Applying Chemical Ideas, complete assessment covering all three inquiry questions across L01–L16. 15 MC questions (auto-marked) + 5 written questions (self-marked). Complete all questions before submitting.
In an acid–base titration, the equivalence point is best described as the point where:
Which property is essential for a substance to be used as a primary standard?
A 2.00 g sample of impure NaCl is dissolved and treated with excess AgNO₃. The dried AgCl precipitate has a mass of 4.52 g. What is the percentage of NaCl in the sample? (M(NaCl) = 58.5, M(AgCl) = 143.5)
A chloride sample is precipitated as AgCl (M = 143.3 g mol⁻¹). If 0.2870 g of dry AgCl is collected, how many moles of chloride were present? (Cl : AgCl is 1 : 1)
In a flame test, lithium compounds produce a flame that is:
A student adds NaOH to an unknown solution and observes a red-brown precipitate. Which cation is most likely present?
According to the Beer–Lambert law, at fixed path length and wavelength the absorbance of a solution is directly proportional to:
In thin-layer chromatography, a spot travels 3.6 cm while the solvent front travels 4.8 cm. The Rf value is:
The Australian Drinking Water Guidelines recommend a pH for drinking water in the range of about:
A Winkler titration of a water sample requires 8.00 mL of 0.0100 mol L⁻¹ Na₂S₂O₃. Using the ratio 1 mol O₂ : 4 mol Na₂S₂O₃, how many moles of O₂ were present?
A sample has an initial dissolved oxygen of 8.4 mg L⁻¹. After 5 days at 20 °C in the dark the dissolved oxygen is 2.1 mg L⁻¹. What is the BOD₅?
The base peak in a mass spectrum is the:
A strong infrared absorption around 1700 cm⁻¹ (1680 to 1750) is characteristic of which bond?
A ¹H NMR spectrum shows a triplet and a quartet in an integration ratio of 3 : 2. This pattern is characteristic of:
For the aspirin synthesis, salicylic acid (M = 138) + acetic anhydride (M = 102) → aspirin (M = 180) + ethanoic acid (M = 60). The atom economy for aspirin is:
A 0.500 g antacid tablet containing CaCO₃ is treated with 50.0 mL of 0.200 mol L⁻¹ HCl, which is in excess. The unreacted excess HCl requires 15.0 mL of 0.150 mol L⁻¹ NaOH for neutralisation. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. (a) Calculate the moles of HCl added and the moles of excess HCl. (b) Calculate the moles of HCl that reacted with the CaCO₃. (c) Calculate the mass and the percentage by mass of CaCO₃ in the tablet. (M(CaCO₃) = 100.1)
In the Winkler titration of a 250.0 mL water sample, the liberated iodine required 8.00 mL of 0.0100 mol L⁻¹ Na₂S₂O₃. The overall relationship is 1 mol O₂ : 4 mol Na₂S₂O₃. (a) Explain why sodium thiosulfate does not titrate the dissolved oxygen directly, but instead titrates iodine. (b) Calculate the moles of O₂ in the sample. (c) Calculate the dissolved oxygen concentration in mg L⁻¹. (M(O₂) = 32.00)
(a) Thiosulfate does not react with dissolved oxygen directly. In the Winkler method the O₂ first oxidises Mn²⁺, and the oxidised manganese then oxidises iodide (I⁻) to iodine (I₂). The thiosulfate titrates this liberated I₂, and the amount of iodine is stoichiometrically linked back to the original dissolved oxygen.
(b) n(Na₂S₂O₃) = 0.00800 × 0.0100 = 8.00×10⁻⁵ mol n(O₂) = 8.00×10⁻⁵ ÷ 4 = 2.00×10⁻⁵ mol (c) m(O₂) = 2.00×10⁻⁵ × 32.00 = 6.40×10⁻⁴ g = 0.640 mg DO = 0.640 mg ÷ 0.2500 L = 2.56 mg L⁻¹ Marks: 1, iodine (not O₂) is titrated | 1, valid reasoning via Mn and iodide | 1, n(thiosulfate) | 1, n(O₂) using 1 : 4 ratio | 1, DO in mg L⁻¹A river downstream of farmland and a factory is being monitored. (a) An AAS test reads mercury below the drinking-water limit, yet fish from the same water contain much higher mercury levels in their tissue. Explain this using bioaccumulation and biomagnification. (b) Fertiliser runoff has caused an algal bloom in the same river. Outline, in order, how excess nutrients lead to a fish kill, and identify the direct chemical cause of death.
