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Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.
76 original questions across 15 selected topic areas. No account or payment required to practise.
Cambridge International AS & A Level Chemistry (9701), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.
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Original English concept practice addressing selected identified gaps. Not complete syllabus coverage or practical competence.
Selected subtopics: Isotope mass spectra · Gas molar mass · Bond-energy accounting · Kp and Contact equilibrium
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Question 1A compound contains only carbon, hydrogen and chlorine. Its molecular-ion peaks at m/z 64 and 66 have approximate abundance ratio 3:1; M+1 relative to the 64 peak is 2.2%. Use carbon count n = 100 x abundance(M+1)/(1.1 x abundance(M)), Ar(C)=12, Ar(H)=1 and chlorine isotope masses 35 and 37. Deduce a molecular formula and explain the isotope evidence. Treat ions as singly charged and ignore overlapping minor isotope contributions.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
A compound contains only carbon, hydrogen and chlorine. Its molecular-ion peaks at m/z 64 and 66 have approximate abundance ratio 3:1; M+1 relative to the 64 peak is 2.2%. Use carbon count n = 100 x abundance(M+1)/(1.1 x abundance(M)), Ar(C)=12, Ar(H)=1 and chlorine isotope masses 35 and 37. Deduce a molecular formula and explain the isotope evidence. Treat ions as singly charged and ignore overlapping minor isotope contributions.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked the official M+1 carbon relation, one-chlorine isotope pattern and independent atom-mass sum; assumptions prevent overidentification from an arbitrary M+2 peak. This is conceptual placement, not completion of the entire referenced subsection.
One chlorine produces two molecular-ion isotopologues two mass units apart. Two chlorines would instead produce a three-peak M/M+2/M+4 pattern, so the given simplified pair is consistent with one.
Apply the M+1 carbon estimate before using the 64 mass balance. Two carbons contribute 24 and 35Cl contributes 35; the residual 5 corresponds to five hydrogens.
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The 3:1 M:M+2 pair supports one chlorine atom with the supplied 35/37 isotopes.
n(C)=2.2/1.1=2.
For the 35Cl molecular ion, hydrogen mass is 64 - 24 - 35 = 5, giving C2H5Cl.
The 37Cl isotopologue has mass 66, matching the second molecular-ion peak.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 2A 0.400 g sample of an ideal gas occupies 244 cm^3 at 298 K and 101 kPa. Use R=8.31 J mol^-1 K−1 to calculate its molar mass. State two ideal-gas assumptions and explain why the approximation is poorer at high pressure and low temperature.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
A 0.400 g sample of an ideal gas occupies 244 cm^3 at 298 K and 101 kPa. Use R=8.31 J mol^-1 K−1 to calculate its molar mass. State two ideal-gas assumptions and explain why the approximation is poorer at high pressure and low temperature.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Recomputed SI volume/pressure conversion and molar mass; explicit gas assumptions are read from 4.1 rather than claiming real-gas accuracy. This is conceptual placement, not completion of the entire referenced subsection.
pV=24.644 J and RT=2476.38 J mol^-1, giving n=0.00995163 mol. The mass is in grams because the requested molar-mass unit is g mol^-1.
Divide 0.400 by the unrounded amount to obtain 40.1945 g mol^-1. A low-density warm gas more closely approaches the point-particle/no-attraction model than a compressed cold gas.
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Use V=2.44 x 10^-4 m^3 and p=101000 Pa; n=pV/(RT)=0.00995 mol.
M=m/n=40.2 g mol^-1, to three significant figures.
An ideal gas neglects molecular volume and intermolecular attractions.
Crowding and stronger relative effects of attractions make these assumptions poorer at high pressure/low temperature.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 3Estimate deltaH for H2(g)+Cl2(g)->2HCl(g) using mean bond energies H-H=436, Cl-Cl=243 and H-Cl=431 kJ mol^-1. Explain why bond breaking alone cannot explain an exothermic reaction and why using mean values gives an estimate.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
Estimate deltaH for H2(g)+Cl2(g)->2HCl(g) using mean bond energies H-H=436, Cl-Cl=243 and H-Cl=431 kJ mol^-1. Explain why bond breaking alone cannot explain an exothermic reaction and why using mean values gives an estimate.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked bond counts, signs and -183 kJ/mol against bond-energy convention; no hazardous reaction procedure supplied. This is conceptual placement, not completion of the entire referenced subsection.
Count one H-H and one Cl-Cl broken, not two of either. Count two product H-Cl bonds because the equation makes two HCl molecules.
Subtract energy released on formation from energy absorbed on breaking. The negative result follows from the entire process, not from a bond breaking spontaneously to release energy.
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Breaking reactant bonds requires 436+243=679 kJ per mole of reaction.
Forming two H-Cl bonds releases 2 x 431=862 kJ.
deltaH is approximately 679-862=-183 kJ mol^-1.
Bond formation releases more than breaking absorbs; mean bond energies may differ from specific molecular values.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 4For 2SO2(g)+O2(g)<=>2SO3(g), an equilibrium mixture at a stated fixed temperature has mole fractions 0.20, 0.10 and 0.70 respectively, and total pressure 200 kPa. Calculate partial pressures and Kp with pressure units kPa. The forward reaction is exothermic. Explain separately the equilibrium effect of compression and heating, and the role of a catalyst.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
For 2SO2(g)+O2(g)<=>2SO3(g), an equilibrium mixture at a stated fixed temperature has mole fractions 0.20, 0.10 and 0.70 respectively, and total pressure 200 kPa. Calculate partial pressures and Kp with pressure units kPa. The forward reaction is exothermic. Explain separately the equilibrium effect of compression and heating, and the role of a catalyst.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked partial pressures sum to 200 kPa, coefficient powers, Kp units and independent temperature/pressure effects in Contact-process equilibrium. This is conceptual placement, not completion of the entire referenced subsection.
Multiply each mole fraction by total pressure, then use stoichiometric coefficients as powers: 140^2/(40.02 x 20.0)=19600/32000.
The pressure powers give kPa^(2-3)=kPa^-1. Do not substitute total pressure for every species or assume Kp itself increases on compression at fixed temperature.
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Partial pressures are 40.0 kPa SO2, 20.0 kPa O2 and 140 kPa SO3.
Kp=p(SO3)^2/[p(SO2)^2 p(O2)] = 0.6125 kPa^-1.
Compression at fixed temperature favours the two-mole product side; heating favours the endothermic reverse direction.
A catalyst changes the rate of approaching equilibrium, not Kp at that temperature.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
