RI 2019 Y5 Promo Solutions

1 2019 Y5 H2 Chemistry Promotion Exam – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer D C A B B C C B A D D A A C B MCQ worked solutions 1 (Ans: D) A : Incorrect. (A mole of substance is the amount of that substance which contains 6.02  1023 elementary entities of that substance.) Since 1 mole of compound (e.g. CO2) contains more than 1 mole of atoms (e.g. 1 mole of C and 2 moles of O atoms), it will not contain the same number of atoms as there are atoms in 12 g of carbon-12 (i.e. 1 mole of 12C). B : Incorrect. Relative molecular mass = (weighted) average mass of 1 molecule of the substance 1 12  mass of 1 atom of carbon-12 C : Incorrect. Relative isotopic mass = mass of 1 atom of the isotope 1 12  mass of 1 atom of carbon-12 D : Correct. Relative atomic mass = (weighted) average mass of 1 atom of the element 1 12  mass of 1 atom of carbon-12 2 (Ans: C) 2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g) Mass of carbon dioxide and water vapour lost = 0.85 g Let the amount of NaHCO3 be x mol. Amount of CO2(g) = amount of H2O(g) = (x/2) mol (x/2)(44) + (x/2)(18) = 0.85  x = 0.02742 % purity by mass = [(0.02742 x 84.0) / 3.50] x 100 = 65.8% Note: The sample of NaHCO3 is impure. After decomposition, the mass of Na 2CO3 in the remaining residue is NOT = (3.50 – 0.85) g due to the impurities present. 3 (Ans: A) No of protons and neutrons in X = 289 + 3(1) – 244 = 48 No of protons in X = 114 + 3(0)  94 = 20 No of neutrons in X = 48  20 = 28

2 4 (Ans: B) CO32− ion exhibits resonance. ⇒ All C–O bonds present in the ion are of the same bond length. CO32− ion has 3 regions of electron density around the central C atom (sp2 hybridisation). Since there are 3 bond pairs of electrons present, the CO32− ion has a trigonal planar shape and a bond angle of 1200. SO32− ion has 4 regions of electron density around the central S atom. Since there are 3 bond pairs and 1 lone pair of electrons present, the SO 32− ion has a trigonal pyramidal shape and a bond angle of approximately 1070. Hence, only options 1 and 2 are correct. 5 (Ans: B) A: Correct There are 2 𝜋 bonds in the C≡C triple bond in ethyne (H–C≡C–H). There are 2 𝜋 bonds in the C≡N triple bond in butanenitrile (CH3CH2CH2C≡N). ⇒ Both molecules have the same number of 𝜋 bonds B: Incorrect. Since butanenitrile is polar (with a net overall dipole moment ), the molecules are held together by permanent dipole-permanent dipole (pd-pd) interactions. On the other hand, ethyne is non-polar (with no net dipole moment) and the molecules are held together by the weaker instantaneous dipole–induced dipole (id-id) interactions. In addition, since butan enitrile has a larger and more polarisable electron cloud than ethyne, its id-id interactions are stronger than that of ethyne. ⇒ Butanenitrile forms stronger intermolecular forces of at