(a) Bioaccumulation is the build-up of mercury within the tissues of a single organism over its lifetime, because mercury is retained rather than excreted. Biomagnification is the increase in mercury concentration at each higher trophic level of the food chain. Together they mean a low concentration in the water can still produce very high concentrations in top predators such as large fish, so a safe water reading does not guarantee safe fish.
(b) Order: nutrient loading (nitrate and phosphate) → rapid algal bloom that shades and kills submerged plants → bacteria decompose the large mass of dead algae and plants, consuming dissolved oxygen and raising the BOD → dissolved oxygen collapses (hypoxia). The direct chemical cause of the fish kill is the depletion of dissolved oxygen (hypoxia), not the algae themselves.
Marks: 1, bioaccumulation defined | 1, biomagnification defined and linked to fish | 1, correct eutrophication sequence | 1, decomposition raises BOD / consumes O₂ | 1, direct cause identified as hypoxiaAn unknown organic compound has a molecular ion peak at m/z = 46 in its mass spectrum. Its infrared spectrum shows a broad O–H absorption (3230 to 3550 cm⁻¹) and a C–O band, but no strong band near 1700 cm⁻¹. Its ¹H NMR spectrum shows three signals in an integration ratio of 3 : 2 : 1, including a triplet and a quartet. (a) State what the molecular ion tells you. (b) Interpret the IR data, explaining what the absence of the ~1700 cm⁻¹ band rules out. (c) Interpret the ¹H NMR pattern. (d) Identify the compound and give its structural formula.
(b) The broad O–H (3230 to 3550 cm⁻¹) with a C–O band indicates an alcohol. The absence of a strong C=O band near 1700 cm⁻¹ rules out carbonyl compounds such as carboxylic acids, aldehydes, ketones and esters.
(c) Three signals in a 3 : 2 : 1 ratio mean three hydrogen environments in that proportion. The triplet (CH₃, split by 2 neighbours) and quartet (CH₂, split by 3 neighbours) are the classic ethyl group signature; the remaining single hydrogen is the O–H.
(d) The compound is ethanol, CH₃CH₂OH. Marks: 1, molecular mass 46 / formula C₂H₆O | 1, IR indicates alcohol | 1, no C=O rules out carbonyl compounds | 1, ethyl group plus O–H from NMR | 1, correct identity and structureA pharmaceutical compound can be made by Route A (85% yield, 45% atom economy, a volatile flammable solvent, two stoichiometric steps) or Route B (70% yield, 80% atom economy, water as solvent, one catalytic step). (a) Define atom economy and E-factor. (b) Evaluate which route is the more sustainable choice, referring to atom economy, yield, waste, solvent safety and energy, and state your judgement.
(b) Route B is the more sustainable choice. Its higher atom economy (80% versus 45%) means far less reactant mass is lost as by-products, lowering waste and the E-factor, even though its yield is lower (70% versus 85%). Route B uses water rather than a volatile flammable solvent, reducing fire risk, toxic emissions and the energy needed for solvent recovery. A single catalytic step also tends to use less energy and generate less waste than two stoichiometric steps. Route A's main advantage is its higher yield, but yield alone does not outweigh the large by-product mass, the solvent hazard and the extra step. A full decision would also weigh reagent cost, scalability and product purity, but on the stated criteria Route B is preferable.
Marks: 1, atom economy defined | 1, E-factor defined | 1, atom economy and yield compared correctly | 1, solvent safety and energy or steps addressed | 1, clear justified judgement for Route B