OCR A Jun 2010 Paper 1 Q1 (with explained solutions) |

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OCR A Jun 2010 Paper 1 Q1

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Answer all the questions. Tin mining was common practice on Dartmoor in pre-Roman times. Most of the tin extracted was mixed with copper to produce bronze. (a) The table below shows the sub-atomic particles of an isotope of tin. isotope protons neutrons electrons 118Sn (i) Complete the table. [1] (ii) In terms of sub-atomic particles, how would atoms of 120Sn differ from atoms of 118Sn?[1] (b) The relative atomic mass of tin is 118.7. Define the term relative atomic mass.[3] (c) A bronze-age shield found on Dartmoor contained 2.08 kg of tin. Calculate the number of tin atoms in this bronze shield. Give your answer to three significant figures.OCR 2010 answer =[2]<br />
(d) Tin ore, known as cassiterite, contains an oxide of tin. This oxide contains 78.8% tin by mass. Calculate the empirical formula of this oxide. You must show your working. answer =[2] [Total: 9]OCR 2010 Turn over<br />

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Marks 1 1 F321 Question 1 a i ii Expected Answers 118Sn 50p 68n 50e Complete row (cid:57) 120 ora 50Sn has (two) more neutrons / 70 neutrons (cid:57) i The (weighted) mean mass of an atom (of an element) OR The (weighted) average mass of an atom (of an element) (cid:57) compared with 1/12th (the mass) (cid:57) of (one atom of) carbon-12 (cid:57) moles of Sn = 2080 = 17.52 (cid:57) 118.7 17.52 (cid:117) 6.02 (cid:117) 1023 = 1.05 (cid:117) 1025 atoms (cid:57) Mark Scheme June 2010 Additional Guidance ALLOW There is a different number of neutrons IGNORE correct reference to protons / electrons DO NOT ALLOW incorrect references to protons / electrons ALLOW ECF for stated number of neutrons from 1a(i) ALLOW average atomic mass DO NOT ALLOW mean mass of an element ALLOW mean mass of isotopes OR average mass of isotopes DO NOT ALLOW the singular; isotope For second and third marking points ALLOW compared with (the mass of) carbon-12 which is 12 ALLOW mass of one mole of atoms (cid:57) compared to 1/12th (cid:57) (mass of) one mole OR 12g of carbon-12 (cid:57) ALLOW mass of one mole of atoms . 1/12th mass of one mole OR 12g of carbon-12 ALLOW 17.5 up to (correctly rounded) calculator value of 17.52316765 DO NOT ALLOW use of 118, which makes moles of Sn = 17.63 ALLOW 105 (cid:117) 1023 atoms DO NOT ALLOW answers which are not to three sig figs for second marking point ALLOW two marks for answer only of 1.05 (cid:117) 1025 ALLOW one mark for answer only if not 3 sig figs up to calculator value of 1.054894693 (cid:117) 1025 Eg 100 (cid:117) 1 ALLOW ECF for any calculated moles of Sn (based on use of any Ar value) (cid:117) 6.02 (cid:117) 1023 if shown to 3 sig figs DO NOT ALLOW mass of Sn (cid:117) 6.02 (cid:117) 1023<br />
F321 Question 1 d Expected Answers 78.8 and 21.2 118.7 16.0 OR = 0.66(4) and = 1.3(25) (cid:57) 0.66(4) = 1 1.325 = 2 0.66(4) 0.66(4) ans = SnO2 (cid:57) Total Mark Scheme June 2010 Marks Additional Guidance ALLOW SnO2 for one mark if no working shown ALLOW use of 118 for this part IGNORE incorrect rounding provided given to two sig figs IGNORE incorrect symbols e.g. T or Ti for Tin, as long as correct Ar of tin (118.7 or 118) used ALLOW Sn2O for 1 mark ECF if both inverted mole calculations are shown ALLOW Sn3O5 with evidence of use of both atomic numbers for one mark ALLOW 2 marks if candidate has adopted the following approach 78.8% of mass = 118.7 100% of mass = 118.7/0.788 = 150.6 (151) 150.6118.7 = 31.9 (32) Both masses would get one mark 31.9/16 = 2<br />

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