RI 2020 Y5 Timed Practice Solutions

1 © Raffles Institution 2020 2020 Year 5 Timed Practice Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer D B C C A A D D A A D B B C C Question 1 (D) A Incorrect. Molar mass has units of g mol–1, while relative molecular mass has no units. B Incorrect. In 12.0 g of carbon-12, there is 1 mole of atoms. However, compounds are made up of more than one type of atoms. Hence, one mole of a compound will contains more than 1 mole of atoms. C Incorrect. The correct expression for the relative atomic mass of fluorine should be the weighted average mass of 1 atom of fluorine 1 12 x mass of 1 atom of carbon-12 D Correct. At room temperature and pressure, 1 mol of a gas occupies 24.0 dm3. Amount of O2(g) = 12.0/24.0 = 0.5 mol Amount of O atoms = 2(0.5) = 1.0 mol Amount of Ar(g) = 24.0/24.0 = 1.0 mol Since the amount of O atoms and amount of Ar amount are the same, they contain the same number of atoms. Question 2 (B) Let the oxidation number of N in HNO2 be x. (+1) + x + 2(-2) = 0 x = +3 The oxidation number of N is N2 is 0. The oxidation number of N decreased from +3 in HNO2 to 0 in N 2 i.e. the change in oxidation is –3. Question 3 (C) Mass of 1 mol of urea = 1 x [2(14.0) + 4(1.0) + 12.0 + 16.0] = 60.0 g Mass of urea reacted = 60.0 – 6.0 = 54.0 g Percentage of urea reacted = (54.0/60.0) x 100 = 90.0% Question 4 (C) There is a large increase between the 4 th and 5th IE i.e. significantly more energy is required to remove the 5 th electron compared to the 4 th electron. ⇒ 5th electron is from an inner electronic shell ⇒ Q has 4 electrons in its valence shell ⇒ Q is a Group 14 element and has valence shell electronic configuration of ns2 np2. ⇒ Q2+ has valence shell electronic configuration of ns2. Question 5 (A) Let pA and nA be the respective number of protons and neutrons of element A and p B and n B be the respective number of protons and neutrons of element B. 1 Correct. For mirror nuclei, pA = nb and pb = nA. Mass number of A = pA + nA = nB + pb = mass number of B. 2 Correct. 11O has 8 protons ⇒ it has 11 – 8 = 3 neutrons 11Li has 3 protons ⇒ it has 11 – 3 = 8 neutrons Since number of protons of 11O = number of neutrons of 11Li and number of protons of 11Li = number of neutrons of 11O, they are mirror nuclei. 3 Incorrect. 40Ar has 18 protons ⇒ it has 40 – 18 = 22 neutrons 40Ca has 20 protons ⇒ it has 40 – 20 = 20 protons Since number of protons of 40Ar is not equals to the number of neutrons of 40Ca, they are not mirror nuclei.

