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STPM Chemistry 962 practice questions

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Question 1Atomic Structure, Electronic Configuration and the Periodic Table

Which statement correctly describes the quantum numbers used to specify an orbital in an atom?

  1. The principal quantum number n determines the orbital's shape, while the azimuthal quantum number l fixes its orientation in space.
  2. The principal quantum number n determines the orbital's size and energy, and the azimuthal quantum number l determines its shape. Correct answer.
  3. The spin quantum number s fixes the orbital's orientation, and m sets its size.
  4. The magnetic quantum number m determines the orbital's shape, and l determines its spin state.

Explanation

The principal quantum number n sets the main energy level and the overall size of the orbital, while l (from 0 to n-1) fixes its shape (s, p, d, f). The magnetic quantum number m gives the orientation of the orbital and the spin quantum number describes electron spin, not shape or size.

Question 2Atomic Structure, Electronic Configuration and the Periodic Table

How many orbitals are present in the 3d subshell, and what is the maximum number of electrons it can hold?

  1. Seven orbitals holding fourteen electrons
  2. Three orbitals holding six electrons
  3. One orbital holding two electrons
  4. Five orbitals holding ten electrons Correct answer.

Explanation

For a d subshell l = 2, and the number of orbitals is 2l + 1 = 5. Each orbital holds two electrons with opposite spins, giving a maximum of 10 electrons.

Question 3Atomic Structure, Electronic Configuration and the Periodic Table

Which of the following is the correct ground-state electronic configuration of a chromium atom (Z = 24)?

  1. [Ar] 3d6
  2. [Ar] 3d4 4s2
  3. [Ar] 3d3 4s3
  4. [Ar] 3d5 4s1 Correct answer.

Explanation

Chromium shows an anomalous configuration: [Ar] 3d5 4s1. The half-filled d subshell together with a half-filled 4s orbital gives extra exchange stability compared with the expected [Ar] 3d4 4s2 arrangement.

Question 4Atomic Structure, Electronic Configuration and the Periodic Table

The first ionisation energies of Mg and Al are 738 and 578 kJ mol-1 respectively. Which explanation accounts for the lower value for Al?

  1. The electron removed from Al comes from a 3p orbital, which is higher in energy and better shielded than the 3s orbital of Mg. Correct answer.
  2. Al has a smaller atomic radius, so its outer electron is held more weakly.
  3. Al has more neutrons in its nucleus, which reduces the effective nuclear charge.
  4. Al has a larger nuclear charge than Mg, which attracts the outer electron more strongly.

Explanation

Aluminium loses a 3p electron whereas magnesium loses a 3s electron. The 3p orbital lies at higher energy and is shielded by the filled 3s subshell, so the 3p electron is easier to remove despite the greater nuclear charge.

Question 5Atomic Structure, Electronic Configuration and the Periodic Table

Across Period 3 from Na to Cl, which trend is observed for atomic radius and for first ionisation energy?

  1. Atomic radius increases and first ionisation energy generally increases.
  2. Atomic radius increases and first ionisation energy generally decreases.
  3. Atomic radius decreases and first ionisation energy generally increases. Correct answer.
  4. Atomic radius decreases and first ionisation energy generally decreases.

Explanation

Across a period the nuclear charge increases while electrons enter the same shell, so the effective nuclear charge rises. This pulls the outer electrons closer (radius decreases) and makes them harder to remove (ionisation energy rises).

Question 6Atomic Structure, Electronic Configuration and the Periodic Table

Which species has the electronic configuration 1s2 2s2 2p6?

  1. O2-
  2. Na+ Correct answer.
  3. F-
  4. Ne+

Explanation

Na+ has lost one electron from sodium (Z = 11), leaving 10 electrons, which gives 1s2 2s2 2p6. This is the same configuration as neon.

Question 7Atomic Structure, Electronic Configuration and the Periodic Table

Which pair of elements would be expected to show the greatest similarity in chemical properties?

  1. Na and Mg
  2. Cl and Ar
  3. K and Rb Correct answer.
  4. Ne and Na

Explanation

Potassium and rubidium are both Group 1 alkali metals, so they share the same valence configuration (ns1) and show very similar chemical behaviour. Elements in the same group have matching outer-shell arrangements.

Question 8Atomic Structure, Electronic Configuration and the Periodic Table

A neutral atom has the configuration [Ar] 3d10 4s2 4p3. In which group and period of the periodic table does this element lie?

  1. Group 5, Period 3
  2. Group 15, Period 4 Correct answer.
  3. Group 17, Period 4
  4. Group 13, Period 4

Explanation

The outer configuration is 4s2 4p3, giving 5 valence electrons and placing the element in Group 15. The highest occupied shell is n = 4, so the element is in Period 4 (this is arsenic).

Question 9Atomic Structure, Electronic Configuration and the Periodic Table

Which statement best explains why the first ionisation energy of chlorine is greater than that of bromine?

