IOE entrance Chemistry · Chapter 1
Physical Chemistry
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123 questions in 8 syllabus topics · 50 are 2-mark questions.
1.1 Chemical arithmetic
16 questions
1. 21 g of magnesium carbonate was treated with 50 mL of 12 N hydrochloric acid. How many molecules of unreacted hydrochloric acid remain?
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Find the HCl used by 0.25 mol MgCO₃ and subtract.
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Answer: B. 6.025 × 10²²
MgCO₃ (M = 84): 21/84 = 0.25 mol, which needs 0.5 mol HCl (MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂). HCl taken = 0.050 L × 12 = 0.6 mol. Left over = 0.1 mol = 0.1 × 6.025 × 10²³ = 6.025 × 10²² molecules.
2. Which postulate of Dalton's atomic theory has been shown to be NOT valid by the discovery of isotopes?
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Think about what isotopes of an element differ in.
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Answer: A. Atoms of the same element are identical in all respects, including mass
Isotopes are atoms of the same element with different masses (e.g. ³⁵Cl and ³⁷Cl), so atoms of one element are not all identical in mass.
3. Carbon forms two oxides, CO and CO₂. For a fixed mass of carbon, the masses of oxygen combined are in the ratio 1 : 2. This illustrates the law of
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Two different compounds of the same two elements are being compared.
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Answer: B. multiple proportions
Two elements form more than one compound, and the masses of one element combining with a fixed mass of the other are in a simple whole-number ratio (16 : 32 = 1 : 2): law of multiple proportions.
4. Pure water obtained from rain, a river or by burning hydrogen always contains hydrogen and oxygen in the mass ratio 1 : 8. This is a statement of the law of
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Only one compound is involved, prepared from different sources.
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Answer: C. definite proportions
A given compound always contains the same elements combined in the same fixed ratio by mass, whatever its source: law of definite (constant) proportions.
5. The number of molecules present in 4.4 g of CO₂ is (Nₐ = 6.022 × 10²³ mol⁻¹)
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Find moles first: n = mass ÷ molar mass.
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Answer: C. 6.022 × 10²²
Molar mass of CO₂ = 44 g/mol, so 4.4 g = 0.1 mol. Molecules = 0.1 × 6.022 × 10²³ = 6.022 × 10²². (1.806 × 10²³ is the number of atoms.)
6. The equivalent mass of H₂SO₄ when it is completely neutralised by NaOH is
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Equivalent mass of an acid = molar mass ÷ basicity.
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Answer: B. 49
H₂SO₄ is dibasic (basicity 2) in complete neutralisation. Equivalent mass = 98/2 = 49.
7. The volume occupied by 8 g of oxygen gas at NTP is
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Use the molar mass of O₂, not of O.
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Answer: A. 5.6 L
Moles of O₂ = 8/32 = 0.25 mol. Volume = 0.25 × 22.4 = 5.6 L. (11.2 L comes from wrongly taking O₂ as 16 g/mol.)
8. Avogadro's hypothesis states that, at the same temperature and pressure,
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Remember the hypothesis was about molecules, needed to explain Gay-Lussac's law.
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Answer: C. equal volumes of all gases contain equal numbers of molecules
Avogadro: equal volumes of all gases under identical T and P contain equal numbers of molecules (not atoms — e.g. He is monatomic, O₂ diatomic).
9. Which pair of compounds has the same empirical formula?
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Divide each formula by the highest common factor of its subscripts.
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Answer: B. C₂H₂ and C₆H₆
Both acetylene (C₂H₂) and benzene (C₆H₆) reduce to the simplest ratio CH.
10. The vapour density of a gas is 32. Its molecular mass is
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Vapour density is measured relative to hydrogen (H₂ = 2).
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Answer: A. 64
Molecular mass = 2 × vapour density = 2 × 32 = 64.
11. An organic compound contains C = 40%, H = 6.67% and O = 53.33% by mass. Its vapour density is 30. The molecular formula of the compound is
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Find the empirical formula, then compare its mass with 2 × vapour density.
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Answer: B. C₂H₄O₂
Mole ratio C : H : O = 40/12 : 6.67/1 : 53.33/16 = 3.33 : 6.67 : 3.33 = 1 : 2 : 1, so empirical formula CH₂O (mass 30). Molar mass = 2 × 30 = 60, n = 60/30 = 2, giving C₂H₄O₂.
12. 4 g of hydrogen is made to react with 16 g of oxygen to form water. The mass of water formed is
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Identify the limiting reactant first.
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Answer: A. 18 g
2H₂ + O₂ → 2H₂O. H₂ = 4/2 = 2 mol, O₂ = 16/32 = 0.5 mol. 2 mol H₂ would need 1 mol O₂, so O₂ is limiting. 0.5 mol O₂ gives 1 mol H₂O = 18 g.
13. 0.5 g of a metal displaces 560 mL of hydrogen gas at NTP from an acid. The equivalent mass of the metal is
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One gram-equivalent of hydrogen (1.008 g) occupies 11.2 L at NTP.
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Answer: D. 10
1 equivalent of H₂ occupies 11.2 L at NTP. Equivalents of H₂ = 0.56/11.2 = 0.05. Equivalent mass of metal = 0.5/0.05 = 10.
14. 25 g of pure calcium carbonate is heated strongly until decomposition is complete. The mass of the solid residue is
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The residue is CaO; CO₂ escapes.
