2024 RI Prelim H2 Chem Paper 1 Suggested solutions

© Raffles Institution 2024 9729/01/S/24 2024 Y6 H2 Chemistry Preliminary Exams Paper 1 – Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer A B B C C D B D C A B A D A A MCQ worked solutions Q1 (Ans: A) At r.t.p, number of moles of molecules in 1 cm3 of CH4 gas = 1 24000 mol number of molecules in 1 cm3 of CH4 gas = 6.02 × 1023 24000 Q2 (Ans: B) CxHy(g) + (x+ y 4 )O2 (g) → xCO2(g) + y 2H2O(l) Volume of CO2 produced = 100 − 40 = 60 cm3 Volume of unreacted O2 = 40 cm3 Volume of O2 reacted = 130 – 40 = 90 cm3 Hence, x = 60 20 = 3. (x+ y 4 ) = (3+ y 4 ) = 90 20 y = 6 Thus, formula of the hydrocarbon is C3H6. Q3 (Ans: B) Option A is incorrect as 18O2− and 19F− each has 10 electrons and 10 neutrons. Option B is correct as both 18O2− and 19F− have 10 neutrons each. Option C is incorrect as both ions have outer electronic configuration of 2s2 2p6. Option D is incorrect as nucleons refer to protons and neutrons. Hence 18O2− has 18 nucleons and 19F− has 19 nucleons. Q4 (Ans: C) After the beta decay, the mass number remains the same at 234 and the resulting species has 90 + 1 = 91 protons, which is protactinium (Pa). Q5 (Ans: C) Option A is incorrect as b oiling involves overcoming intermolecular forces (hydrogen bonding) and does not involve the breaking of covalent bonds. Furthermore, the F−H bond is stronger than the O−H bond. Option B is incorrect as both H2O and HF are isoelectronic (same number of electrons). Option C is correct as, on average, each H2O can form 2 hydrogen bonds per molecule whereas each HF only forms 1 hydrogen bond per molecule. Option D is incorrect as the statement is not a major reason for the higher boiling point of H 2O since hydrogen bonding is the predominant intermolecular force of attraction. Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer D B C A D B D B C C D D B C A

© Raffles Institution 2024 9729/01/S/24 Q6 (Ans: D) Option A is incorrect as ice is less dense than water due to its open structure, preventing the molecules from getting too close to one another . This also causes the same mass of H 2O to occupy a larger volume. Option B is incorrect as covalent O−H bonds are stronger than hydrogen bonds. Option C is incorrect as each O atom forms 2 hydrogen bonds, hence 4 electrons (2 electron pairs) are involved. Option D is correct as e ach H2O molecule is tetrahedrally bonded to four H atoms via covalent and hydrogen bonds. Q7 (Ans: B) C C C Cl Cl H H C C C Cl H H Cl C C C H Cl H Cl C C CH2Cl Cl C C CHCl2 H (1) (2) (3) (4) (5) Option 1 is correct to adhere to the given molecular formula and non-cyclic structure. Option 2 is incorrect as isomers 2 and 3 are optically active as they lack an internal plane of symmetry. However, both do not contain any chiral C atom. Option 3 is correct as there is a total of five non-cyclic isomers, including stereoisomers. Q8 (Ans: D) pV = nRT Option A is incorrect as p ∝ T, in K , at constant V . Since units of T on x -axis is °C, when p = 0, T = −273 °C. The graph should not pass through the origin. Option B is incorrect as pV = nRT = constant at constant T. Hence, the graph should be a vertical line. Option C is incorrect as V ∝ 1/p at constant T. H ence, the graph should resemble a y = 1 x hyperbolic graph, with asymptotes along x and y axes. Option D is correct as p/T = nR/V = constant at constant V. Hence, the graph should be a horizontal line. Q9 (Ans: C) Atomic radius decreases across a period from sodium to chlorine. Electronegativity increases across the period from sodium to chlorine. The first ionisation energy of elements generally increase across a period from sodium to chlorine. Q10 (Ans: A) Only MgO dissolves sparingly in water to form Mg(OH)2. SiO2 and Al2O3 are insoluble while Na2O and SO2 dissolves completely in water. Hence, options A and B are possible answers. MgCl2 solution has a pH of 6.5, which is higher than that of A l2Cl3 solution (pH of 3). Hence, option A is correct. NaCl has a pH of 7, and hence option B is incorrect.

