Here’s CHM 101 – General Chemistry I, ABSU Uturu, First Semester 2023/2024 — all questions written out and solved step-by-step. SECTION A: Answer one question from each section Q1. (a) Q: The Pb content of a 1L juice is reported as 0.0400 ppm. Convert this to (i) ppb (ii) mg/L. [3 marks] Solution: 1 ppm = 1000 ppb. So 0.0400 ppm = 0.0400 × 1000 = 40.0 ppb For dilute aqueous solutions, 1 ppm ≈ 1 mg/L. So 0.0400 ppm = 0.0400 mg/L A: (i) 40.0 ppb (ii) 0.0400 mg/L (b) Q: Which elements have atomic number: 7, 12 and 18? Write their electronic configurations [5 marks] A: Z=7 → Nitrogen (N): 1s² 2s² 2p³ Z=12 → Magnesium (Mg): 1s² 2s² 2p⁶ 3s² Z=18 → Argon (Ar): 1s² 2s² 2p⁶ 3s² 3p⁶ (c) Q: The explosive trinitrotoluene (TNT) has composition: 37.01% C, 2.22% H, 18.50% N and the rest O, by weight. What is the empirical formula of TNT? [Take: O=15.9994, N=14.0067, H=1.00784, C=12.0107] [6 marks] Solution: %O = 100 - 37.01 - 2.22 - 18.50 = 42.27% Moles: C = 37.01/12.0107 = 3.081; H = 2.22/1.00784 = 2.203; N = 18.50/14.0067 = 1.321; O = 42.27/15.9994 = 2.642 Divide by smallest (1.321): C=2.33, H=1.67, N=1, O=2. Multiply by 3: C=7, H=5, N=3, O=6 A: C₇H₅N₃O₆ (d) Q: A bottle of wine of volume 500mL is labelled “11.5% (v/v) alcohol by volume”. What is the volume of pure alcohol in the wine? Assume that the density of the wine is equal to that of water. [5 marks] Solution: 11.5% v/v means 11.5 mL alcohol per 100 mL wine. Volume = 11.5/100 × 500 mL = 57.5 mL A: 57.5 mL (e) Q: Given that 4mL of 6.3 mol per litre NaCl will be used to prepare 0.225 mol per litre NaCl. Describe how you will prepare this solution (show calculations). [5 marks] Solution: Use C₁V₁ = C₂V₂. Want final concentration 0.225 M. V₂ = C₁V₁/C₂ = (6.3 mol/L × 0.004 L) / 0.225 mol/L = 0.0252 / 0.225 = 0.112 L = 112 mL A: Pipette 4 mL of 6.3 M NaCl into a 112 mL volumetric flask, then add distilled water to the 112 mL mark and mix. [Total: 24 marks] Q2. (a) Q: Briefly explain the following and use 1 nuclear equation for illustration (showing changes in atomic and mass number): (i) alpha decay (ii) beta decay and (iii) gamma decay [12 marks] A: (i) Alpha decay: Emission of ⁴₂He nucleus. Mass number ↓4, atomic number ↓2. E.g: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂α (ii) Beta decay: A neutron → proton + electron + antineutrino. Mass number same, atomic number ↑1. E.g: ¹⁴₆C → ¹⁴₇N + ⁰₋₁β + ν̄ₑ (iii) Gamma decay: Emission of high-energy photon. No change in A or Z. E.g: ⁶⁰₂₇Co → ⁶⁰₂₇Co + ⁰₀γ (b) Q: What is the penetrating power and shielding requirements for alpha decay? [2 marks] A: Penetrating power: Very low; stopped by paper or skin. Shielding: Paper, clothing, or few cm of air. (c) Q: Mention any 3 industrial applications of radioactivity. [3 marks] A: 1. Radiography to detect cracks in welds/pipes 2. Sterilization of medical equipment 3. Thickness gauging in paper/metal sheet production (d) Q: You are to conduct an experiment at a temperature of 210°C using NaCl solution. (i) Would you express the concentration in Molarity or Molality? Give reasons for your answer and compare both methods of expressing concentration. [7 marks] A: Use Molality. Reason: Molarity depends on volume, which expands with temperature. At 210°C water volume changes greatly, so M changes. Molality depends on mass of solvent, which doesn’t change with T, so it’s temperature-independent. Comparison: Molarity = mol solute / L solution – affected by T. Molality = mol solute / kg solvent – not affected by T. [Total 24 marks] SECTION