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Answer: D. 14 g
CaCO₃ → CaO + CO₂. Moles CaCO₃ = 25/100 = 0.25 mol, giving 0.25 mol CaO = 0.25 × 56 = 14 g. (11 g is the mass of CO₂ lost.)
15. The number of oxygen atoms in 0.5 mol of washing soda, Na₂CO₃·10H₂O, is
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Do not forget the oxygen in the water of crystallisation.
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Answer: D. 6.5 Nₐ
Each formula unit has 3 O (in CO₃²⁻) + 10 O (in 10 H₂O) = 13 O atoms. In 0.5 mol: 0.5 × 13 Nₐ = 6.5 Nₐ.
16. A metal oxide contains 60% metal by mass. The equivalent mass of the metal is
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Equivalent mass is the mass that combines with 8 parts by mass of oxygen.
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Answer: D. 12
In 100 g oxide, 60 g metal combines with 40 g oxygen. Equivalent mass = (mass of metal/mass of oxygen) × 8 = (60/40) × 8 = 12.
1.2 States of matter
15 questions
17. At constant temperature, the pressure on a fixed mass of an ideal gas is doubled. Its volume becomes
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Use Boyle's law.
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Answer: B. half
Boyle's law: PV = constant at constant T, so doubling P halves V.
18. The value of the universal gas constant R in J K⁻¹ mol⁻¹ is
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Each number is R in some unit; pick the SI one.
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Answer: D. 8.314
R = 8.314 J K⁻¹ mol⁻¹. The other numbers are R in L atm K⁻¹ mol⁻¹ (0.0821), cal K⁻¹ mol⁻¹ (1.987) and L mmHg K⁻¹ mol⁻¹ (62.36).
19. Under identical conditions of temperature and pressure, which gas diffuses fastest?
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Rate of diffusion depends on molar mass.
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Answer: C. H₂
Graham's law: rate ∝ 1/√M. H₂ has the lowest molar mass (2), so it diffuses fastest.
20. According to the kinetic theory of gases, the average kinetic energy of the molecules of an ideal gas depends only on
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Recall KE = (3/2)kT.
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Answer: B. its absolute temperature
Average KE per molecule = (3/2)kT, which depends only on absolute temperature.
21. As the temperature of a liquid is raised, its surface tension
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Think about intermolecular forces at higher temperature.
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Answer: C. decreases
Higher temperature increases molecular kinetic energy and weakens intermolecular attraction, so surface tension falls (becoming zero at the critical temperature).
22. A liquid boils at the temperature at which
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Why does water boil below 100 °C on a mountain?
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Answer: C. its vapour pressure becomes equal to the external pressure
Boiling occurs when vapour pressure equals the external (atmospheric) pressure; 760 mm Hg applies only to the normal boiling point at sea level.
23. Which of the following is a covalent (network) solid?
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Look for a giant structure held only by covalent bonds.
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Answer: D. Diamond
In diamond each C is covalently bonded to four others in a 3-D network. NaCl is ionic, ice is a molecular (H-bonded) solid and copper is metallic.
24. The number of atoms per unit cell in a body-centred cubic (bcc) lattice is
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A corner atom is shared by 8 unit cells.
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Answer: B. 2
Corners: 8 × 1/8 = 1; body centre: 1. Total = 2. (9 comes from counting every atom fully.)
25. A real gas behaves most like an ideal gas at
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Which conditions make intermolecular forces and molecular volume unimportant?
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Answer: C. high temperature and low pressure
At high T and low P, intermolecular attractions are negligible and the molecular volume is small compared with the container volume.
26. A gas occupies 300 mL at 27 °C and 1 atm. Its volume at 127 °C and 2 atm will be
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Convert temperatures to kelvin and use P₁V₁/T₁ = P₂V₂/T₂.
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Answer: D. 200 mL
V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 300 × (1/2) × (400/300) = 200 mL. (705 mL results from using °C instead of K.)
27. A container holds 8 g of O₂ and 14 g of N₂ at a total pressure of 3 atm. The partial pressure of oxygen is
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Partial pressure = mole fraction × total pressure.
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Answer: D. 1 atm
n(O₂) = 8/32 = 0.25 mol, n(N₂) = 14/28 = 0.5 mol, total 0.75 mol. x(O₂) = 0.25/0.75 = 1/3. p(O₂) = (1/3) × 3 = 1 atm.
28. Under identical conditions, a gas takes three times as long as hydrogen to effuse through the same pinhole. The molar mass of the gas is
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Effusion time is inversely related to rate; rate ∝ 1/√M.
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Answer: D. 18 g/mol
Time of effusion ∝ √M. t/t(H₂) = 3 = √(M/2), so M/2 = 9 and M = 18 g/mol.
29. The temperature at which the rms speed of oxygen molecules becomes double its value at 27 °C is
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u(rms) = √(3RT/M); work in kelvin.
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Answer: D. 927 °C
u(rms) ∝ √T. Doubling u needs T to become 4 times: 4 × 300 K = 1200 K = 927 °C.
30. A gas has a density of 2.6 g/L at 2 atm and 300 K. Taking R = 0.082 L atm K⁻¹ mol⁻¹, its molar mass is approximately
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Rearrange the ideal gas equation as M = dRT/P.
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Answer: B. 32 g/mol
From PV = (m/M)RT, M = dRT/P = (2.6 × 0.082 × 300)/2 = 63.96/2 ≈ 32 g/mol. (64 comes from forgetting to divide by P.)
