SRJC H1 CHEM P1 Worked Solutions

SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 1 CHEMISTRY 8873/01 JC2 Preliminary Examination 20 September 2018 Paper 1 Multiple Choice 1 hour 1 B 6 B 11 A 16 C 21 A 26 D 2 B 7 A 12 B 17 B 22 A 27 C 3 D 8 A 13 D 18 B 23 D 28 A 4 C 9 C 14 D 19 B 24 A 29 D 5 D 10 A 15 B 20 C 25 B 30 A WORKED SOLUTIONS 1 Use of the Data Booklet is relevant to this question. What is the number of atoms in 800 cm 3 of hydrogen gas under room temperature conditions? A 2.0 x 1022 B 4.0 x 1022 C 2.0 x 1025 D 4.0 x 1025 Ans: B nH= ଴.଼ ଶସ × 2 number of H atoms = ଴.଼ ଶସ × 2 × 6.02 × 10 ଶଷ ൌ 4.0 x 1022 2 Use of the Data Booklet is relevant to this question. The relative abundances of the isotopes of a sample of carbon are shown in the table below. What is the relative atomic mass of carbon in this sample? relative isotopic mass 12 13 14 relative abundance 100 1.08 0.01 A 12.00 B 12.01 C 12.10 D 12.15 Ans: B Relative atomic mass = ሺଵଶ ×ଵ଴଴ሻାሺଵଷ ×ଵ.଴଼ሻାሺଵସ ×଴.଴ଵሻ ଵ଴଴ାଵ.଴଼ା଴.଴ଵ = 12.01

3 Four substances E, F, G and H have physical properties as shown. substance melting point/ °C boiling point/ °C electrical conductivity of solid of liquid E 17 45 Poor Poor F 64 759 Good Good G 1132 1950 Poor Good H 3550 3825 Good Unknown What could be the identities of E, F, G and H? E F G H A PCl5 SO3 Na 2O Al2O3 B PCl5 A l2O3 K Na 2O C SO3 PCl5 C (graphite) Na 2O D SO3 K Na 2O C (graphite) Ans: D substance melting point/ °C boiling point/ °C electrical conductivity Structure of solid of liquid E 17 45 Poor Poor SMS PCl5 / SO3 F 64 759 Good Good GMLS K G 1132 1950 Poor Good GILS Na2O H 3550 3825 Good Unknown GMS Graphite

4 10 cm3 of hydrogen peroxide was diluted to 250 cm 3 using a standard flask. 25.0 cm 3 of the diluted solution was reacted with 10 cm3 of concentrated potassium iodide in the presence of dilute sulfuric acid. The resulting mixture was then titrated with 0.10 mol dm –3 sodium thiosulfate. It was found that 21.00 cm3 of thiosulfate was required to reach end-point. The following shows the reactions described above. H 2O2 + 2I– + 2H+ → I2 + 2H2O I2 + 2S2O32– → 2I– + S4O62– What is the initial concentration of the hydrogen peroxide used? A 0.04 mol dm–3 B 0.08 mol dm–3 C 1.05 mol dm–3 D 2.10 mol dm–3 Ans: C Mole ratio 2S2O32– : 1 I2 : 1H2O2 n S2O32– = ଶଵ.଴଴ ଵ଴଴଴ × 0 . 1 0 = 2.10 × 10–3mol n H2O2in 25 cm3 = 1.05 × 10–3 mol n H2O2in 250 cm3 = 1.05 × 10–2 mol initial concentration of H2O2 = 1.05 × 10–2 / (10/1000) = 1.05 mol dm–3 5 Which of the following is a redox reaction? A NaOH + HBr → NaBr + H2O B FeCl3 + 6H2O → [Fe(H2O)6]3+ + 3Cl– C H2SO4 + Ca(OH)2 → CaSO4 + 2 H2O D Cu + NH3 + O2 + H2O → [Cu(NH3)4](OH)2 Ans: D There are no change in oxidation numbers of any species for options A, B and D. Option C: Cu is oxidised from 0 to +2 O is reduced from 0 to –2

