e) Calculate:

M1xV1 = M2xV2 n1 n2 27.5 × 0.10 = 25 × M2 [3 for LHS: 3 for RHS] 6 1 1 M2 = 0.11 moles per litre 3

(f) How: Pour into an evaporating dish // 3 State: Evaporate to dryness 3


ORDINARY LEVEL

QUESTION 3

(a) Describe: fi lter a known (named) volume of water through weighed fi lter paper 3×3 // dry and reweigh fi lter paper // subtract weighings

(b) Diagram:

3

[at least one label required]

Describe: evaporate to dryness a known volume // of fi ltered water (fi ltered water 3 from previous experiment) in a weighed evaporating dish (beaker, other) 3 // cool and reweigh dish // subtract weighings [any 3 ] 3

(c) Find: (i) 0.03g in 500 0.06 g in 1 litre 3 0.06 × 1000 = 60 mg / l or 60 ppm 3

(ii) 0.13g in 200 0.65g in 1 litre 3 0.65 × 1000 = 650 mg / l or 650 ppm 3

Describe: platinum (nichrome) wire (probe) / soaked splint (lollipop stick) // 4×2 dip wire (splint) into salt // place salt in (into, on, at edge of, at top of) fl ame // note (observe) colour

Colour: Lilac 3

(d) How: add solution of silver nitrate // 3 white colour (precipitate) formed 3

Evaporating DishGauze

Tripod

Bunsen burner


ORDINARY LEVEL

Section B QUESTION 4 Eight items to be answered. Six marks to be allocated to each item and one additional mark to be added to each of the fi rst two items for which the highest marks are awarded.

(a) 2 8 8 1 6

(b) Sharing of // electron pairs 2 × 3

(c) Tetrahedral 6

(d) Kills bacteria / sterilize / disinfectant 6

(e) Endothermic 6

(f) 28/80 = 0.35 3 0.35 × 100 = 35% [Addition must be shown for error to be treated as slip (– 1)] 3

(g) platinum / Pt // palladium / Pd // rhodium / Rh 6

(h) Secondary Treatment 6

(i) C olour changes from blue // (brick) red 2 × 3

(j) Bomb calorimeter [allow 3 for calorimeter] 6

(k) A Sulfur Dioxide / SO2 6

B Calcium // Sodium 2 × 3


ORDINARY LEVEL

QUESTION 5

(a) State: (i) O = 2 2 (ii) H = 1 2

Explain: Oxygen requires 2 electrons // to have 2 × 2 the same electron confi guration as neon (noble gas) Hydrogen needs one electron // to have 2 × 2 the same electron confi guration as helium

(b) Use:

6 valence oxygen electrons // 3 2 hydrogen electrons // 3 two lone pairs in molecule // 3 two bond pairs in molecule [any 3]

What: V planar / planar / angular / v-shaped / bent 3

(c) Defi ne: relative (measure of) attraction / number expressing (giving) 3 attraction // for a shared pair of electrons / for electrons in a covalent bond. 3

Predict: Electronegativity diff erence (O) 3.44 – 2.20 (H) = 1.24 3 Less than 1.7 therefore polar covalent 3

(d) Distinguish: mass no: number of protons and neutrons / number of nucleons / 3 mass of nucleus / mass of particular isotope

rel at mass: average mass of atom(s) of an element / average of isotopes 3 taking abundance into account // based on (relative to, compared with) 1/12 mass of carbon-12 atom (isotope) 3

Why: they are mixtures of isotopes / average masses of isotopes 5


ORDINARY LEVEL

QUESTION 6

(a) What: measure of (indication of, showing, giving) tendency (likelihood) to 6 auto-ignite (knock, pink, pre-ignite, ignite early, ignite before spark) /

number representing ability (tendency) of fuel to resist auto-igniting (knocking, pinking, pre-igniting, igniting early, igniting before spark)

or based on a scale where 2,2,4-trimethylpentane (iso-octane) is assigned a

rating of 100 and heptane (n-heptane) a value of 0. or percentage by volume of 2,2,4-trimethylpentane (iso-octane) in a blend

(mix) with heptane (n-heptane) that matches the behaviour of the fuel in terms of auto-ignition

[If (6) not given, allow (3) for mention of ‘auto-igniting (knocking, etc….see above)’]

Name: isomerisation / catalytic cracking / dehydrocyclisation (reforming) / add 6 lead / add oxygenate / add methanol (ethanol, MTBE, methyl tert-butyl

ether] / use short chain molecules [any one]

Name: fractional distillation / fractionation [no marks for distillation on its own] 6

(b) Explain: Contains a benzene ring 5

Name: Ethylbenzene 3 Octane no. High octane number 3

Why: Contains a ring / cyclic / aromatic 3 (c) Name: Alkenes 3 A = Ethene 3 B = Propene 3 C = Butene 3

Copy: C2H4 + 3O2 2CO2 + 2H2O 6 [Formulas (3) correctly balanced (3)]


ORDINARY LEVEL

QUESTION 7

(a) Defi ne: – log10

[H+] / [H

3O

+] 3

Defi ne: (i) gives hydrogen (hydronium) ions (H+, H3O+) in aqueous solution / 3 proton (hydrogen ion, H+) donor // (ii) gives hydroxyl (hydroxide) ions (OH¯) in aqueous solution / 3 proton (hydrogen ion, H+) acceptor