31. In a cubic crystal, atoms of A occupy all the corners and atoms of B occupy the centres of all the faces. The formula of the compound is
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Count each atom's share in the unit cell: corner 1/8, face 1/2.
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Answer: A. AB₃
A: 8 corners × 1/8 = 1 atom per cell. B: 6 faces × 1/2 = 3 atoms per cell. Formula AB₃.
1.3 Atomic structure and periodic classification
16 questions
32. Any p-orbital can accommodate up to
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One orbital, not the whole subshell.
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Answer: D. two electrons with anti-parallel spins
Each orbital (one of px, py, pz) holds at most two electrons, and by Pauli's exclusion principle they must have opposite spins. Six is the capacity of the whole p subshell.
33. The maximum number of electrons that can be accommodated in the shell with principal quantum number n = 3 is
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Use 2n².
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Answer: C. 18
Maximum electrons in a shell = 2n² = 2 × 9 = 18. (9 is the number of orbitals, n².)
34. An orbital described by n = 3 and l = 1 is
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Match l to the subshell letter.
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Answer: A. 3p
l = 0, 1, 2, 3 correspond to s, p, d, f. n = 3, l = 1 is a 3p orbital.
35. Which set of quantum numbers (n, l, m, s) is NOT permissible for an electron?
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Check the allowed range of l for each n.
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Answer: A. 2, 2, 0, +½
l can take values 0 to (n − 1) only. For n = 2, l cannot be 2.
36. The rule stating that pairing of electrons in orbitals of the same subshell does not begin until each orbital is singly occupied is
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This rule explains why nitrogen has three unpaired 2p electrons.
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Answer: D. Hund's rule of maximum multiplicity
This is Hund's rule. Pauli limits an orbital to two electrons of opposite spin; Aufbau gives the order of filling by energy.
37. The ground-state electronic configuration of chromium (Z = 24) is
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Chromium is an exception to the Aufbau order.
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Answer: A. [Ar] 3d⁵ 4s¹
A half-filled 3d⁵ subshell is extra stable, so one 4s electron shifts to 3d: [Ar] 3d⁵ 4s¹.
38. Which of the following elements has the highest first ionisation energy?
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Consider the stability of a half-filled p subshell.
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Answer: D. N
IE generally increases across period 2, but N (half-filled 2p³, stable) has a higher IE than O (2p⁴). Order: N > O > C > B.
39. Which element has the most negative electron gain enthalpy (highest electron affinity)?
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The most electronegative element is not the one with the highest electron affinity.
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Answer: C. Cl
Cl has the highest electron affinity. F is smaller and its compact 2p subshell causes strong electron–electron repulsion, so its electron affinity is less than that of Cl.
40. The most electronegative element in the periodic table is
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Electronegativity increases across a period and decreases down a group.
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Answer: C. fluorine
Fluorine has the highest electronegativity (4.0 on the Pauling scale).
41. The number of unpaired electrons in the Fe³⁺ ion (Z of Fe = 26) is
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Electrons are removed from 4s before 3d.
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Answer: C. 5
Fe: [Ar] 3d⁶ 4s². Fe³⁺ loses both 4s and one 3d electron: [Ar] 3d⁵, with 5 unpaired electrons (Hund's rule).
42. The radius of the first Bohr orbit of the hydrogen atom is 0.529 Å. The radius of the second Bohr orbit of Li²⁺ (Z = 3) is
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Bohr radius varies as n²/Z.
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Answer: D. 0.705 Å
rₙ = 0.529 × n²/Z Å = 0.529 × 4/3 = 0.705 Å. (2.116 Å ignores Z.)
43. Taking the energy of the electron in the nth orbit of hydrogen as −13.6/n² eV, the energy of the photon emitted when the electron falls from n = 3 to n = 2 is
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ΔE = 13.6(1/n₁² − 1/n₂²) eV.
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Answer: B. 1.89 eV
ΔE = 13.6(1/2² − 1/3²) = 13.6 × 5/36 ≈ 1.89 eV (the first Balmer line). 10.2 eV is 2→1 and 12.09 eV is 3→1.
44. Given that the ionisation energy of the hydrogen atom is 13.6 eV, the energy needed to remove the electron from a ground-state He⁺ ion is
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Energy scales as Z² for hydrogen-like ions.
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Answer: D. 54.4 eV
For hydrogen-like species, E = −13.6 Z²/n² eV. For He⁺ (Z = 2, n = 1): IE = 13.6 × 4 = 54.4 eV.
45. When electrons in a sample of hydrogen atoms fall from the n = 5 level to the ground state, the maximum number of different spectral lines that can be observed is
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Count all possible pairs of levels between n = 5 and n = 1.
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Answer: D. 10
Number of lines = n(n − 1)/2 with n = 5: 5 × 4/2 = 10. (4 counts only a single electron's step-by-step path.)
46. The correct order of increasing ionic radius among the isoelectronic ions O²⁻, F⁻, Na⁺ and Mg²⁺ is
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Same number of electrons; compare nuclear charges.
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Answer: C. Mg²⁺ < Na⁺ < F⁻ < O²⁻
All have 10 electrons. The greater the nuclear charge (Mg 12 > Na 11 > F 9 > O 8), the stronger the pull and the smaller the ion.
47. An electron in a hydrogen atom has energy −1.51 eV (energy of the nth level = −13.6/n² eV). Its orbital angular momentum according to Bohr's theory is
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Find n first from the energy, then use mvr = nh/2π.