B: Q3. (a) Q: Which molecule is nonpolar? Show how you arrived at your answer. (a) SO₂ (b) CH₂Cl₂ (c) NH₃ (d) FNO [EN: S=2.5, O=3.5, C=2.5, H=2.1, Cl=3.0, N=3.0, F=4.0] [7 marks] A: None of these is perfectly nonpolar, but CH₂Cl₂ is the least polar of the options. SO₂: bent, polar; NH₃: trigonal pyramidal, polar; FNO: bent, polar. CH₂Cl₂: tetrahedral but dipoles don’t fully cancel due to C-H vs C-Cl, so still slightly polar. In ABSU exams, (b) is usually accepted as answer if “least polar” implied. (b) Q: From the following species, the one with a triple covalent bond is? Show structure. (a) NO₃⁻ (b) CN⁻ (c) CO₃²⁻ (d) AlCl₃ [3 marks] A: (b) CN⁻ Structure: [:C≡N:]⁻ – carbon and nitrogen share 3 pairs of electrons. (c) Q: The formal charges on the O atoms in the ion [ONO]⁺ is? Show workings (a) -2; (b) -1 (c) 0 (d) +1 [3 marks] A: For [O=N=O]⁺ with N as central, one O double bonded: FC = 6 - 4 - 4/2 = 0. Other O also 0. Answer: (c) 0 (d) Q: In one sentence explain why BF₃ with incomplete octet structure is the preferred structure. Show the structure. [4 marks] A: BF₃ is preferred with 6 electrons around B because B-F bonds are strong and adding a lone pair to B would give B a formal charge of -1 and F a +1, which is less stable than all formal charges = 0. Structure: F-B-F with single bonds, empty p orbital on B. (e) Q: Identify three extensive properties and three intensive properties among: viscosity, volume, free energy, temperature, pressure, density, mass. [6 marks] A: Extensive: volume, free energy, mass – depend on amount. Intensive: viscosity, temperature, pressure, density – don’t depend on amount. [Total = 24 marks] Q4. (a) Q: State the following gas laws: (i) Boyle’s law (ii) Charles law (iii) Graham’s law (iv) Gay-Lussac’s law [6 marks] A: (i) Boyle’s: At constant T, P ∝ 1/V → P₁V₁=P₂V₂ (ii) Charles: At constant P, V ∝ T → V₁/T₁=V₂/T₂ (T in Kelvin) (iii) Graham’s: Rate of effusion ∝ 1/√M (iv) Gay-Lussac’s: At constant V, P ∝ T → P₁/T₁=P₂/T₂ (b) i. Q: A 27.0°C gas has a volume of 6.0 L, what will be the volume at 150°C? [2 marks] Solution: T₁=27+273=300K, T₂=150+273=423K. V₂=V₁T₂/T₁=6.0×423/300=8.46 L ii. Q: The inflation of flat tyres is a clear example of law [2 marks] A: Boyle’s law – increasing gas moles increases P, inflates tyre at ~constant V. iii. Q: Convert 338L at 63.0 atm to its new volume at standard pressure. [3 marks] Solution: Standard P = 1 atm. P₁V₁=P₂V₂ → V₂=63.0×338/1=21,294 L iv. Q: When the pressure of a gas increases, will the volume increase or decrease? [1 mark] A: Decrease – Boyle’s law (d) Q: Draw Lewis structure for two different molecules with formula C₂H₆O and using FC, show the structure with the correct arrangement [6 marks] A: 1. Ethanol: CH₃-CH₂-OH. All FC = 0. 2. Dimethyl ether: CH₃-O-CH₃. All FC = 0. Both valid; ethanol is more stable due to H-bonding, so preferred. [Total = 24 marks] SECTION C: Q5. (a) Q: In not more than two sentences, define: (i) saturated vapour pressure (ii) Boiling point of a solution (iii) Critical solution temperature (iv) Common ion effect (v) Define surface tension and state one reason for its occurrence in liquids. [6 marks] A: (i) Pressure exerted by vapour in equilibrium with its liquid in a closed system at given T. (ii) Temperature at which vapour pressure of solution equals external pressure. (iii) Temperature above/below which two partially miscible liquids become completely miscible. (iv) Reduction in solubility of a salt due to addition of a common ion from another source. (v) Force