Noesis
Level 2 — RecallMatter, Measurement & the Mole

Matter, Measurement & the Mole

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level 2 Test Paper (Recall & Standard Problems)

Time Limit: 30 minutes
Total Marks: 40

Instructions: Answer all questions. Show working for numerical problems. Use NA=6.022×1023 mol−1N_A = 6.022 \times 10^{23}\ \text{mol}^{-1} and molar volume of an ideal gas at STP =22.4 L mol−1= 22.4\ \text{L mol}^{-1}.

Q1. Define the following, giving one example of each: (a) homogeneous mixture, (b) compound. (3 marks)

Q2. State whether each is a physical or chemical change: (a) melting of ice, (b) rusting of iron, (c) dissolving sugar in water, (d) burning of magnesium ribbon. (4 marks)

Q3. Name the most suitable separation technique for each of the following mixtures, giving a one-line reason: (a) sand and water, (b) two miscible liquids with different boiling points, (c) coloured pigments in ink, (d) ammonium chloride and common salt. (4 marks)

Q4. Express the following to the required significant figures: (a) 0.0045080.004508 to 2 s.f., (b) 1234512345 to 3 s.f., (c) Round 2.675+1.42.675 + 1.4 and give the answer to the correct number of decimal places. (3 marks)

Q5. Calculate the number of moles and the number of molecules in 8.8 g8.8\ \text{g} of CO2\text{CO}_2. (Molar mass =44 g mol−1= 44\ \text{g mol}^{-1}) (4 marks)

Q6. The density of a certain gas is 1.25 g L−11.25\ \text{g L}^{-1} at STP. Determine its molar mass and identify the gas. (4 marks)

Q7. A compound contains 40.0%40.0\% carbon, 6.7%6.7\% hydrogen and 53.3%53.3\% oxygen by mass. Its molar mass is 180 g mol−1180\ \text{g mol}^{-1}. Determine its empirical and molecular formula. (5 marks)

Q8. State the Law of Multiple Proportions. Carbon forms two oxides: in CO, 12 g of carbon combines with 16 g of oxygen; in CO2_2, 12 g of carbon combines with 32 g of oxygen. Show that these data illustrate this law. (4 marks)

Q9. Calculate the molarity of a solution prepared by dissolving 4.0 g4.0\ \text{g} of NaOH (molar mass =40 g mol−1= 40\ \text{g mol}^{-1}) in enough water to make 500 mL500\ \text{mL} of solution. (3 marks)

Q10. What volume of 12 M12\ \text{M} concentrated HCl is needed to prepare 250 mL250\ \text{mL} of 0.50 M0.50\ \text{M} HCl solution? (3 marks)

End of paper

Answer keyMark scheme & solutions

Q1. (3 marks)

  • (a) Homogeneous mixture: a mixture with uniform composition throughout, single visible phase — e.g. salt solution / air. (1½)
  • (b) Compound: a pure substance formed by chemical combination of two or more elements in a fixed ratio, separable only by chemical means — e.g. water (H2_2O). (1½) Why: definitions must include "fixed ratio / chemical combination" for compound to distinguish from mixture.

Q2. (4 marks — 1 each)

  • (a) Physical change (no new substance; reversible).
  • (b) Chemical change (new substance, iron oxide, forms).
  • (c) Physical change (sugar recoverable by evaporation).
  • (d) Chemical change (MgO formed, energy released).

Q3. (4 marks — 1 each)

  • (a) Filtration — sand is insoluble solid, separated from liquid.
  • (b) (Fractional) distillation — separates miscible liquids by boiling-point difference.
  • (c) Chromatography — separates dissolved coloured components by differential adsorption.
  • (d) Sublimation — NH4_4Cl sublimes; salt does not.

Q4. (3 marks — 1 each)

  • (a) 0.00450.0045 (1)
  • (b) 1.23×1041.23 \times 10^4 (1)
  • (c) 2.675+1.4=4.075→4.12.675 + 1.4 = 4.075 \to 4.1 (least decimals = 1) (1) Why: addition rule uses least number of decimal places.

Q5. (4 marks)

  • Moles =8.844=0.20 mol= \dfrac{8.8}{44} = 0.20\ \text{mol} (2)
  • Molecules =0.20×6.022×1023=1.2044×1023≈1.20×1023= 0.20 \times 6.022\times10^{23} = 1.2044\times10^{23} \approx 1.20\times10^{23} (2)

Q6. (4 marks)

  • Molar mass =density×molar volume=1.25×22.4=28 g mol−1= \text{density} \times \text{molar volume} = 1.25 \times 22.4 = 28\ \text{g mol}^{-1} (3)
  • Gas: N2_2 (or CO, C2_2H4_4; N2_2 most likely). (1) Why: at STP, M=ρ×VmM = \rho \times V_m.

Q7. (5 marks)

  • Moles ratio: C =40.0/12=3.33= 40.0/12 = 3.33; H =6.7/1=6.7= 6.7/1 = 6.7; O =53.3/16=3.33= 53.3/16 = 3.33. (2)
  • Divide by smallest (3.33): C:H:O = 1:2:11:2:1 → empirical formula CH2_2O. (1)
  • Empirical mass =12+2+16=30= 12+2+16 = 30. (1)
  • n=180/30=6n = 180/30 = 6 → molecular formula C6H12O6C_6H_{12}O_6. (1)

Q8. (4 marks)

  • Statement: when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio. (2)
  • Fixing carbon at 12 g: oxygen masses are 16 g (CO) and 32 g (CO2_2). Ratio 16:32=1:216:32 = 1:2, a simple whole-number ratio → illustrates the law. (2)

Q9. (3 marks)

  • Moles NaOH =4.0/40=0.10 mol= 4.0/40 = 0.10\ \text{mol} (1)
  • Volume =0.500 L= 0.500\ \text{L}; Molarity =0.10/0.500=0.20 M= 0.10/0.500 = 0.20\ \text{M} (2)

Q10. (3 marks)

  • M1V1=M2V2M_1V_1 = M_2V_2: 12×V1=0.50×25012 \times V_1 = 0.50 \times 250 (1)
  • V1=125/12=10.4 mLV_1 = 125/12 = 10.4\ \text{mL} (2)
[
  {"claim":"Q5 moles of CO2 in 8.8 g = 0.20","code":"result = (8.8/44 == 0.20)"},
  {"claim":"Q6 molar mass = density x molar volume = 28","code":"result = (1.25*22.4 == 28.0)"},
  {"claim":"Q7 empirical CH2O mass=30 and n=6 gives C6H12O6","code":"emp=12+2+16\nresult = (emp==30) and (180/emp==6)"},
  {"claim":"Q10 dilution volume = 10.42 mL approx","code":"V1=0.50*250/12\nresult = (round(V1,2)==10.42)"}
]