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Answer: A. 3h/2π
13.6/n² = 1.51 gives n² = 9, n = 3. Bohr: mvr = nh/2π = 3h/2π.
1.4 Oxidation, reduction and equilibrium
15 questions
48. The oxidation number of chromium in K₂Cr₂O₇ is
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The sum of oxidation numbers in a neutral compound is zero.
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Answer: C. +6
2(+1) + 2x + 7(−2) = 0 → 2x = 12 → x = +6. (+12 is the total for both Cr atoms.)
49. The oxidation number of oxygen in OF₂ is
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Fluorine always has oxidation number −1.
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Answer: A. +2
Fluorine, the most electronegative element, is always −1. x + 2(−1) = 0 → x = +2.
50. According to the electronic concept, oxidation is
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Remember: OIL RIG.
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Answer: C. loss of electrons
Oxidation is the loss of electrons (and so an increase in oxidation number); reduction is the gain of electrons.
51. In the reaction Zn + CuSO₄ → ZnSO₄ + Cu, the oxidising agent is
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The oxidising agent is the species that is itself reduced.
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Answer: B. CuSO₄
Cu²⁺ (from CuSO₄) gains electrons and is reduced to Cu, so it oxidises Zn. Zn is the reducing agent.
52. Which of the following can act as both an oxidising agent and a reducing agent?
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Look for an element in an intermediate oxidation state.
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Answer: C. H₂O₂
In H₂O₂ oxygen is −1, between 0 and −2, so it can be oxidised or reduced. Mn (+7) in KMnO₄ and F₂ can only be reduced; Na can only be oxidised.
53. For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the relation between Kp and Kc is
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Δn = moles of gaseous products − moles of gaseous reactants.
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Answer: A. Kp = Kc(RT)⁻²
Kp = Kc(RT)^Δn with Δn = 2 − (1 + 3) = −2, so Kp = Kc(RT)⁻².
54. For the exothermic reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium yield of ammonia is increased by
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Apply Le Chatelier's principle to moles of gas and to heat.
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Answer: D. increasing the pressure and lowering the temperature
Forward reaction reduces gas moles (4 → 2), so high pressure favours it; it is exothermic, so low temperature favours it (Le Chatelier). A catalyst does not shift equilibrium.
55. Adding a catalyst to a system at equilibrium
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A catalyst speeds up both directions equally.
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Answer: D. does not change the equilibrium constant, but equilibrium is reached faster
A catalyst lowers the activation energy of forward and reverse reactions equally, so K and equilibrium composition are unchanged; only the time to reach equilibrium decreases.
56. For which of the following equilibria is Kp equal to Kc?
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Use Kp = Kc(RT)^Δn.
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Answer: C. H₂(g) + I₂(g) ⇌ 2HI(g)
Kp = Kc when Δn(gas) = 0. For H₂ + I₂ ⇌ 2HI, Δn = 2 − 2 = 0.
57. The oxidation number of sulphur in peroxymonosulphuric acid, H₂SO₅, which contains one peroxide (–O–O–) linkage, is
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Peroxide oxygen has oxidation number −1.
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Answer: A. +6
Of the 5 O atoms, 2 are peroxide (−1 each) and 3 are normal (−2 each). 2(+1) + x + 3(−2) + 2(−1) = 0 → x = +6. (+8 comes from taking all O as −2, which is impossible for S.)
58. When the equation MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O is balanced with the smallest whole-number coefficients, the coefficient of H⁺ is
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Balance electrons first, then O with H₂O and H with H⁺.
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Answer: D. 8
Mn: +7 → +2 (gains 5e⁻); Fe: +2 → +3 (loses 1e⁻), so 1 MnO₄⁻ : 5 Fe²⁺. The 4 O of MnO₄⁻ form 4 H₂O, needing 8 H⁺: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
59. Acidified potassium dichromate oxidises iodide ions to iodine (Cr₂O₇²⁻ → Cr³⁺). The number of moles of I₂ liberated by one mole of Cr₂O₇²⁻ is
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Count electrons gained by two Cr atoms.
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Answer: C. 3
Each Cr goes from +6 to +3, so one Cr₂O₇²⁻ gains 6 electrons. 2I⁻ → I₂ + 2e⁻, so 6 electrons come from 6 I⁻, giving 3 I₂.
60. For H₂(g) + I₂(g) ⇌ 2HI(g), a 1 L vessel at equilibrium contains 0.2 mol H₂, 0.2 mol I₂ and 1.6 mol HI. The value of Kc is
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Raise each concentration to the power of its coefficient.
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Answer: A. 64
Kc = [HI]²/([H₂][I₂]) = (1.6)²/(0.2 × 0.2) = 2.56/0.04 = 64. (40 results from not squaring [HI].)
61. The equilibrium constant Kc for A ⇌ B is 4. The equilibrium constant for 2B ⇌ 2A at the same temperature is
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Reversing inverts K; multiplying the equation by n raises K to the power n.
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Answer: C. 1/16
Reversing the reaction gives 1/K = 1/4; doubling the coefficients squares it: (1/4)² = 1/16.
62. N₂O₄ is 20% dissociated into NO₂ at a total pressure of 1 atm according to N₂O₄(g) ⇌ 2NO₂(g). The value of Kp is about
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Find mole fractions using the total moles at equilibrium.