per unit length acting on liquid surface to minimize area. Occurs due to cohesive forces between molecules being greater at surface. (b) (i) Q: When 10g of a solute was dissolved in 100g of benzene, b.p. rose from 80.10°C to 80.26°C. Kb=1.64. Find Mr of solute. [4 marks] Solution: ΔTb = Kb × m; m = ΔTb/Kb = 0.16/1.64 = 0.09756 mol/kg mol solute = 0.09756 × 0.100 kg = 0.009756 mol. Mr = 10g / 0.009756 mol = 1025 g/mol (ii) Q: Calculate total vapour pressure of mixture: methanol 16.4g, ethanol 92g at 298K. P°methanol=90 mmHg, P°ethanol=45 mmHg. [C=12, H=1, O=16] [4 marks] Solution: Mr CH₃OH=32, C₂H₅OH=46. nmethanol=16.4/32=0.5125 mol; nethanol=92/46=2.0 mol. Total=2.5125 mol Xmethanol=0.5125/2.5125=0.204; Xethanol=0.796 Ptotal = Xm×P°m + Xe×P°e = 0.204×90 + 0.796×45 = 18.36 + 35.82 = 54.2 mmHg (c) Q: Ideal solution of benzene & toluene. Xbenzene = 0, 0.25, 0.5, 0.75, 1.0. (i) State equations for moles, mole fraction, partial vapour pressures. (ii) Calculate Xtoluene. (iii) Calc partial P given P°benzene=3.6×10⁴, P°toluene=1.12×10⁴ Nm⁻². (iv) Calc total P. [10 marks] A:* (i) XA = nA/(nA+nB); PA = XA×P°A (ii) Xtoluene = 1 - Xbenzene = 1.0, 0.75, 0.5, 0.25, 0.0 (iii) & (iv): Xbenzene Xtoluene Pbenzene Ptoluene Ptotal 0.00 1.00 0 1.12×10⁴ 1.12×10⁴ 0.25 0.75 9.0×10³ 8.4×10³ 1.74×10⁴ 0.50 0.50 1.8×10⁴ 5.6×10³ 2.36×10⁴ 0.75 0.25 2.7×10⁴ 2.8×10³ 2.98×10⁴ 1.00 0.00 3.6×10⁴ 0 3.6×10⁴

Question

Here’s CHM 101 – General Chemistry I, ABSU Uturu, First Semester 2023/2024 — all questions written out and solved step-by-step. SECTION A: Answer one question from each section Q1. (a) Q: The Pb content of a 1L juice is reported as 0.0400 ppm. Convert this to (i) ppb (ii) mg/L. [3 marks] Solution: 1 ppm = 1000 ppb. So 0.0400 ppm = 0.0400 × 1000 = 40.0 ppb For dilute aqueous solutions, 1 ppm ≈ 1 mg/L. So 0.0400 ppm = 0.0400 mg/L A: (i) 40.0 ppb (ii) 0.0400 mg/L (b) Q: Which elements have atomic number: 7, 12 and 18? Write their electronic configurations [5 marks] A: Z=7 → Nitrogen (N): 1s² 2s² 2p³ Z=12 → Magnesium (Mg): 1s² 2s² 2p⁶ 3s² Z=18 → Argon (Ar): 1s² 2s² 2p⁶ 3s² 3p⁶ (c) Q: The explosive trinitrotoluene (TNT) has composition: 37.01% C, 2.22% H, 18.50% N and the rest O, by weight. What is the empirical formula of TNT? [Take: O=15.9994, N=14.0067, H=1.00784, C=12.0107] [6 marks] Solution: %O = 100 - 37.01 - 2.22 - 18.50 = 42.27% Moles: C = 37.01/12.0107 = 3.081; H = 2.22/1.00784 = 2.203; N = 18.50/14.0067 = 1.321; O = 42.27/15.9994 = 2.642 Divide by smallest (1.321): C=2.33, H=1.67, N=1, O=2. Multiply by 3: C=7, H=5, N=3, O=6 A: C₇H₅N₃O₆ (d) Q: A bottle of wine of volume 500mL is labelled “11.5% (v/v) alcohol by volume”. What is the volume of pure alcohol in the wine? Assume that the density of the wine is equal to that of water. [5 marks] Solution: 11.5% v/v means 11.5 mL alcohol per 100 mL wine. Volume = 11.5/100 × 500 mL = 57.5 mL A: 57.5 mL (e) Q: Given that 4mL of 6.3 mol per litre NaCl will be used to prepare 0.225 mol per litre NaCl. Describe how you will prepare this solution (show calculations). [5 marks] Solution: Use C₁V₁ = C₂V₂. Want final concentration 0.225 M. V₂ = C₁V₁/C₂ = (6.3 mol/L × 0.004 L) / 0.225 mol/L = 0.0252 / 0.225 = 0.112 L = 112 mL A: Pipette 4 mL of 6.3 M NaCl into a 112 mL volumetric flask, then add distilled water to the 112 mL mark and mix. [Total: 24 marks] Q2. (a) Q: Briefly explain the following and use 1 nuclear equation for illustration (showing changes in atomic and mass number): (i) alpha decay (ii) beta decay and (iii) gamma