  1. The outer electron in chlorine is in a shell closer to the nucleus and is less shielded, so it is held more strongly. Correct answer.
  2. Chlorine has a larger atomic radius than bromine, so its electron is removed more easily.
  3. Chlorine has more electron shells than bromine, increasing its effective nuclear charge.
  4. Bromine has a lower nuclear charge than chlorine, which makes its electron harder to remove.

Explanation

Down a group the outer electron occupies a higher shell, farther from the nucleus and more shielded by inner electrons. In chlorine the 3p electron is closer and less shielded than the 4p electron in bromine, so it is held more strongly and requires more energy to remove.

Question 10Atomic Structure, Electronic Configuration and the Periodic Table

What is the shape of the p orbital and how many p orbitals make up a p subshell?

  1. Dumbbell-shaped, with three orbitals Correct answer.
  2. Tetrahedral, with two orbitals
  3. Dumbbell-shaped, with five orbitals
  4. Spherical, with one orbital

Explanation

A p orbital has a dumbbell (two-lobed) shape. For l = 1, the number of orbitals is 2l + 1 = 3, oriented along the x, y and z axes.

Question 11Atomic Structure, Electronic Configuration and the Periodic Table

Which set of quantum numbers (n, l, m, s) is allowed for an electron in an atom?

  1. (2, 0, 1, +1/2)
  2. (2, 1, -2, -1/2)
  3. (3, 2, -1, +1/2) Correct answer.
  4. (3, 3, 0, +1/2)

Explanation

For n = 3, l may be 0, 1 or 2, and for l = 2 the value of m may be -2, -1, 0, +1 or +2. The set (3, 2, -1, +1/2) satisfies l < n and -l <= m <= +l, so it is allowed.

Question 12Atomic Structure, Electronic Configuration and the Periodic Table

The successive ionisation energies of an element show a very large jump between the fourth and fifth ionisations. To which group does the element most likely belong?

  1. Group 14 Correct answer.
  2. Group 2
  3. Group 1
  4. Group 13

Explanation

A large jump occurs when an electron is removed from a new, inner shell after the valence electrons are exhausted. A jump after the fourth ionisation means the element has four valence electrons, placing it in Group 14.

Question 13Atomic Structure, Electronic Configuration and the Periodic Table

Which isoelectronic series is arranged in order of increasing ionic radius?

  1. F- < Na+ < Mg2+ < Al3+
  2. Mg2+ < Al3+ < F- < Na+
  3. Al3+ < Mg2+ < Na+ < F- Correct answer.
  4. Na+ < F- < Al3+ < Mg2+

Explanation

All four ions have 10 electrons. As nuclear charge increases from F (9) to Al (13), the same number of electrons is pulled in more strongly, so radius decreases. Increasing radius therefore follows Al3+ < Mg2+ < Na+ < F-.

Question 14Atomic Structure, Electronic Configuration and the Periodic Table

Which statement about the emission spectrum of a hydrogen atom is correct?

  1. It consists of a single line because hydrogen has only one electron.
  2. It is identical to the absorption spectrum of hydrogen in line positions.
  3. It is continuous because the electron can occupy any energy value.
  4. It consists of discrete lines because electrons move between quantised energy levels. Correct answer.

Explanation

Electrons in a hydrogen atom occupy quantised energy levels. When an electron falls from a higher to a lower level, a photon of fixed energy (and hence fixed frequency) is emitted, producing discrete lines rather than a continuum.

Question 15Atomic Structure, Electronic Configuration and the Periodic Table

Which element has the highest first ionisation energy?

  1. Si Correct answer.
  2. Na
  3. Mg
  4. Al

Explanation

Across Period 3 the effective nuclear charge increases while the outer electrons remain in the same shell, so ionisation energy generally rises. Silicon lies furthest right among these four, so it has the highest first ionisation energy.

Question 16Atomic Structure, Electronic Configuration and the Periodic Table

The ionic radii of Na+ and K+ are 102 pm and 138 pm respectively. Which explanation is correct?

  1. K+ has a smaller nuclear charge than Na+, so its electrons are held less tightly.
  2. Na+ has more electron shells than K+, making it smaller.
  3. K+ has an extra electron shell, so its outer electrons are further from the nucleus. Correct answer.
  4. Both ions have the same number of shells, but K+ has more neutrons.

Explanation

Potassium is below sodium in Group 1, so K+ has electrons in the n = 4 shell in addition to the inner shells. The extra shell places the outer electrons further from the nucleus, giving a larger ionic radius.

Question 17Atomic Structure, Electronic Configuration and the Periodic Table

Which of the following ground-state configurations violates the Pauli exclusion principle?

  1. 1s2 2s1
  2. 1s2 2s2 2p6
  3. 1s2 2s2 2p7 Correct answer.
  4. 1s2 2s2 2p6 3s1

Explanation

A p subshell contains only three orbitals, which can hold a maximum of six electrons. Placing seven electrons in 2p requires two electrons with identical quantum numbers in one orbital, violating the Pauli exclusion principle.

Question 18Atomic Structure, Electronic Configuration and the Periodic Table

Which factor best explains why the second ionisation energy of calcium is much greater than its first?

