ALS1A - Section 1/ Atoms, Molecules and Stoichiometry...

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Section 1/ Atoms, Molecules and Stoichiometry /Suggested solutions/ SKH Bishop Baker Secondary School A-Level Chemistry Section 1 Atoms, Molecules and Stoichiometry 80 IA Q.4 (d) IO 3 - (aq) + 8I - (aq) + 6H + (aq) → 3I 3 - (aq) + 3H 2 O(l) [1] [Note that iodine is slightly soluble in water but dissolves in potassium iodide solution: IO 3 - (aq) + 5I - (aq) + 6H + (aq) 3I 2 (aq) + 3H 2 O(l); I 2 (aq) + I - (aq) I 3 - (aq)] 80 IB Q.5 (b) (i) Compartment A is the ionization chamber to produce high energy electrons to collide with atoms or molecules to form positive ions. [1] (ii) (I) X has two isotopes of masses 203 and 205, their relative abundances are 30.05% and 69.95% respectively. [1] (II) Relative atomic mass of X is (203)(30.05%) + (205)(69.95%) = 204.4 [1] (iii) By inspection, 2(12.000) + 6(1.008) + 1(15.995) = 46.043 [1] The possible formula of Y is C 2 H 6 O. [1] 81 IA Q.1 (a) (i) Propanone should be injected into the syringe air space through the rubber cap. [1] (ii) Weigh the hypodermic syringe before and after the injection. [1] (iii) Rinse the hypodermic syringe with propanone. [1] Air bubbles must be expelled from the hypodermic syringe. [1] Await until a steady temperature before injecting propanone. [1] Await steady volume reading, ensuring thereby the complete vaporisation of the propanone. [1] ( ANY TWO OF THE ABOVE ) 81 IB Q.9 (a) (i) Points correctly marked. [1] Correct shape [1] (ii) Before the equivalence point, AgCl is precipitated and Cl - is removed from solution .[1] & (iii) Cl - is effectively replaced by NO 3 - . [1] Since conductivity decreases before the equivalence point. NO 3 - must have a lower molar conductivity than Cl - . [1] After the equivalence point, no ions are being removed from the solution, and the added Ag + and NO 3 - ions increase the conductivity . [1] (iv) From the graph, the equivalence point is reached when 10 cm 3 of 0.1 M AgNO 3 is added. [1] 10 cm 3 of 0.1 M AgNO 3 10 cm 3 of KCl Molarity of 0.1 M KCl = 10 x 0.1 / 100 M [1] = 0.01 M [1] If a wrong value for the equivalence point is found, give full mark for correct calculation 82 IA Q.2 (a) Given: 1 atm. = 101 325 N m -2 1
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Section 1/ Atoms, Molecules and Stoichiometry /Suggested solutions/ R = 82.05 cm 3 atm K - 1 mol - 1 = 82.05 x 10 - 6 x 101325 m 3 N m - 2 K - 1 mol - 1 = 8.314 N m K - 1 mol - 1 = 8.314 J K - 1 mol - 1 [1] 82 IIB Q.8 (a) (i) Cu : S = 70/63.5 : 25/32.1 = 1.10 : 0.78 = 1.4: 1.0 approximate empirical formula is Cu 1.4 S. [2] If write Cu 3 S 2 or Cu 1.5 S with above calculation. [1 ½ ] If convert 1.4: 1 to 1.5: 1 (and write wrong empirical formula) [1] Possible compounds are CuS and Cu 2 S. [1] (Write one compound still earn zero mark.) air [1] [1] C, CO, H 2 [1] (ii) CuS → CuO → Cu [3] or atm. O 2 or O 2 / heat heat 82 IIB Q.9 (b) (i) Excess IO 3 - and I - added [1] to a known volume of the acid solution in question. [ ½ ] Use standardised [ ½ ] S 2 O 3 2 - [1] , or S 2 - or Sn 2+ to titrate I
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ALS1A - Section 1/ Atoms, Molecules and Stoichiometry...

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