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Answer: D. 0.167 atm
Start with 1 mol N₂O₄: equilibrium 0.8 mol N₂O₄ and 0.4 mol NO₂, total 1.2 mol. p(NO₂) = 0.4/1.2 = 1/3 atm, p(N₂O₄) = 0.8/1.2 = 2/3 atm. Kp = (1/3)²/(2/3) = 1/6 ≈ 0.167 atm. (0.2 atm ignores the change in total moles.)
1.5 Volumetric analysis
12 questions
63. What is the equivalent mass of H₂SO₄ when it is completely neutralised by NaOH?
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Divide the molar mass by the basicity.
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Answer: A. 49
H₂SO₄ is dibasic, so its n-factor is 2. Equivalent mass = molar mass/2 = 98/2 = 49.
64. A 0.2 M solution of H₃PO₄ is used in a reaction where all three of its hydrogens are replaced. What is its normality?
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N = M × n-factor.
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Answer: B. 0.6 N
Normality = molarity × n-factor = 0.2 × 3 = 0.6 N.
65. 25 mL of a H₂SO₄ solution exactly neutralises 20 mL of 0.25 N NaOH. What is the strength of the acid in g/L?
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Find the normality first, then multiply by the equivalent mass.
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Answer: C. 9.8 g/L
N₁V₁ = N₂V₂ gives N(acid) = (0.25 × 20)/25 = 0.2 N. Strength = N × equivalent mass = 0.2 × 49 = 9.8 g/L.
66. Which of the following can be used as a primary standard in acid–base titration?
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Think of a pure, stable, non-hygroscopic solid.
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Answer: C. Anhydrous Na₂CO₃
A primary standard must be pure, stable and non-hygroscopic. Anhydrous sodium carbonate meets these conditions. NaOH absorbs moisture and CO₂, HCl is volatile and KMnO₄ is not obtained pure (it is also used in redox, not acid–base, work).
67. Which indicator is most suitable for titrating acetic acid against sodium hydroxide?
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Is the equivalence point acidic, neutral or basic?
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Answer: A. Phenolphthalein
A weak acid–strong base titration has its equivalence point in the alkaline range (pH ≈ 8–9), which matches the colour-change range of phenolphthalein (8.3–10).
68. Molality of a solution is defined as the number of
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Molality uses mass, not volume.
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Answer: D. moles of solute per kilogram of solvent
Molality (m) = moles of solute/mass of solvent in kg. Because it uses mass, it does not change with temperature.
69. 9 g of glucose (C₆H₁₂O₆) is dissolved in 250 g of water. What is the molality of the solution?
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Molar mass of glucose is 180 g/mol; convert the solvent mass to kg.
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Answer: A. 0.2 m
Moles of glucose = 9/180 = 0.05 mol. Molality = 0.05/0.250 kg = 0.2 m.
70. What is the equivalent mass of KMnO₄ (molar mass 158) when it acts as an oxidising agent in acidic medium?
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Find the change in oxidation state of Mn.
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Answer: D. 31.6
In acidic medium MnO₄⁻ → Mn²⁺, a change of 5 electrons. Equivalent mass = 158/5 = 31.6.
71. 200 mL of 0.1 M HCl is mixed with 300 mL of 0.2 M HCl. What is the molarity of the resulting solution?
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M = (M₁V₁ + M₂V₂)/(V₁ + V₂).
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Answer: D. 0.16 M
Total moles = 0.2 × 0.1 + 0.3 × 0.2 = 0.02 + 0.06 = 0.08 mol in 0.5 L. Molarity = 0.08/0.5 = 0.16 M.
72. How much water must be added to 200 mL of a 0.5 N solution to make it exactly 0.2 N?
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Find the final volume first.
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Answer: C. 300 mL
N₁V₁ = N₂V₂: 0.5 × 200 = 0.2 × V₂, so V₂ = 500 mL. Water to add = 500 − 200 = 300 mL.
73. What is the normality of a 0.1 M Na₂CO₃ solution when it is titrated to complete neutralisation with HCl?
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How many H⁺ ions does one CO₃²⁻ take up?
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Answer: D. 0.2 N
CO₃²⁻ accepts two H⁺ ions on complete neutralisation, so the n-factor is 2. N = 0.1 × 2 = 0.2 N.
74. 1.06 g of anhydrous Na₂CO₃ is dissolved in water and the solution is made up to 250 mL. What is its normality?
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Equivalent mass of Na₂CO₃ = 106/2.
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Answer: D. 0.08 N
Equivalent mass of Na₂CO₃ = 106/2 = 53. Gram equivalents = 1.06/53 = 0.02. Normality = 0.02/0.250 L = 0.08 N.
1.6 Ionic equilibrium, acids, bases and salts
16 questions
75. Which of the following is a Lewis acid but NOT a Brønsted–Lowry acid?
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A Brønsted acid must have a proton to donate.
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Answer: A. BF₃
BF₃ has an incomplete octet and accepts an electron pair (Lewis acid), but it has no H⁺ to donate, so it is not a Brønsted acid.
76. What is the conjugate base of HSO₄⁻?
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Remove one proton.
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Answer: D. SO₄²⁻
A conjugate base is formed when an acid loses one H⁺: HSO₄⁻ → H⁺ + SO₄²⁻.
77. What is the pH of a 0.001 M HCl solution at 25 °C?
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pH = −log[H⁺].
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Answer: D. 3
HCl is a strong acid, so [H⁺] = 10⁻³ M and pH = −log(10⁻³) = 3.
