JPJC 2026 The Mole Concept & Stoichiometry Revision Answer

1 2026 JC1 H2 Chemistry (9476) Topical Revision for Physical Chemistry (Topic 2: The Mole Concept & Stoichiometry) Suggested Answers FOUNDATION KIT 1 [PJC MYE 2013/P1/Q1] Ans: B A Ar of Si =                  91.1 7.9 1.0 100 100 100×28 + ×29 + ×30 = 28.099 B Ar of Si =              92.2 4.7 3.1 100 100 100×28 + ×29 + ×30 = 28.109 C Ar of Si =                  95.0 3.2 1.8 100 100 100×28 + ×29 + ×30 = 28.068 D Ar of Si =                  96.3 0.3 3.4 100 100 100×28 + ×29 + ×30 = 28.071 2 [ACJC Prelim2020/P1/Q2] Ans: D A Density =                  88.8 10.2 1.0 100 100 100×2.0 + × 4.0 + ×39.9 = 2.583 B Density =                  80.0 19.0 1.0 100 100 100×2.0 + × 4.0 + ×16.0 = 2.520 C Density =                  92.3 7.2 0.5 100 100 100×2.0 + × 4.0 + ×16.0 = 2.214 D Density =                  82.5 15.2 2.3 100 100 100×2.0 + × 4.0 + ×16.0 = 2.626 3 [PJC MYE 2013/P1/Q3] Ans: D 1 ppm = 1 mg  50 ppm = 50 mg = 50  10–3 g Mass of CO in 1 kg of air = 50  10–3 g n(CO) present in 1 kg of air = − 350 10 28.0 mol no. of CO molecules present = − 350 10 28.0  L = − 350 10 28.0  6.02  1023

2 4 [IJC MYE 2011/P1/Q1] Ans: D n(O2) = 16 32.0 = 0.5 mol O2 : O 1 : 2 n(O) = 2  0.5 = 1.0 mol  No. of O atoms = 1.0 L where L: Avogadro’s constant A : n(NH3) = 17 17.0 = 1.0 mol NH3 : atoms 1 : 4 n(atoms) = 4  1.0 = 4.0 mol No. of atoms = 4.0 L B : n(He) = 2 4.0 = 0.5 mol No. of He atoms = 0.5 L C : n(CH4) = 8 16.0 = 0.5 mol CH4 : atoms 1 : 5 n(atoms) = 5  0.5 = 2.5 mol No. of atoms = 2.5 L D : n(N2) = 14 28.0 = 0.5 mol N2 : N 1 : 2 n(N) = 2  0.5 = 1.0 mol No. of N atoms = 1.0 L 5 [ACJC Prelim2020/P1/Q1] Ans: B Volume of 109 g of water = 109 cm3 = 106 dm3 Amount of 10 g of THF = 10 72.0 = 0.1389 mol [THF] = 6 0.1389 10 = 1.39  10–7 mol dm–3 6 Ans: C Let x be the oxidation no. of S in the compound. S8 8x = 0  x = 0 SO2 x + 2(–2) = 0 x = +4 SO2Cl2 x + 2(–2) + 2(–1) = 0 x = +6 Na2S2O3 2(+1) + 2x + 3(–2) = 0 x = +2

3 7 [PJC MYE 2012/P1/Q1] Ans: B A CH3CHO + 5/2O2 → 2CO2 + 2H2O  1 mol of organic cpd requires 2.5 mol of O2 B CH3CH2OH + 3O2 → 2CO2 + 3H2O  1 mol of organic cpd requires 3 mol of O2 C CH3CH3 + 7/2O2 → 2CO2 + 3H2O  1 mol of organic cpd requires 3.5 mol of O2 D CH3CO2H + 2O2 → 2CO2 + 2H2O 1 mol of organic cpd requires 2 mol of O2 8 [PJC MYE 2012/P1/Q3] Ans: B 1 mol of any gas occupies 24 dm3 at r.t.p. n(O2) at r.t.p = 1 24 000 mol no. of O2 molecules present = 1 24 000  6.02  1023 9 [PJC MYE 2011/P1/Q2] Ans: C CS2(g) + 3O2(g) → CO2(g) + 2SO2(g) Apply Avogadro’s Law: 20 cm3 60 cm3 20 cm3 40 cm3 reacted formed (For gas, mole ratio = volume ratio) All CS2 completely burnt in O2. Volume of O2 left unreacted = 100 – 60 = 40 cm3 Final volume of gas after combustion = 40 + 20 + 40 = 100 cm3 After treatment with excess aqueous alkali, acidic gaseous CO2 and SO2 are absorbed by alkali via acid-base reaction.  volume of gas dissolves in the alkali = 20 + 40 = 60 cm3  % of this final volume dissolves in the alkali = 60 100% 100 = 60 %

