TJC Prelim P2 Solutions

TJC H2 Chemistry Paper 2 2008 (Answers) [Turn over 2 2008 TJC H2 Prelim P2 Answers 1 (a) For decades, scientists were puzzled by why dinosaurs suddenly became extinct 65 million years ago. In studying core samples of rock dating back to that period, scientists found an unusual high level of iridium. This possibly came from an iridium- rich asteroid that struck the Earth’s surface and the mystery to dinosaurs’ extinction was solved. Iridium exists as two naturally occurring isotopes, iridium-191 and iridium-193. For Examiner’s Use (i) Define the term relative atomic mass. · Relative atomic mass is the average mass of one atom of an element on a scale in which one atom of 12C isotope of carbon has a mass of 12 units. (ii) The relative atomic mass of iridium is 192.2. Calculate the natural abundance, in percentage, of each isotope. Let y be the percentage of iridium-191 · 191 x 100 )y100(193 100 y -´+ = 192.2 191y + 19300 – 193y = 19220 y = 40 · Iridium-191 (40%) , Iridium-193 (60%) [3] (b) Aqua regia (a mixture of 75% nitric acid and 25% hydrochloric acid by volume) is highly corrosive. Only noble metals like iridium are inert to this solution. A 5 g sample of platinum-iridium alloy required 24.6 cm3 of aqua regia for complete reaction. Platinum was completely oxidised to platinum(IV) ions by nitric acid and 0.5 g of metal was recovered. (i) Find the percentage of each metal in the alloy. · Since 0.5 g of iridium metal was recovered, Percentage of iridium = 1005 5.0 ´ = 10 % Percentage of platinum = 90%

TJC H2 Chemistry Paper 2 2008 (Answers) [Turn over 3 (ii) The concentration of nitric acid used to make aqua regia is 5.0 mol dm-3. Assuming that the reaction between the sample and aqua regia is complete, construct the balanced equation for the reaction. · No. of moles of platinum = 195 5.05 - = 0.0231 mol No. of moles of nitrate = 51000 4 36.24 ´ ´ = 0.0923 mol · (1 mark working for construction of balanced equation) Pt ºººº 4e- ºººº Pt4+ No. of moles of electrons transferred = 0.0231 x 4 = 0.0924 mol No. of moles of electrons transferred per mole of nitrate = 0.0924 / 0.0923 = 1 NO3 - is reduced to NO2 NO3 - + 2H+ + e- ®®®® NO2 + H2O Pt ®®®® Pt4+ + 4e- · Pt + 4NO3 - + 8H+ ®®®® Pt4+ + 4NO2 + 4H2O [4] [Total: 7]

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