Interleaved — Phase 1

Chemistry interleaved practice

printable — key stays hidden on paper

Instructions: Solve all problems. Each mixes a different concept, so identify the correct method before calculating. Show all working. Use $N_A = 6.022\times10^{23}\ \text{mol}^{-1}$ , and constants as needed. Report answers to the correct number of significant figures. Marks shown in [ ].

1. A sample of copper contains two isotopes: $^{63}\text{Cu}$ (mass $62.93$ u, abundance $69.2\%$ ) and $^{65}\text{Cu}$ (mass $64.93$ u). Calculate the relative atomic mass of copper. [3]

2. Classify each as a physical or chemical change, giving a one-line reason: (a) dry ice sublimes, (b) iron rusts, (c) salt dissolves in water, (d) magnesium burns in air. [4]

3. For the hydrogen atom, calculate the radius of the $n=3$ orbit (in Å) and the energy of an electron in the $n=2$ level (in eV) using the Bohr model. [4]

4. Round and express correctly: (a) evaluate $\dfrac{4.51 \times 3.6672}{2.09}$ to the proper number of significant figures; (b) state how many significant figures are in $0.004030$ . [3]

5. A $250\ \text{mL}$ solution contains $4.00\ \text{g}$ of $\text{NaOH}$ . Calculate its molarity. Then find the volume of this stock needed to prepare $500\ \text{mL}$ of a $0.100\ \text{M}$ solution. [4]

6. Two oxides of nitrogen contain, per gram of nitrogen, $1.14\ \text{g}$ and $2.28\ \text{g}$ of oxygen respectively. Show that these data illustrate a specific chemical law and name it. [3]

7. Suggest and briefly justify a separation technique for each mixture: (a) sand and water, (b) ethanol and water, (c) coloured pigments in ink, (d) iodine mixed with sand. [4]

8. A compound contains $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen by mass. Its molar mass is $180\ \text{g mol}^{-1}$ . Determine its empirical and molecular formulas. [5]

9. When $6.3\ \text{g}$ of sodium hydrogen carbonate is heated in a sealed vessel, it decomposes. If the vessel is instead open and $2.0\ \text{g}$ of gas escapes, comment on what the "missing" mass tells us and which law applies. Also compute the number of molecules in $6.3\ \text{g}$ of $\text{NaHCO}_3$ ( $M = 84\ \text{g mol}^{-1}$ ). [4]

10. At STP a gas occupies $5.6\ \text{L}$ and has a mass of $11.0\ \text{g}$ . Calculate (a) the number of moles, (b) the molar mass, and (c) identify a plausible gas. (Molar volume at STP $= 22.4\ \text{L mol}^{-1}$ .) [4]

Answer keyMark scheme & solutions

1. (Subtopic 1.2.6 — atomic mass from isotopic abundance) Abundance of $^{65}\text{Cu} = 100 - 69.2 = 30.8\%$ . $A = 0.692(62.93) + 0.308(64.93) = 43.55 + 20.00 = 63.55\ \text{u}$ Answer: $63.55$ u. Why: Given isotope masses + abundances → weighted average. Method choice cued by "isotopes + abundance."

2. (Subtopic 1.1.4 — physical vs chemical change) (a) Physical — sublimation is a change of state, $\text{CO}_2$ unchanged. (b) Chemical — new substance (iron oxide) formed. (c) Physical — dissolution, ions dispersed but no new compound; reversible by evaporation. (d) Chemical — MgO formed, energy released, irreversible. Why: Test = does composition/identity change? Not a calculation.

3. (Subtopic 1.2.7 — Bohr model) Radius: $r_n = 0.529\,\dfrac{n^2}{Z}$ Å, with $Z=1$ , $n=3$ : $r_3 = 0.529(9) = 4.76\ \text{Å}$ Energy: $E_n = -13.6\,\dfrac{Z^2}{n^2}$ eV, $n=2$ : $E_2 = -\frac{13.6}{4} = -3.40\ \text{eV}$ Answers: $r_3 = 4.76$ Å, $E_2 = -3.40$ eV. Why: Direct formula substitution; choose radius vs energy formula per quantity requested.