6 Which of the following statements is untrue about the second ionisation energies of Period 2 elements? A The general trend is similar to that of first ionisation energies. B Increase in shielding effect outweighs the increase in nuclear charge. C The anomalies lie between group 13 and 14, and between 16 and 17 elements. D There is increasing electrostatic forces of attraction between the nucleus and valance electrons. Ans: B Across the period, shielding effect is similar as electrons are added to the same quantum shell. Thus, B is untrue. 7 When 4He nuclei was pass through an electric field, it was deflected 4°. Which of the following ions would be deflected half as much as the 4He nuclei in the opposite direction? A C3– B N3– C Na+ D Mg2+ Ans: A Concept: ࢋ ࢓ α angle of deflection ࢋ ࢓of 4He nuclei is +½ ( You need to know that the nucleus is positively charged) Thus, only options A and B are possible, as they will deflect in the opposite direction. ࢋ ࢓of C3–is –¼ Hence, will be deflected half as much as the 4He nuclei in the opposite direction.

8 Which of the following statements is true for Al3+? A There are no d orbitals occupied. B There are a total of five subshells occupied. C There are a total of three s orbitals occupied. D The last electron removed to obtain Al3+ came from the 3p orbital. Ans: A Al3+ : 1s22s22p6 • No d orbitals is occupied. • There are 3 subshells occupied. • There are two s orbitals occupied. • The first 3 electrons removed obtain A l3+ came from the the 3p orbital first, followed by 3s orbital. 9 A chloride and an oxide of the elements in the third period of the Periodic Table were dissolved in two separate portions of water to form aqueous solutions. Both of the resulting solutions could dissolve Al2O3 but only one of the two could dissolve P4O10. Which of the following pairs could be the chloride and the oxide used? 1 NaCl SO3 2 MgCl2 MgO 3 SiCl4 Na2O A 2 only B 1 and 3 C 2 and 3 D 1, 2 and 3 Ans: C You should work these out first: In aqueous solution Compound pH Compound pH NaCl 7 (neutral) SO3 3 (acidic) MgCl2 6.5 (slightly acidic) MgO 12 (basic) SiCl4 2 (acidic) Na2O 13 (basic) Since both of the resulting solutions can be used to dissolve Al2O3 (amphoteric) but only one of the two can be used to dissolve P4O10, one of the solution has to be acidic while the other basic. Only options 2 and 3 gives both an acidic and basic solution.

10 The diagram represents the melting points of four consecutive elements in the third period of the Periodic Table. The sketches below represent another two properties of the elements. What are properties J and K? property J property K A first ionisation energy electronegativity B number of valence electrons melting point C ionic radius nuclear charge D boiling point atomic radius Ans: A The four consecutive elements are aluminium (conductor), silicon (semiconductor), phosphorus (non-conductor) and sulphur (non-conductor) respectively. Hence, property J can represent the 1 st IE trend only and K can represent electronegativity or nuclear charge.     proton number 0 electrical conductivity     proton number 0 property J     proton number 0 property K

11 Which one of the following cannot be explained using hydrogen bonding? A CH3CHO has a higher boiling point than C2H6. B CH3COOH has a Mr of 120 in benzene. C HF has a higher boiling point than HCl. D CH3COCH3 is miscible with water. Ans: A Intermolecular H-bonds form btw molecules with N, O or F atoms directly bonded to H atom. Aldehyde groups such as CH3CHO do not have intermolecular H-bonds. CH3CHO has intermolecular pd-pd forces of attraction , while C 2H6 has intermolecular id-id forces of attraction. H-bond does not account for this difference. 12 The thyroid gland concentrates iodine and uses it to produce thyroxine, which is a hormone that controls the metabolic rate. HO I I O I I CC H H C N H OH O H H Thyroxine What are the values of the bond angles p, q and r? p q r A 105° 107 ° 180 ° B 105° 107 ° 120 ° C 109° 120 ° 180 ° D 109° 120 ° 120 ° Ans: B OH OC H C H N H C OH O I I I I H H Bent about O Trigonal pyramidal about N Trigonal planar about C p q r

13 The following graph shows the changes in pH of 20.0 cm3 of 1.0 mol dm-3 CH3COOH when excess 1.0 mol dm-3 alkali solution is added gradually. Which one of the following alkali solution with a suitable indicator could have resulted in the graph below? Alkali Indicator A NaOH Methyl orange B