© Raffles Institution 2024 9729/01/S/24 Q11 (Ans: B) Option A is incorrect as the boiling point of X2 increases down the group due to stronger instantaneous dipole-induced dipole interactions between the molecules . Hence, the volatility decreases down the group. Option B is correct as the bond length of X 2 increases down the group due to increase in the size of the atom down the group. Option C is incorrect as the bond energy of X2 decreases down the group as the electron cloud of the atoms get more diffused down the group and the effectiveness of orbital overlap decreases down the group. Option D is incorrect as the oxidizing power of X2 decreases down the group since the tendency of X2 to gain electrons decreases down the group. Q12 (Ans: A) ∆H = −2×(+91) + 2×(+34) + (−58) = −172 kJ mol–1 Q13 (Ans: D) Reaction Sign of ∆H ∆H1 enthalpy change of reaction between sodium and hydrochloric acid 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) ∆Hr < 0 OH−(aq) + H+(aq) → H2O(l) ∆Hr < 0 negative ∆H2 enthalpy change of combustion of sodium negative ∆H3 enthalpy change of reaction between sodium oxide and hydrochloric acid Cannot be determined from the question ∆H4 enthalpy change of combustion of hydrogen negative Option A is incorrect as ∆H1 + ∆H2 is always negative. Option B is incorrect as ∆H3 + ∆H4 is not always positive since ∆H4 is negative. Option C is incorrect, by Hess’ Law, ∆H3 – ∆H4 – ∆H1 = –∆H2, and –∆H2 is positive. Option D is correct, by Hess’ Law, ∆H2 + ∆H3 – ∆H1 = ∆H4 and ∆H4 is negative. Q14 (Ans: A) Option A is correct. When [CH 3CHO] = c, the active sites of the enzymes become saturated with CH3CHO. Further increase in [CH3CHO] will not have any effect on the reaction rate. Option B is incorrect. When [CH3CHO] = c, the order of reaction with respect to [CH3CHO] is zero. Option C is incorrect. When [CH 3CHO] = c, the reaction rate remains constant as long as the active sites of the enzymes remain saturated with CH3CHO. Option D is incorrect. When [CH3CHO] = c, the reaction rate remains constant but non-zero. 2NO(g) + O2(g) N2O4(g) N2(g) + 2O2(g) 2×(+91) ∆H 2NO2(g) −58 2×(+34)

© Raffles Institution 2024 9729/01/S/24 Q15 (Ans: A) For X(g) ⇌ Y(g), Kc = [Y] [X] and Kp = pY pX Statement 1 is correct. Given that X and Y behave as ideal gases, pXV = nXRT and hence pX = [X]RT. Thus, Kp = pY pX = [Y]RT [X]RT = [Y] [X] = Kc. Statement 2 is correct. Under the same conditions, Kp is constant regardless if the equilibrium was achieved from 100% of X(g) or 100% of Y(g). Thus, pY pX = Kp remains constant. Statement 3 is correct. Adding more X(g) increases the concentration of X(g) and the position of equilibrium of X(g) ⇌ Y(g) shifts to the right to partially offset this increase. However, at the same temperature, Kp is constant and hence the partial pressures of both gases must increase so that Kp = pY pX = constant. Q16 (Ans: D) NH3(aq) + HNO3(aq) → NH4+(aq) + NO3−(aq) When HNO3 was titrated against NH3, NH3 is added from the burette to HNO3 in the conical flask. vol. of NH3 added / cm3 remarks 10 The conical flask contains unreacted HNO3 and the product, NH4+, which does not act as a buffer. 20 All the HNO3 is completely neutralised and the conical flask contains the product, NH4+, which does not act as a buffer. 30 20 cm3 of NH3 is used for neutralisation and the remaining 10 cm 3 is in excess. The conical flask contains excess NH3 and the product, NH4+, in the ratio 1 : 2 and hence can act as a buffer but does not have maximum buffering capacity 40 20 cm3 of NH3 is used for neutralisation and the remaining 20 cm3 is in excess. The conical flask contains excess NH3 and the product, NH4+, in the ratio 1 : 1 and hence c