4. (Subtopic 1.1.6 — significant figures) (a) $\dfrac{4.51 \times 3.6672}{2.09} = \dfrac{16.539}{2.09} = 7.9134$ . Least sig figs among inputs = 3 (from $4.51$ and $2.09$ ) → $7.91$ . (b) $0.004030$ : leading zeros not significant; trailing zero after decimal counts → 4 sig figs. Why: Multiply/divide → answer limited by fewest sig figs; recognize significant zeros.

5. (Subtopics 1.1.15 molarity + 1.1.16 dilution) Moles NaOH $= 4.00/40.0 = 0.100\ \text{mol}$ . $M = \frac{0.100}{0.250} = 0.400\ \text{M}$ Dilution: $M_1V_1 = M_2V_2 \Rightarrow V_1 = \dfrac{(0.100)(500)}{0.400} = 125\ \text{mL}$ . Answers: $0.400$ M; $125$ mL of stock. Why: First part is a concentration definition; second cues dilution formula ("prepare from stock").

6. (Subtopic 1.1.10 — law of multiple proportions) For fixed 1 g N, ratio of O masses $= \dfrac{2.28}{1.14} = \dfrac{2}{1}$ , a small whole-number ratio. This illustrates the Law of Multiple Proportions (Dalton). Why: Same two elements, fixed mass of one → simple whole-number ratio distinguishes this from definite proportions (which involves a single compound).

7. (Subtopic 1.1.3 — separation techniques) (a) Filtration — insoluble solid in liquid. (b) Distillation (fractional) — miscible liquids with different boiling points. (c) Chromatography — separates soluble coloured components by differential adsorption. (d) Sublimation — iodine sublimes; sand does not. Why: Match technique to the physical property that differs (solubility, b.p., adsorption, volatility).

8. (Subtopic 1.1.14 — empirical/molecular formula) Assume 100 g: C $=40.0/12=3.33$ , H $=6.7/1=6.7$ , O $=53.3/16=3.33$ mol. Divide by $3.33$ : C $_1$ H $_2$ O $_1$ → empirical $\text{CH}_2\text{O}$ , mass $=30$ . $n = 180/30 = 6$ → molecular formula $\text{C}_6\text{H}_{12}\text{O}_6$ . Answers: empirical $\text{CH}_2\text{O}$ ; molecular $\text{C}_6\text{H}_{12}\text{O}_6$ . Why: %→moles→ratio→scale by $M$ ; cued by "% composition + molar mass."

9. (Subtopics 1.1.8 conservation of mass + 1.1.12 mole/counting) In the sealed vessel total mass is conserved (Law of Conservation of Mass, Lavoisier). In the open vessel, $2.0\ \text{g}$ of $\text{CO}_2$ / $\text{H}_2\text{O}$ gas escapes — mass is not "lost," it left the system as gas. Molecules: moles $=6.3/84 = 0.075\ \text{mol}$ ; $N = 0.075 \times 6.022\times10^{23} = 4.5\times10^{22}$ molecules. Answers: apparent loss = escaped gas; $4.5\times10^{22}$ molecules. Why: Conceptual conservation reasoning + straightforward mole → molecule count.

10. (Subtopics 1.1.7 molar volume + 1.1.13 molar mass) (a) $n = 5.6/22.4 = 0.25\ \text{mol}$ . (b) $M = 11.0/0.25 = 44\ \text{g mol}^{-1}$ . (c) $M=44$ → $\text{CO}_2$ (or $\text{N}_2\text{O}$ /propane). Answers: $0.25$ mol, $44$ g/mol, $\text{CO}_2$ . Why: Molar volume gives moles; mass/moles gives molar mass → identity.

[
  {
    "claim": "Cu relative atomic mass = 63.55 u",
    "code": "m = 0.692*62.93 + 0.308*64.93; result = round(m,2) == 63.55"
  },
  {
    "claim": "Bohr r3 = 4.761 A and E2 = -3.40 eV",
    "code": "r3 = 0.529*(3**2)/1; E2 = -13.6*(1**2)/(2**2); result = round(r3,2)==4.76 and round(E2,2)==-3.40"
  },
  {
    "claim": "Glucose molecular formula factor n=6 for CH2O (mass 30) to reach 180",
    "code": "emp = 12+2*1+16; n = 180/emp; result = int(round(n))==6"
  }
]