IJC H1 CHEM P2 solution

INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 2 in preparation for General Certificate of Education Advanced Level Higher 1 CANDIDATE NAME CLASS INDEX NUMBER CHEMISTRY Paper 2 Structured and Free Response Questions Section A: Structured Candidates answer Section A on the Question Paper Section B: Free Response Additional Materials: Writing Paper Data Booklet 8872/02 2 Sept 2015 2 hours READ THESE INSTRUCTIONS FIRST Write your index number, name and civics group on all the work you hand in. Write in dark blue or black pen. You may use pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A: Structured Questions (40m) Answer all questions in the space provided. Section B: Free Response Questions (40m) Answer two questions on separate writing papers. You are advised to show all working in calculations. You are reminded of the need for good English and clear presentation in your answers. You are reminded of the need for good handwriting. Your final answers should be in 3 significant figures. You may use a calculator. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 13 printed pages and 1 blank page. Innova Junior College [Turn over For Examiner’s Use Section A 1 12 2 8 3 10 4 10 Section B 20 20 Significant figures Handwriting Total 80

2  IJC 2015 Prelim 2/8872/02 [Turn over For Examiner’s Use Section A Answer ALL questions on the spaces provided. 1 Chromium is a transition metal. Chromium compounds are highly valued as pigments for their vivid green, yellow, red and orange colours. (a) The element chromium has four main naturally occurring isotopes. isotope relative abundance / % 50Cr 4.33 52Cr 83.8 53Cr 9.50 54Cr 2.37 Use the relative abundance data to calculate the relative atomic mass of Cr, showing your working clearly. A r = 2.389.5083.74.35 54(2.38)53(9.5)52(83.7)50(4.35)   = 52.0 [1] (b) Chromium(III) chloride dissolves in water to form a green solution. Draw a labelled diagram to illustrate the interaction between a Cr3+ ion and a water molecule. [2] (c) Chromium(IV) oxide, CrO 2 is a black solid that disproportionates into a mixture of Cr3+ and Cr2O7 2– in acidic solutions. The half equation for the reduction of CrO 2 to Cr3+ is as follows: CrO2 + 4H+ + e Cr3+ + 2H2O To determine the percentage purity of a sample of chromium( IV) oxide, a student dissolves 1.5 g of impure sample in 20 cm 3 of sulfuric acid and make up the total volume to 100 cm 3 with distilled water. He then found that 25 cm 3 of the resulting solution, containing Cr 2O7 2– ions, required 16.00 cm 3 of 0.200 moldm -3 of iron( II) sulfate solution for complete reaction. (i) Construct the half equation for the oxidation of CrO2 to Cr2O7 2–. [O]: 2CrO2 + 3H2O Cr 2O7 2– + 6H+ + 4e (ii) Hence, construct the overall equation for the disproportionation of CrO 2 to Cr3+ and Cr2O7 2–. Cr3+ : ion-dipole interaction

3  IJC 2015 Prelim 2/8872/02 [Turn over For Examiner’s Use 4CrO2 + 16H+ + 12e 4Cr3+ + 8H2O 2CrO2 + 3H2O Cr 2O7 2– + 6H+ + 4e 6CrO2 + 10H+ Cr 2O7 2– + 4Cr3+ + 5H2O (iii) Calculate the amount of Fe2+ used in the reaction. No. of moles of Fe2+ = 1000 16.00 x 0.200 = 0.0032 (iv) Hence, calculate the amount of Cr2O7 2– present in 100 cm3 of the solution. Fe2+: Cr2O7 2– = 6: 1 No. of moles of Cr2O7 2– in 25.0 cm3 = 6 1 x 0.0032 = 5.333 x 10-4 No. of moles of Cr2O7 2– in 100 cm3 = 25.0 100 x 5.333 x 10-4 = 2.13 x 10-3 (v) Calculate the mass of CrO 2 present in the sample and hence, the percentage purity of the sample. [You may assume a mole ratio of CrO 2 : Cr2O7 2– of 6:1 if you were unable to derive the overall equation in (ii).] No. of moles of CrO2 = 6 x 2.133 x 10-3 = 0.0128 Mass of CrO 2 = 0.0128 x 84.0 = 1.075 g Percentage purity of sample = 1.5 1.075 x 100% = 71.7 % [7] (d) Beams of charged particles are deflected by an electric field. lf the particles are all travelling at the same speed, through an electric field of constant strength, the angle of deflection is proportional to their charge/mass ratio. ln a particular experimental set-up, protons are deflected through an angle of +15 o. (i) Assuming an identical set of experiment al conditions, by what angle will Cr 3+ be deflected? m c of 1H+ = 1 k = 15 m c of 52Cr3+ = 52 3 Angle of deflection = +15( 52 3 ) = +0.866o (ii) Under identical conditions, a beam of particles, A, each having 12 times the mass of a proton, was deflected by an angle of +5o.

4  IJC 2015 Prelim 2/8872/02 [Turn over For Examiner’s Use Suggest the overall charge on a particle of A. 5 = 15( 12 charge ) charge = +4 [2] [Total: 12] 2 (a) The table below shows the lattice energies for the sodium halides and magnesium oxide. compound lattice energy / kJ mol 1 NaCl 781  NaBr 743  NaI 699  MgO 3933  (i) Define the term lattice energy. Lattice energy is the enthalpy change when 1 mole of solid ionic compound is formed from its separate gaseous ions at standard conditions of 1 atm and 298 K. (ii) By quoting appropriate data from the Data Booklet , explain why the lattice energy of MgO is considerably larger than those of the sodium halides. Ion Na+ Mg2+ Ionic radius /nm 0.095 0.065 Both Mg2+ ions and O2- ions have a higher charge than Na+ ions and halide ions. In addition, the ionic radii of Mg2+ and O2 are smaller than that of Na+ and the halide ions Since .. qqLE rr      , thus magnitude of lattice energy of MgO is considerably larger than those of sodium halides. [4] (b) The 1,2-dichloroethene molecule can exist in either of the following forms, B or C as shown:

5  IJC 2015 Prelim 2/8872/02 [Turn over For Examiner’s Use By considering the polarity of the molecule, predict which form, B or C, has a lower boiling point. Explain your answer. C has lower boiling point than B. This is because B is polar covalent simple molecule as the dipole moments do not cancel out each other and has stronger intermolecular permanent dipole - permanent dipole interactions than C that has intermolecular temporary dipole - induced dipole interactions. [2] (c) Account for the following observation: At around 30 oC, the relative molecular mass of hydrogen fluoride appears to be 40.0 and above 60 oC, it is about 20.0. At low temperatures, hydrogen fluoride exists in the form of dimers as there are strong/significant intermolecular hydrogen bonds. At higher temperature, the intermolecular hydrogen bonds are overcomed and thus, hydrogen fluoride appears as single HF molecules. [2] [Total: 8] 3 (a) Using blood as an example, explain what is meant by an acidic buffer solution. [5] An acidic buffer solution (with pH between 4 and 7) is one that contains a weak acid and the salt of that acid , it is a solution