4 10 [JJC Prelim 2018/P1/Q5] Ans: D CxHy(g) + (x + y 4 ) O2(g) → x CO2(g) + y 2 H2O(l) at r.t.p. 10 cm3 10(x + y 4 ) cm3 10x cm3 - First contraction = 10 + 10(x + y 4 ) - 10x = 35 10 + 10x + 2.5y - 10x = 35 2.5y = 25  y = 10 Second contraction = Volume of CO2 = 40 cm3 10x = 40  x = 4 OR, First contraction = x y 2 2C H CO O reactedV + V - V 35 = 10 + 2O reactedV - 40 Volume of O2 reacted = 65 cm3 CxHy(g) + (x + y 4 ) O2(g) → x CO2(g) + y 2 H2O(l) at r.t.p. Volume/cm3 10 65 40 - mole ratio 1 : 6.5 : 4 1 : (x + y 4 ) : x  x = 4 and (4 + y 4 ) = 6.5  y = 10  the molecular formula of the hydrocarbon is C4H10.

5 11 [PJC Prelim 2014/P1/Q2] Ans: B (Na2CO3)x•y(H2O2) : Na2CO3 : H2O2 1 : x : y n((Na2CO3)x•y(H2O2)) = 5.0 0.200 1000 = 0.001 mol n(Na2CO3) in the solution = 0.001x mol Na2CO3 in the solution reacts with HCl to give CO2. n(CO2) produced at r.t.p. = 48 24 000 = 0.002 mol From the equation, Na2CO3 + 2HCl → 2NaCl + CO2 + H2O 0.001x mol 0.002 mol 0.001x = 0.002 x = 2 n(H2O2) in the solution = 0.001y mol H2O2 in the solution reacts with acidified MnO4–. To find the mole ratio between H2O2 & MnO4–, (+7) (+2) [R] MnO4– + 5e– → Mn2+ (x2) 2(-1) (0) [O] H2O2 → O2 + 2e– (x5) n(MnO4–) required = 12.0 0.100 1000 = 0.0012 mol 22 - 4 n(H O ) 5 = 2n(MnO )  0.001 5 = 0.0012 2y  y = 3  ratio of x : y = 2 : 3 12 [PJC Promo 2013/P1/Q2] Ans: B Concentration of HCN that is sufficient to kill a person within 1 hour = 300 mg in 1 m3 = 300 mg in 103 dm3 = − 3 3 300 10 10 g in 1 dm3 n(HCN) in 1 dm3 = − 33300 10 27.010 mol No. of HCN molecules in 1 dm3 = −  3 3 300 10 10 27.0  6.02  1023 MnO4– : H2O2 2 : 5 MnO4– : H2O2 0.0012 : 0.001y 2 : 5

6 13 [PJC Promo 2014/P1/Q2] Ans: D n(H2O) produced = 7.20 18.0 = 0.4 mol NH3 + HCl → NH4Cl n(NH3) liberated = n(HCl) reacted = 200 1.0 1000 = 0.2 mol NxOy(g) + 3 + 2 2 xy H2(g) → x NH3(g) + y H2O(l) 0.1 mol 0.2 mol 0.4 mol NxOy : NH3 : H2O 0.1 : 0.2 : 0.4 1 : 2 : 4 1 : x : y  x = 2 and y = 4  the formula of the oxide of nitrogen is N2O4. 14 [JJC Prelim 2012/P1/Q1] Ans: B Co(s) + 2Co3+(aq) → 3Co2+(aq) (mass and charge must be balanced) Initial/mol : x y 0 Final/mol : 0 z 3z moles of Co used = x moles of Co3+(aq) = y resultant mixture A 1 2 3 mol Co2+ only B 1 3 3 mol Co2+ + 1 mol Co3+(left) C 1 5 3 mol Co2+ + 3 mol Co3+(left) D 2 3 4.5 mol Co2+ + 0.5 mol Co (left) Illustration for B: Co(s) + 2Co3+(aq) → 3Co2+(aq) Initial/mol : 1 3 0 Ch