2024 Prelims RVHS H2 Chem P1 (Ans)

2024 JC 2 H2 Chemistry Prelim Exam Paper 1 Worked Solutions 1 D The decomposition involves: n → p+ + e− where the newly -formed proton stays in the nucleus. As the proton number of the atom increases by 1 following the decomposition, the element will no longer be the same. Options A and C are eliminated. Option D involves an increase in proton number by 1 going from K to Ca, hence it is the correct answer. 2 C Option A: By definition, one mole of a substance contains exactly 6.02  1023 (or Avogadro number) elementary entities. Option A is incorrect because of the phrase “same number of atoms as there are in 12.000 g of carbon-12”. Option B: By definition, relative isotopic mass (Ar) is the mass of one mole of atoms of an isotope (of a certain element) relative to 1 12 the mass of one mole of 12C atoms. Option B is incorrect because in the formula given, the numerator used is “average mass of all isotopes of lithium”. Option C: By definition, relative atomic mass ( Ar) is the average mass of one mole of atoms of an element relative to 1 12 the mass of one mole of 12C atoms. Note that some textbooks define relative atomic mass as “the average mass of one atom of an element relative to 1 12 the mass of one 12C atom”. Hence, option C is correct. Option D: By definition, relative molecular ( Mr) mass is the average mass of one mole of molecules relative to 1 12 the mass of one mole of 12C atoms. Option D is incorrect because in the formula given, the numerator used is “average mass of one atom of E”.

3 C Amount of CO2 = 48.0 24 000 = 0.00200 mol Amount of sodium percarbonate = 10.00.100 1000 = 0.00100 mol (Na2CO3)xy(H2O2) : CO2 x : 1 0.00200 : 0.00100 2 : 1 Therefore, x = 2 Amount of H2O2 = 0.00100y mol Amount of KMnO4 = 24.00.0500 1000 = 0.00120 mol Given the ratio is 2 : 5 0.00120 2 0.00100y 5= y = 3 Therefore, y3 x2= 4 D A: Due to the overlap of unhybridised 2p orbitals, mobile delocalised electrons are found in the lattice structure. B: Each carbon atom forms 3 sigma bonds with 3 other carbon atoms. C: Instantaneous dipole -induced dipole interactions exist between each graphite plane. D: Conduction of electricity occurs due to the overlap of unhybridised 2p orbitals. The overlapping occurs perpendicular to the axis of the unhybridised 2p orbitals.

5 A A: There is an unpaired electron on N atom in NO2 molecule. B: The Cu+ ion has an electronic configuration of 1s2 2s2 2p6 3s2 3p6 3d10 4s0. C: The Li+ ion has an electronic configuration of 1s2 2s0. D: After heterolytic fission, Cl+ and Cl ions are formed, which do not have a single unpaired electron. 6 D Using Boyle’s Law, p1V1 = p2V2. Upon connecting, the total volume of the container becomes 3 dm3. Helium: Neon: (2)(1) = p2(3) (1)(2) = p2(3) p2 = 2 3 kPa p2 = 2 3 kPa Final pressure = 2 3 + 2 3 = 4 3 kPa 7 B Option B is correct: Using Hess’ Law, −1561 = − (−85) + 2(−394) + 3