78. What is the pH of a 0.005 M Ba(OH)₂ solution at 25 °C, assuming complete dissociation?
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Count the OH⁻ ions per formula unit.
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Answer: B. 12
Each Ba(OH)₂ gives 2 OH⁻, so [OH⁻] = 0.01 M. pOH = 2, so pH = 14 − 2 = 12.
79. Pure water is heated from 25 °C to 60 °C. Which statement is correct?
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Ionisation of water absorbs heat; neutral means [H⁺] = [OH⁻].
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Answer: A. Kw increases and pH falls below 7, but the water is still neutral
Ionisation of water is endothermic, so Kw increases with temperature, [H⁺] rises and pH falls below 7. Since [H⁺] still equals [OH⁻], the water remains neutral.
80. Which of the following mixtures acts as a buffer solution?
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Look for a weak acid paired with its conjugate base.
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Answer: B. CH₃COOH + CH₃COONa
A buffer is a weak acid with its salt of a strong base (or a weak base with its salt). Acetic acid and sodium acetate form an acidic buffer; the others contain a strong acid or base with its own salt.
81. A solution is 0.1 M in CH₃COOH and 1.0 M in CH₃COONa. If pKa of acetic acid is 4.74, what is the pH of the solution?
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Use pH = pKa + log([salt]/[acid]).
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Answer: C. 5.74
Henderson equation: pH = pKa + log([salt]/[acid]) = 4.74 + log(1.0/0.1) = 4.74 + 1 = 5.74.
82. The solubility product of CaF₂ is 3.2×10⁻¹¹ at a certain temperature. What is its molar solubility?
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For an AB₂ salt, Ksp = 4s³.
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Answer: B. 2×10⁻⁴ M
CaF₂ ⇌ Ca²⁺ + 2F⁻; Ksp = s × (2s)² = 4s³. So s³ = 3.2×10⁻¹¹/4 = 8×10⁻¹², giving s = 2×10⁻⁴ M.
83. Ksp of AgCl is 1×10⁻¹⁰. What is the solubility of AgCl in a 0.01 M NaCl solution?
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Take [Cl⁻] from NaCl as fixed and solve for [Ag⁺].
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Answer: A. 1×10⁻⁸ M
Because of the common ion, [Cl⁻] ≈ 0.01 M. Then s = Ksp/[Cl⁻] = 10⁻¹⁰/10⁻² = 10⁻⁸ M, much lower than in pure water (10⁻⁵ M).
84. Solid NH₄Cl is added to an aqueous ammonia solution. What happens?
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Common ion effect.
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Answer: D. The pH decreases
NH₄⁺ is a common ion; it shifts NH₄OH ⇌ NH₄⁺ + OH⁻ to the left, lowering [OH⁻] and hence lowering the pH.
85. An aqueous solution of which salt is acidic?
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Identify the parent acid and base of each salt.
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Answer: C. NH₄Cl
NH₄Cl is a salt of a strong acid and a weak base; NH₄⁺ hydrolyses to give H₃O⁺, making the solution acidic.
86. An aqueous solution of ammonium acetate has a pH of about 7. Why?
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For a weak acid–weak base salt, pH depends on Ka and Kb.
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Answer: D. Ka of CH₃COOH is nearly equal to Kb of NH₄OH
Both ions hydrolyse, but because Ka(CH₃COOH) ≈ Kb(NH₄OH) ≈ 1.8×10⁻⁵, the H⁺ and OH⁻ produced balance and the pH stays near 7.
87. In an aqueous solution of pH 4 at 25 °C, what is the hydroxide ion concentration?
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Use Kw = [H⁺][OH⁻] = 10⁻¹⁴.
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Answer: A. 1×10⁻¹⁰ M
pOH = 14 − 4 = 10, so [OH⁻] = 10⁻¹⁰ M (because [H⁺][OH⁻] = 10⁻¹⁴).
88. 50 mL of 0.2 M HCl is mixed with 50 mL of 0.1 M NaOH. What is the pH of the mixture? (log 2 = 0.3)
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Find the excess H⁺ and divide by the total volume.
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Answer: B. 1.3
H⁺ = 50 × 0.2 = 10 mmol; OH⁻ = 50 × 0.1 = 5 mmol. Excess H⁺ = 5 mmol in 100 mL, so [H⁺] = 0.05 M. pH = −log(5×10⁻²) = 2 − 0.7 = 1.3.
89. What is the pH of a 0.1 M solution of a weak monobasic acid with Ka = 1×10⁻⁵?
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For a weak acid, [H⁺] = √(Ka C).
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Answer: B. 3
[H⁺] = √(Ka × C) = √(10⁻⁵ × 10⁻¹) = √10⁻⁶ = 10⁻³ M, so pH = 3.
90. Which species can act both as a Brønsted acid and as a Brønsted base?
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It must both have a proton to lose and be able to gain one.
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Answer: D. HCO₃⁻
HCO₃⁻ can donate H⁺ (to form CO₃²⁻) or accept H⁺ (to form H₂CO₃), so it is amphiprotic.
1.7 Electrochemistry
13 questions
91. What mass of copper will be deposited by passing 2 faradays of electricity through a solution of a copper(II) salt?
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How many faradays does one mole of Cu²⁺ need?
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Answer: C. 63.5 g
Cu²⁺ + 2e⁻ → Cu: 2 F deposit 1 mol of copper = 63.5 g.
92. According to Faraday's first law (m = ZIt), what is the electrochemical equivalent Z of a substance equal to? (E = equivalent mass, F = Faraday constant)
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Mass deposited by 1 coulomb.
