Year 11 ChemistryModule 2Module Quiz⏱ ~35 min40 marks
Module 2 Quiz
Introduction to Quantitative Chemistry, complete assessment covering all four inquiry questions across L01–L20. 15 MC questions (auto-marked) + 5 written questions (self-marked). Complete all questions before submitting.
IQ1
Chemical Reactions & Stoichiometry
IQ2
The Mole Concept & Formulas
IQ3
Concentration & Molarity
IQ4
Gas Laws (PV = nRT)
Progress
0 / 15 MC answered
Section A, Multiple Choice
15 questions · 1 mark each · 15 marks
Q1, L01 Mole Concept
How many molecules are in 0.500 mol of CO₂? (Nₐ = 6.022×10²³)
Q2, L02 Molar Mass
What mass of H₂O contains 2.00 mol? (H = 1.008, O = 15.999)
Q3, L03 Empirical Formula
A compound is 40.0% C, 6.72% H, and 53.3% O by mass. Its empirical formula is:
Q4, L04 Gas Volume
What volume does 3.00 mol of N₂ occupy at RTP (25 °C, 100 kPa)?
Q5, L06 Concentration
0.250 mol of NaOH is dissolved to make 500 mL of solution. What is the concentration?
Q6, L07 Dilution
50.0 mL of 2.00 mol/L HCl is diluted to 400 mL. What is the new concentration?
Q7, L09 Gravimetric Analysis
AgNO₃ + NaCl → AgCl↓ + NaNO₃. If 25.0 mL of AgNO₃ produces 0.717 g of AgCl, what is [AgNO₃]? (MM AgCl = 143.32)
Q8, L10 Titration
25.0 mL NaOH is titrated with 0.100 mol/L HCl. Titres: Rough = 24.5, T1 = 22.4, T2 = 22.3, T3 = 22.5 mL. What is [NaOH]?
Q9, L11 Mole Ratios
In 2Al + 3Cl₂ → 2AlCl₃, if 6.00 mol of Al reacts completely, how many moles of Cl₂ are consumed?
Q10, L12 Mass–Mass
In 2H₂ + O₂ → 2H₂O, what mass of water forms from 4.00 g of H₂? (H=1.008, O=15.999)
Q11, L13 Limiting Reagent
In Fe + S → FeS, 11.2 g of Fe and 8.00 g of S are mixed. Which is the limiting reagent? (Fe=55.845, S=32.06)
Q12, L14 Percentage Yield
Theoretical yield = 25.0 g, actual yield = 19.0 g. What is the percentage yield?
Q13, L15 Gas Stoichiometry
CaCO₃ → CaO + CO₂. What volume of CO₂ at STP (0 °C, 100 kPa) forms from 50.05 g of CaCO₃? (Ca=40.078, C=12.011, O=15.999; STP Vₘ = 22.71 L mol⁻¹)
Q14, L16 Solution Stoichiometry
In AgNO₃ + NaCl → AgCl + NaNO₃, 40.0 mL of 0.100 mol/L AgNO₃ is used. What mass of AgCl forms? (MM AgCl = 143.32)
Q15, L17 Back Calculation
25.0 mL H₂SO₄ is titrated with 0.200 mol/L NaOH. Titre = 20.0 mL. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. What is [H₂SO₄]?
Section B, Written Questions
5 questions · 5 marks each · 25 marks
Q16, L02 & L03: Molar Mass + Empirical Formula5 MARKS
A compound has the following percentage composition: 54.55% C, 9.16% H, 36.33% O. (a) Determine the empirical formula. (b) The molar mass of the compound is 88.0 g/mol. Determine the molecular formula. (c) Calculate the percentage by mass of carbon in the compound. (C = 12.011, H = 1.008, O = 15.999)
Model Answer:(a) n(C) = 54.55÷12.011 = 4.542; n(H) = 9.16÷1.008 = 9.088; n(O) = 36.33÷15.999 = 2.271Ratio ÷2.271: C = 2.00, H = 4.00, O = 1.00 → Empirical formula = C₂H₄O(b) MM(C₂H₄O) = 44.05; n = 88.0÷44.05 = 2.00 → Molecular formula = C₄H₈O₂ (e.g. butanoic acid or an ester such as ethyl ethanoate)(c) %C = (4×12.011)÷88.11 × 100 = 54.5%Marks: 1, n(C) and n(O) | 1, empirical formula | 1, n multiplier | 1, molecular formula | 1, %C
Q17, L07 & L08: Dilution + Purity5 MARKS
