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 answ