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Answer: B. E/F
One faraday deposits one gram equivalent, so 1 coulomb deposits E/F grams. Hence Z = E/96500.
93. A current of 2 A is passed through CuSO₄ solution for 965 s. What mass of copper is deposited at the cathode?
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Convert charge to faradays; Cu²⁺ needs 2 electrons.
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Answer: A. 0.635 g
Q = 2 × 965 = 1930 C = 0.02 F. Cu²⁺ + 2e⁻ → Cu, so 0.02 F deposits 0.01 mol = 0.01 × 63.5 = 0.635 g.
94. AgNO₃ and CuSO₄ solutions are electrolysed in series. If 1.08 g of silver is deposited, what mass of copper is deposited?
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Same charge → masses proportional to equivalent masses.
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Answer: C. 0.3175 g
By Faraday's second law, masses are in the ratio of equivalent masses: m(Cu) = 1.08 × (31.75/108) = 0.3175 g.
95. What volume of hydrogen gas at NTP is liberated by passing 1 faraday of electricity through acidified water?
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How many electrons does one H₂ molecule need?
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Answer: D. 11.2 L
2H⁺ + 2e⁻ → H₂: 2 F gives 1 mol (22.4 L) of H₂, so 1 F gives 11.2 L.
96. Which statement about the Daniell cell (Zn | Zn²⁺ || Cu²⁺ | Cu) is correct?
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In a galvanic cell, oxidation occurs at the anode.
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Answer: C. Zinc is the anode and is the negative electrode
Zinc is oxidised (Zn → Zn²⁺ + 2e⁻), so it is the anode; in a galvanic cell the anode is negative. Electrons flow from Zn to Cu outside the cell.
97. Given E°(Zn²⁺/Zn) = −0.76 V and E°(Ag⁺/Ag) = +0.80 V, what is the standard emf of the cell Zn | Zn²⁺ || Ag⁺ | Ag?
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E°cell = E°(right) − E°(left), using reduction potentials.
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Answer: C. 1.56 V
E°cell = E°cathode − E°anode = 0.80 − (−0.76) = 1.56 V. Electrode potentials are not multiplied by the stoichiometric coefficient of 2 for Ag.
98. Which metal cannot liberate hydrogen from dilute hydrochloric acid?
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Check the sign of the standard reduction potential.
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Answer: D. Cu
Only metals placed above hydrogen in the electrochemical series (negative E°) can reduce H⁺ to H₂. Copper has E° = +0.34 V, so it cannot.
99. What value is assigned by convention to the standard electrode potential of the standard hydrogen electrode?
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It is the reference point of the series.
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Answer: C. 0.00 V
The SHE (1 atm H₂, 1 M H⁺, 298 K) is the reference electrode and its potential is taken as exactly zero.
100. Which of the following is the strongest reducing agent in aqueous solution?
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The more negative E°, the stronger the reducing agent.
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Answer: C. Li
Lithium has the most negative standard reduction potential (about −3.04 V), so it loses electrons most readily in water.
101. E°(Fe²⁺/Fe) = −0.44 V and E°(Cu²⁺/Cu) = +0.34 V. Can a CuSO₄ solution be stored in an iron vessel?
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Find E°cell for iron reducing Cu²⁺ and check its sign.
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Answer: B. No; iron displaces copper, E°cell = +0.78 V
For Fe + Cu²⁺ → Fe²⁺ + Cu, E°cell = 0.34 − (−0.44) = +0.78 V. A positive emf means the reaction is spontaneous, so the iron vessel would corrode.
102. How long must a current of 5 A be passed through molten Al₂O₃ (dissolved in cryolite) to deposit 2.7 g of aluminium?
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Al³⁺ needs 3 faradays per mole.
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Answer: D. 5790 s
2.7 g Al = 0.1 mol; Al³⁺ + 3e⁻ → Al needs 0.3 F = 0.3 × 96500 = 28950 C. t = Q/I = 28950/5 = 5790 s.
103. During electrolysis of concentrated aqueous NaCl with inert electrodes, what is formed at the cathode?
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Compare the reduction of Na⁺ with that of water.
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Answer: C. H₂ gas
Water (H⁺) is reduced in preference to Na⁺ because its reduction potential is much less negative, so H₂ is liberated at the cathode and NaOH remains in solution.
1.8 Energetics, kinetics and chemical bonding
20 questions
104. Hess's law of constant heat summation is a direct consequence of
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Think about why ΔH cannot depend on the path.
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Answer: D. the law of conservation of energy
Enthalpy is a state function; if ΔH depended on the path, energy could be created in a cycle, violating conservation of energy.
105. C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ; CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ. What is ΔH for C(s) + ½O₂(g) → CO(g)?
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Reverse the second equation and add it to the first.
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Answer: A. −110.5 kJ
Subtract the second equation from the first: ΔH = −393.5 − (−283.0) = −110.5 kJ.
106. Bond energies (kJ/mol): H–H = 436, Cl–Cl = 242, H–Cl = 431. What is ΔH for H₂(g) + Cl₂(g) → 2HCl(g)?
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ΔH = Σ(bonds broken) − Σ(bonds formed).
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Answer: C. −184 kJ
ΔH = bonds broken − bonds formed = (436 + 242) − 2 × 431 = 678 − 862 = −184 kJ (−92 kJ is per mole of HCl, not for the equation as written).