A student has 100.0 mL of concentrated HCl at 12.0 mol/L. (a) She dilutes 20.0 mL of this stock solution to 500.0 mL. Calculate the concentration of the diluted solution. (b) The concentrated HCl is 37.0% HCl by mass (density = 1.18 g/mL). What mass of pure HCl is in 20.0 mL of the concentrated solution? (MM HCl = 36.46)
Model Answer:(a) c₁V₁ = c₂V₂; 12.0 × 20.0 = c₂ × 500.0c₂ = (12.0 × 20.0) ÷ 500.0 = 0.480 mol/L(b) m(solution) = 20.0 mL × 1.18 g/mL = 23.6 gm(pure HCl) = 23.6 × 0.370 = 8.73 gn(HCl) = 8.73 ÷ 36.46 = 0.239 mol — consistent with 12.0 mol/L × 0.0200 L = 0.240 mol, confirming the data reconcileMarks: 1, formula applied correctly | 1, c₂ = 0.480 | 1, m(solution) = 23.6 g | 1, purity applied | 1, m(pure HCl) = 8.73 g
Q18, L13 & L14: Limiting Reagent + Yield5 MARKS
In 2Al + 3Cl₂ → 2AlCl₃, 8.10 g of Al reacts with 14.2 g of Cl₂. (a) Identify the limiting reagent, showing the full comparison. (b) Calculate the theoretical yield of AlCl₃. (c) If 13.4 g of AlCl₃ is collected, calculate the percentage yield. (Al = 26.982, Cl = 35.453)
Model Answer:(a) n(Al) = 8.10÷26.982 = 0.3002; ÷2 = 0.1501n(Cl₂) = 14.2÷70.906 = 0.2003; ÷3 = 0.06677
Cl₂ has smaller quotient (0.0668 < 0.1501) → Cl₂ is the limiting reagent(b) Cl₂:AlCl₃ = 3:2; n(AlCl₃) = 0.2003 × (2÷3) = 0.1335 molMM(AlCl₃) = 133.34; m = 0.1335 × 133.34 = 17.8 g (theoretical)(c) % yield = (actual ÷ theoretical) × 100 = (13.4 ÷ 17.8) × 100 = 75.3% The actual yield (13.4 g) is below the theoretical maximum (17.8 g), as expected. If a measured value ever exceeded 100%, the product would be wet or impure, or actual and theoretical swapped in the formula.Marks: 1, both moles and coefficients | 1, LR identified with comparison | 1, n(AlCl₃) from ratio | 1, theoretical yield | 1, % yield formula applied
Q19, L16: Solution Stoichiometry5 MARKS
Barium chloride reacts with sodium sulfate in solution: BaCl₂ + Na₂SO₄ → BaSO₄↓ + 2NaCl. 40.0 mL of 0.250 mol/L BaCl₂ is mixed with 60.0 mL of 0.150 mol/L Na₂SO₄. (a) Identify the limiting reagent, showing your full comparison. (b) Calculate the mass of BaSO₄ precipitate. (c) Calculate the concentration of the excess reagent remaining in the 100.0 mL solution. (Ba = 137.33, S = 32.06, O = 15.999)
Model Answer:(a) n(BaCl₂) = 0.250 × 0.0400 = 0.01000 mol; ÷1 = 0.01000n(Na₂SO₄) = 0.150 × 0.0600 = 0.009000 mol; ÷1 = 0.009000
Na₂SO₄ has smaller quotient → Na₂SO₄ is the limiting reagent; BaCl₂ is in excess(b) Na₂SO₄:BaSO₄ = 1:1; n(BaSO₄) = 0.009000 molMM(BaSO₄) = 233.39; m = 0.009000 × 233.39 = 2.10 g(c) n(BaCl₂ consumed) = n(Na₂SO₄) = 0.009000 moln(BaCl₂ excess) = 0.01000 − 0.009000 = 0.001000 molc(BaCl₂ excess) = 0.001000 ÷ 0.100 = 0.0100 mol/LMarks: 1, both n values and LR identified | 1, n(BaSO₄) | 1, m(BaSO₄) | 1, n(BaCl₂ excess) | 1, c(excess) using total volume
Q20, L17: Primary Standard + Titration Back-Calculation5 MARKS
A student prepares a standard solution by dissolving 0.795 g of anhydrous Na₂CO₃ (MM = 105.99) in water and making up to 150.0 mL. She titrates 25.0 mL aliquots against HCl solution. Titres: Rough = 18.9 mL, T1 = 20.78 mL, T2 = 20.82 mL, T3 = 20.80 mL. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (a) Calculate the average concordant titre. (b) Calculate [HCl].
Model Answer:
(a) Discard rough (18.9). T1=20.78, T2=20.82, T3=20.80, range = 0.04 mL ≤ 0.10 mL; all concordant. Average = (20.78+20.82+20.80)÷3 = 20.80 mL