2 © Raffles Institution 2020 Question 6(A) The carbon -nitrogen triple bond is made up of 1  and 2  bonds. There are 3 also B–H  bonds. The dative bond from carbon to boron is a single bond and hence is also 1  bond. Hence, there is a total of 5  bonds and 2  bonds. Question 7(D) To answer this question, students must be able to 1. Draw the dot-&-cross diagram / structural formula for each species quickly. 2. Deduce the electron pair geometry and angle around the central atom. 3. Deduce the shape (molecular geometry) and bond angle around central atom. 2 bond pairs + 2 lone pairs (similar to that of H2O) ⇒ bond angle = ~105° 4 bond pairs + no lone pairs ⇒ bond angle = 109.5° 3 bond pairs + no lone pairs ⇒ bond angle = 120° 3 bond pairs + 1 lone pair (similar to that of NH3) ⇒ bond angle = ~107° Question 8(D) 1 Correct. As a result of the polar O –H bonds and the lone pair of electrons on each O atom, water molecules are capable of forming intermolecular hydrogen bonds. More energy is required to overcome the stronger hydrogen bonds between the H 2O molecules than the id-id interaction between CH 4 molecules, leading to the higher boiling point of H2O. 2 Incorrect. While the O–H bond is stronger than the C –H bond, boiling does not involve breaking the O–H nor the C–H bonds. 3 Incorrect. H2O contains 8 + 2(1) = 10 e – while CH4 contains 6 + 4(1) = 10 e–. Therefore, they have very similar electron cloud sizes and have similar strengths of id -id interactions. This does not help to explain the higher boiling point of H2O. Question 9(A) molar mass pV nRT masspV RT = = Since pV is the x-axis and there is no pressure term on the right side of the equation, constantmolar mass massx RT== The graph would be a vertical line, intersecting the x-axis at a value corresponding to molar mass mass RT . For the same mass, if X has a higher molar mass than Y, then the value of molar mass mass RT is smaller for X than Y i.e. the x-intercept for X is smaller than that of Y. Question 10(A) The increase in pressure when gas B was added is due to the presence of gas B. Partial pressure of A in the mixture remains at 1.00 atm since the volume of the vessel is fixed. partial pressure of B + partial pressure of A = final pressure partial pressure of B = 2.50 – 1.00 = 1.50 atm From solutions to question 9, molar mass ()molar mass = masspV RT mass RT pV =

3 © Raffles Institution 2020 molar mass of A : B ( ) ( )= : : 2.00 3.00:1.00 1.50 1:1 AB AB AB AB mass RT mass RT p V p V mass mass pp= = = Question 11(D) 1 Correct. Since the temperature rose after mixing, the reaction is exothermic i.e. heat is released from the reaction. This means that products are at a lower energy and are more energetically stable than the reactants. 2 Incorrect. Heat loss to the surroundings is accounted for by extrapolating to the time of addition of the solid i.e. 2.0 min. 3 Incorrect. From the graph. T = +14.5 °C Heat change = mcT = (250)(4.18)(14.5) = 15150 J = +151 kJ Question 12(B) H = 6BE(P–P) + 3BE(O=O) – 12BE(P–O) = 6(200) + 3(496) – 12(340) = –1392 kJ mol–1 Question 13(B) The following energy level diagram can be drawn from the information provided, taking into account the fact the combustion of both give the same product. The energy level of each form of sulfur can be deduced by taking reference from the energy level of the common product. 1 Correct. From the above diagram, since S(rhombic) is at a lower energy level than S(monoclinic), S(rhombic) is more stable. 2 Correct. Since S(rhombic) is at a lower energy level than S(monoclinic), energy is required to conver t S(rhombic) to S(monoclinic) i.e. it is an endothermic process. 3 Incorrect. Since S(rhombic) is at a lower energy level than S(monoclinic) and atomisation is an endothermic process which produces S(g) as a common product, the following energy level dia gram can be drawn. From the diagram, less energy is required to atomise 1 mol of monoclinic sulfur.

4 © Raffles Institution 2020 Question 14(C) A negative change in entropy would involve a process where there is a decrease in disorder. A Incorrect. There is an increase in disorder since liquid water is more disordered than solid water. B Incorrect. There is an increase in disorder since the ions are no longer held in the giant ionic lattice and are free to move in the aqueous state. C Correct. There is a decrease in disorder since the liquid reactant, with higher disorder, was consumed t o formed a solid product, with lower disorder. D Incorrect. There is an increase in disorder due to the production of highly disordered gaseous products, H2. Question 15(C) Rearranging G = H – TS gives G = ( –S)T + H which is of the form y = mx + c where m = gradient = –S c = y-intercept = H 1 Correct. Since the y -intercept is below the origin, H is a negative value i.e. the reaction is exothermic. 2 Incorrect. Since the gradient of the graph is positive, gradient = (–S) is positive i.e. S is negative. There is decrease in di