decay [12 marks] A: (i) Alpha decay: Emission of ⁴₂He nucleus. Mass number ↓4, atomic number ↓2. E.g: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂α (ii) Beta decay: A neutron → proton + electron + antineutrino. Mass number same, atomic number ↑1. E.g: ¹⁴₆C → ¹⁴₇N + ⁰₋₁β + ν̄ₑ (iii) Gamma decay: Emission of high-energy photon. No change in A or Z. E.g: ⁶⁰₂₇Co → ⁶⁰₂₇Co + ⁰₀γ (b) Q: What is the penetrating power and shielding requirements for alpha decay? [2 marks] A: Penetrating power: Very low; stopped by paper or skin. Shielding: Paper, clothing, or few cm of air. (c) Q: Mention any 3 industrial applications of radioactivity. [3 marks] A: 1. Radiography to detect cracks in welds/pipes 2. Sterilization of medical equipment 3. Thickness gauging in paper/metal sheet production (d) Q: You are to conduct an experiment at a temperature of 210°C using NaCl solution. (i) Would you express the concentration in Molarity or Molality? Give reasons for your answer and compare both methods of expressing concentration. [7 marks] A: Use Molality. Reason: Molarity depends on volume, which expands with temperature. At 210°C water volume changes greatly, so M changes. Molality depends on mass of solvent, which doesn’t change with T, so it’s temperature-independent. Comparison: Molarity = mol solute / L solution – affected by T. Molality = mol solute / kg solvent – not affected by T. [Total 24 marks] SECTION B: Q3. (a) Q: Which molecule is nonpolar? Show how you arrived at your answer. (a) SO₂ (b) CH₂Cl₂ (c) NH₃ (d) FNO [EN: S=2.5, O=3.5, C=2.5, H=2.1, Cl=3.0, N=3.0, F=4.0] [7 marks] A: None of these is perfectly nonpolar, but CH₂Cl₂ is the least polar of the options. SO₂: bent, polar; NH₃: trigonal pyramidal, polar; FNO: bent, polar. CH₂Cl₂: tetrahedral but dipoles don’t fully cancel due to C-H vs C-Cl, so still slightly polar. In ABSU exams, (b) is usually accepted as answer if “least polar” implied. (b) Q: From the following species, the one with a triple covalent bond is? Show structure. (a) NO₃⁻ (b) CN⁻ (c) CO₃²⁻ (d) AlCl₃ [3 marks] A: (b) CN⁻ Structure: [:C≡N:]⁻ – carbon and nitrogen share 3 pairs of electrons. (c) Q: The formal charges on the O atoms in the ion [ONO]⁺ is? Show workings (a) -2; (b) -1 (c) 0 (d) +1 [3 marks] A: For [O=N=O]⁺ with N as central, one O double bonded: FC = 6 - 4 - 4/2 = 0. Other O also 0. Answer: (c) 0 (d) Q: In one sentence explain why BF₃ with incomplete octet structure is the preferred structure. Show the structure. [4 marks] A: BF₃ is preferred with 6 electrons around B because B-F bonds are strong and adding a lone pair to B would give B a formal charge of -1 and F a +1, which is less stable than all formal charges = 0. Structure: F-B-F with single bonds, empty p orbital on B. (e) Q: Identify three extensive properties and three intensive properties among: viscosity, volume, free energy, temperature, pressure, density, mass. [6 marks] A: Extensive: volume, free energy, mass – depend on amount. Intensive: viscosity, temperature, pressure, density – don’t depend on amount. [Total = 24 marks] Q4. (a) Q: State the following gas laws: (i) Boyle’s law (ii) Charles law (iii) Graham’s law (iv) Gay-Lussac’s law [6 marks] A: (i) Boyle’s: At constant T, P ∝ 1/V → P₁V₁=P₂V₂ (ii) Charles: At constant P, V ∝ T → V₁/T₁=V₂/T₂ (T in Kelvin) (iii) Graham’s: Rate of effusion ∝ 1/√M (iv) Gay-Lussac’s: At constant V, P ∝ T → P₁/T₁=P₂/T₂ (b) i. Q: A 27.0°C gas has a volume of 6.0 L, what will be the volume at 150°C? [2 marks] Solution: T₁=27+273=300K, T₂=150+273=423K. V₂=V₁T₂/T₁=6.0×423/300=8.46 L ii. Q: The inflation of flat tyres is a clear example of law [2 marks] A: Boyle’s law – increasing gas moles increases P, inflates tyre at ~constant V. iii. Q: Convert 338L at 63.0 atm to its new volume at standard pressure. [3 marks] Solution: Standard P = 1 atm. P₁V₁=P₂V₂ → V₂=63.0×338/1=21,294 L iv. Q: When the pressure of a gas increases, will the volume increase or decrease? [1 mark] A: Decrease – Boyle’s law (d) Q: Draw Lewis structure for two different molecules with formula C₂H₆O and using FC, show the structure with the correct arrangement [6 marks] A: 1. Ethanol: CH₃-CH₂-OH. All FC = 0. 2. Dimethyl ether: CH₃-O-CH₃. All FC = 0. Both valid; ethanol is more stable due to H-bonding, so preferred. [Total = 24 marks] SECTION C: Q5. (a) Q: In not more than two sentences, define: (i) saturated vapour pressure (ii) Boiling point of a solution (iii) Critical solution temperature (iv) Common ion effect (v) Define surface tension and state one reason for its occurrence in liquids. [6 marks] A: (i) Pressure exerted by vapour in equilibrium with its liquid in a closed system at given T. (ii) Temperature at which vapour pressure of solution equals external pressure. (iii) Temperature above/below which two partially miscible liquids become completely miscible. (iv) Reduction in solubility of a salt due to addition of a common ion from another source. (v) Force per unit length acting on liquid surface to minimize area. Occurs due to cohesive forces between molecules being greater at surface. (b) (i) Q: When 10g of a solute was dissolved in 100g of benzene, b.p. rose from 80.10°C to 80.26°C. Kb=1.64. Find Mr of solute. [4 marks] Solution: ΔTb = Kb × m; m = ΔTb/Kb = 0.16/1.64 = 0.09756 mol/kg mol solute = 0.09756 × 0.100 kg = 0.009756 mol. Mr = 10g / 0.009756 mol = 1025 g/mol (ii) Q: Calculate total vapour pressure of mixture: methanol 16.4g, ethanol 92g at 298K. P°methanol=90 mmHg, P°ethanol=45 mmHg. [C=12, H=1, O=16] [4 marks] Solution: Mr CH₃OH=32, C₂H₅OH=46. nmethanol=16.4/32=0.5125 mol; nethanol=92/46=2.0 mol. Total=2.5125 mol Xmethanol=0.5125/2.5125=0.204; Xethanol=0.796 Ptotal = Xm×P°m + Xe×P°e = 0.204×90 + 0.796×45 = 18.36 + 35.82 = 54.2 mmHg (c) Q: Ideal solution of benzene & toluene. Xbenzene = 0, 0.25, 0.5, 0.75, 1.0. (i) State equations for moles, mole fraction, partial vapour pressures. (ii) Calculate Xtoluene. (iii) Calc partial P given P°benzene=3.6×10⁴, P°toluene=1.12×10⁴ Nm⁻². (iv) Calc total P. [10 marks] A:* (i) XA = nA/(nA+nB); PA = XA×P°A (ii) Xtoluene = 1 - Xbenzene = 1.0, 0.75, 0.5, 0.25, 0.0 (iii) & (iv): Xbenzene Xtoluene Pbenzene Ptoluene Ptotal 0.00 1.00 0 1.12×10⁴ 1.12×10⁴ 0.25 0.75 9.0×10³ 8.4×10³ 1.74×10⁴ 0.50 0.50 1.8×10⁴ 5.6×10³ 2.36×10⁴ 0.75 0.25 2.7×10⁴ 2.8×10³ 2.98×10⁴ 1.00 0.00 3.6×10⁴ 0 3.6×10⁴

Asked on April 18, 2026English

This English question involves literary analysis, grammar, or writing skills. The detailed response below provides a well-structured answer with supporting evidence and clear explanations.

Accepted Answer · 2026-04-18T07:48:11.534Z

Here's a review of your solutions, with corrections and formatting applied. Q1. a) Step 1: Convert ppm to ppb. $$1 \text{ ppm} = 1000 \text{ ppb}$$ $$0.0400 \text{ ppm} = 0.0400 \times 1000 \text{ ppb} = \boxed{\text{40.0 ppb}}$$ Step 2: Convert ppm to mg/L. For dilute aqueous solutions, $1 \text{ ppm} \approx 1 \text{ mg/L}$. $$0.0400 \text{ ppm} = \boxed{\text{0.0400 mg/L}}$$ *

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