107. The enthalpy of neutralisation of HCl by NaOH is −57.3 kJ/mol. The enthalpy of neutralisation of CH₃COOH by NaOH is
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A weak acid must first ionise.
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Answer: B. slightly less negative, because some energy is used to ionise the weak acid
Acetic acid is only partly ionised; part of the heat released is used to ionise it, so the value is less negative (about −55.9 kJ/mol).
108. 200 mL of 0.5 M HCl is mixed with 300 mL of 0.2 M NaOH. How much heat is released? (Enthalpy of neutralisation = −57.3 kJ/mol)
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Only the limiting reagent decides the moles of water formed.
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Answer: D. 3.44 kJ
HCl = 0.1 mol, NaOH = 0.06 mol. NaOH is limiting, so 0.06 mol water forms. Heat = 0.06 × 57.3 ≈ 3.44 kJ.
109. What is the unit of the rate constant of a first-order reaction?
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Use (mol L⁻¹)¹⁻ⁿ s⁻¹.
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Answer: D. s⁻¹
For order n, the unit of k is (mol L⁻¹)¹⁻ⁿ s⁻¹; for n = 1 this is s⁻¹.
110. A first-order reaction has a half-life of 10 minutes. What percentage of the reactant has decomposed after 30 minutes?
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Count the half-lives that have passed.
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Answer: B. 87.5%
30 min = 3 half-lives, so the fraction left = (½)³ = 1/8 = 12.5%. Decomposed = 100 − 12.5 = 87.5%.
111. For the reaction with rate = k[A]²[B], [A] is doubled and [B] is halved at the same time. The rate becomes
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Substitute the new concentrations into the rate law.
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Answer: A. 2 times
New rate = k(2[A])²(½[B]) = 4 × ½ × k[A]²[B] = 2 × the original rate.
112. The half-life of a first-order reaction
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Write the expression for t½ of a first-order reaction.
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Answer: C. is independent of the initial concentration
For first order, t½ = 0.693/k, which contains no concentration term.
113. The rate constant of a first-order reaction is 6.93×10⁻³ s⁻¹. What is its half-life?
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t½ = 0.693/k.
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Answer: A. 100 s
t½ = 0.693/k = 0.693/(6.93×10⁻³) = 100 s.
114. Which statement about a catalyst is NOT true?
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Does a catalyst change K?
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Answer: D. It shifts the equilibrium position towards the products
A catalyst provides a path of lower activation energy and speeds both directions equally, so equilibrium is reached faster but its position (and K) is unchanged.
115. For a reaction, the activation energy of the forward step is 80 kJ/mol and ΔH = −30 kJ/mol. What is the activation energy of the backward reaction?
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ΔH = Ea(f) − Ea(b).
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Answer: D. 110 kJ/mol
ΔH = Ea(forward) − Ea(backward), so Ea(backward) = 80 − (−30) = 110 kJ/mol. For an exothermic reaction the backward barrier is higher.
116. For a zero-order reaction A → products, a plot of [A] against time is
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Write the integrated rate law for zero order.
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Answer: C. a straight line with slope −k
Integrated zero-order law: [A] = [A]₀ − kt, a straight line with intercept [A]₀ and slope −k.
117. According to VSEPR theory, what is the shape of the NH₃ molecule?
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Count bond pairs and lone pairs on N.
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Answer: D. Trigonal pyramidal
N has 3 bond pairs and 1 lone pair (4 electron pairs, tetrahedral arrangement); the lone pair makes the molecular shape trigonal pyramidal.
118. What is the hybridisation of sulphur in SF₆?
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Six electron pairs need six hybrid orbitals.
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Answer: B. sp³d²
S forms six bonds with no lone pair, needing six hybrid orbitals: sp³d², giving an octahedral shape.
119. Which of the following molecules has zero dipole moment?
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Symmetric shapes cancel bond dipoles.
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Answer: A. BF₃
BF₃ is trigonal planar and symmetric, so the three B–F bond dipoles cancel. The others are pyramidal, bent or unsymmetrical.
120. Which of the following species contains a coordinate (dative) bond?
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Look for a lone pair donated to an electron-deficient species.
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Answer: A. NH₄⁺
In NH₄⁺ the lone pair of N is donated to H⁺, forming a coordinate bond (all four N–H bonds become identical afterwards).
121. Water boils at 100 °C whereas H₂S, which has a higher molar mass, is a gas at room temperature. This is mainly due to
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Think of intermolecular forces involving O–H.
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Answer: A. hydrogen bonding between water molecules
O is small and highly electronegative, so water molecules are held together by strong intermolecular hydrogen bonds; S is not electronegative enough for this.
122. How many σ and π bonds are present in acrylonitrile, CH₂=CH–C≡N?
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Every bond has one σ; a double bond adds one π, a triple adds two.
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Answer: C. 6 σ and 3 π
σ bonds: 3 C–H + 1 (C=C) + 1 (C–C) + 1 (C≡N) = 6. π bonds: 1 in C=C and 2 in C≡N = 3.
123. Which pair of species has the same shape (is isostructural)?
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Count bond pairs and lone pairs on each central atom.
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Answer: A. NH₃ and H₃O⁺
NH₃ and H₃O⁺ each have 3 bond pairs and 1 lone pair, so both are trigonal pyramidal. BF₃ is planar vs pyramidal NH₃; XeF₂ is linear vs bent H₂O; SF₄ is see-saw vs tetrahedral CH₄.