Explain: reaction between acid and base // giving salt and water only 3 + 3

Name: Hydrochloric acid/HCl 3

(b) Words: 1 = magnesium 4 2 = temporary 4 3 = hydrogen-carbonate 4 4 = boiling 4 5 = permanent 4 6 = sulfates 4 7 = ion exchange 4 8 = sodium 4


ORDINARY LEVEL

QUESTION 8

(a) What: Forward and reverse reactions // have equal rates 3+2or

concentrations of reactants and products // constant State: if a system at equilibrium is disturbed (experiences a stress) // it tends 3 to minimise (oppose, relieve) the disturbance (stress)

or if a system is disturbed (experiences a stress) // it tends to minimise 3 (oppose, relieve) it

(b) State: Blue 3 Pink 3 Blue 3

(c) What: Contact process 6

(d) Write: Kc = [SO3]2 6 [SO2]2[O2] [square brackets required. Inverted eq. = 3]

Predict: (i) Low Temperature 3 (ii) High Pressure 3

Reason: Low Temperature favours the exothermic reaction 3 High Pressure favours the side with the least number of moles 3

What: Equilibrium will be reached faster / 3 It favours neither side 3


ORDINARY LEVEL

QUESTION 9

Name: (i) A = ethyne 3 B = ethanol 3 C = methanoic acid 3 D = chloroethane 3 E = benzene 3 F = ethanal 3 G = butane 3 H = cyclohexane 3

Which: (ii) G/butane 3

Which: (iii) E/benzene 3

Which: (iv) C/methanoic acid 3

Which: (v) E/B/H/benzene/ethanol/cyclohexane 3

Which: (vi) D/ chloroethane 4

Which: (vii) F/ ethanal 3

Which: (viii) B/ ethanol 3

Which: (ix) A/ ethyne 4


ORDINARY LEVEL

QUESTION 10

(a) Write: 1: The Greeks 5 2: Dalton 5 3: Davy 5 4: Mendeleev 5 5: Bohr 5

(b) What: Catalyst 4

Plot:

Graph = Correctly labelled // correctly scaled axes. 3+3 [Allow 3 marks only if axes are reversed]

All points plotted correctly 6 [Allow 3 marks if 6 to 8 points plotted correctly]

Curve drawn correctly 3 [three marks deducted if graph paper not used]

Time 11 cm3 of oxygen produced in 1.5 to 1.7 min. 6 [accept 1 min 30 sec to 1 min 42 sec] [full marks if reading corresponding to 11 cm3 from students graph]

(c) Words: 1 = Volumes 3 2 = Temperature 3 3 = Pressure [2 and 3 can be reversed for full marks] 3 4 = Atomic 5 = number [both must be correct] 3 6 = Atomic 7 = Mass /(Mass Number) [both must be correct] 3 8 = Pressure 3 9 = Volume 3 10 = Temperature 3 [First correct answer gets 4 otherwise 3 for each]

0

2

8

13

16

17.4

Rate of Reaction

Time / Min

Volu

me

of O

2/cm

3

18 18.4 18.5 18.5

4

8

6

10

12

14

16

18

20

01 2 3 4 5 6 7 8


ORDINARY LEVEL

Question 11

(a) How: (i) (4/40) = 0.1 moles [Addition must be shown for error to be treated as slip (– 1)] 2+2

How: (ii) ratio = 2:1 therefore 0.05 moles 3+3

Copy: (iii) 1.6 g 3 0.3 × 1023 molecules / 3 × 1022 3 1.12 litres / 1120 cm3 3 What: (iv) 56 × 0.1 // = 5.6 g [Addition must be shown for error to be treated as slip (– 1)] 3+3

(b) Name: Becquerel 4 What: spontaneous disintegration (breakdown) of nucleus of atom // emitting radiation 6

Name: alpha / α / beta / β / gamma / γ 3

Describe: Alpha poor (stopped by a sheet of paper) & positive / Beta medium (stopped by a thin sheet of aluminium) & negative / Gamma high

(stopped by a block of lead) & none [must be matched] 3+3

Give: Co-60 / Am-238 / C-14 3 cancer treatment / smoke alarms / carbon dating [must be matched] 3

(c) Formula: Ozone = O3 3

State: Screens (blocks out) uv-light coming from the sun 3

How: Electrical storms / lightning / legumes / nitrogen fi xing bacteria / 6 [Any one suitable example] What: Blocking (preventing escape, absorption or reabsorption) of radiation 4+3 (heat, energy, infrared) // by gases in the atmosphere

Name: Carbon dioxide / water / methane / CFC(s) / HCFC(s) / HFC(s) 3×2 / chloromethane / chloroethane / dinitrogen oxide (nitrogen (I) oxide, nitrous oxide) / PFC(s) / ozone / sulphur hexafl uoride (Any two)

(c) Give: iron oxide / magnetite / haematite / (FeO, Fe2O3, Fe3O4) / iron sulfi de 5 (FeS) / pyrites, [any one]

Name: coke / limestone 5 What: slag 5

Name: steel (cast iron, pig iron) 5

Which: chromium / nickel 5

NOTES

NOTES

89F Lagan Road, Dublin Industrial Estate, Dublin 11

T: 01 808 1494

F: 01 836 2739

E: info@examcra .ie

W: www.examcra .ie

Documents

*WMS12* CHEMISTRY

WMS12 CHEMISTRY