  1. The second ionisation involves removing a neutron from the nucleus.
  2. The second electron is removed from the same shell but with greater shielding.
  3. Calcium has a larger atomic radius after the first ionisation, so the electron is harder to remove.
  4. The second electron is removed from a shell closer to the nucleus, where it experiences stronger attraction. Correct answer.

Explanation

After the first electron is removed, Ca+ has the configuration [Ar] 4s1. Removing the second electron leaves the stable [Ar] core, and the electron is drawn from a shell much closer to the nucleus, so far more energy is required.

Question 19Atomic Structure, Electronic Configuration and the Periodic Table

An element X forms an ion X2- with the configuration 1s2 2s2 2p6 3s2 3p6. What is the ground-state configuration of the neutral atom X?

  1. 1s2 2s2 2p6 3s2 3p5
  2. 1s2 2s2 2p6 3s2 3p4 Correct answer.
  3. 1s2 2s2 2p6 3s2 3p6
  4. 1s2 2s2 2p6 3s2 3p6 4s2

Explanation

The ion X2- has gained two electrons compared with the neutral atom. Removing those two electrons from the 3p subshell leaves 1s2 2s2 2p6 3s2 3p4, which is the configuration of sulfur.

Question 20Atomic Structure, Electronic Configuration and the Periodic Table

Which property increases across Period 3 from sodium to chlorine?

  1. Atomic radius
  2. Metallic character
  3. Number of occupied electron shells
  4. Electronegativity Correct answer.

Explanation

Electronegativity increases across a period because the effective nuclear charge rises while the outer electrons stay in the same shell, so atoms attract bonding electrons more strongly. Sodium is the least and chlorine the most electronegative of these elements.

Question 21Chemical Bonding and Intermolecular Forces

Which statement correctly describes the shape of a molecule of boron trifluoride, BF₃, according to VSEPR theory?

  1. T-shaped, because two fluorine atoms occupy axial positions
  2. Trigonal pyramidal, because the boron atom carries a lone pair
  3. Bent, because one B–F bond is a double bond
  4. Trigonal planar, because the three bonding pairs around boron repel each other equally Correct answer.

Explanation

Boron in BF₃ has three valence electrons and forms three single bonds, giving three bonding pairs and no lone pair on boron. Three electron domains arrange themselves at 120° in a plane to minimise repulsion, so the molecule is trigonal planar with bond angles of 120°.

Question 22Chemical Bonding and Intermolecular Forces

Which of the following species has a bond angle closest to 180°?

  1. H₂O
  2. CO₂ Correct answer.
  3. NH₃
  4. SO₂

Explanation

Carbon dioxide has two electron domains around the central carbon and no lone pairs on carbon, so the domains adopt a linear arrangement with an O–C–O angle of 180°. The other species contain lone pairs on the central atom that compress the bond angles well below 180°.

Question 23Chemical Bonding and Intermolecular Forces

Which type of intermolecular force is primarily responsible for the relatively high boiling point of water compared with hydrogen sulfide, H₂S?

  1. London dispersion forces
  2. Dipole–dipole attractions
  3. Hydrogen bonding Correct answer.
  4. Ionic bonding

Explanation

Oxygen is highly electronegative and small, so the O–H bonds allow water molecules to form hydrogen bonds with each other. These hydrogen bonds require significant energy to break, raising the boiling point. Sulfur is larger and less electronegative, so H₂S forms only weaker dipole–dipole and dispersion forces.

Question 24Chemical Bonding and Intermolecular Forces

Which of the following molecules is non-polar overall despite containing polar bonds?

  1. CHCl₃
  2. CCl₄ Correct answer.
  3. H₂O
  4. NH₃

Explanation

In CCl₄ the four identical C–Cl bond dipoles are arranged tetrahedrally, so they cancel by symmetry and the molecule has no net dipole moment. The other molecules have bond dipoles that do not cancel because of their geometry or because the bonds are not identical.

Question 25Chemical Bonding and Intermolecular Forces

Which statement best explains why the melting point of magnesium oxide, MgO, is much higher than that of sodium chloride, NaCl?

  1. MgO has a lower lattice enthalpy because the ions are smaller
  2. MgO has stronger London dispersion forces between its molecules
  3. MgO contains ions of higher charge, giving stronger electrostatic attraction Correct answer.
  4. MgO has a giant covalent structure, whereas NaCl is ionic

Explanation

Both MgO and NaCl are ionic lattices, but Mg²⁺ and O²⁻ carry double the charge of Na⁺ and Cl⁻, and Mg²⁺ and O²⁻ are also smaller. The electrostatic attraction in the lattice is therefore much stronger, so more energy is needed to break the lattice.

Question 26Chemical Bonding and Intermolecular Forces

Which of the following substances is held together by a giant covalent network, giving it a very high melting point?

  1. Iodine
  2. Silicon dioxide Correct answer.
  3. Sodium chloride
  4. Carbon dioxide

Explanation

Silicon dioxide has a continuous three-dimensional network of covalent Si–O bonds, so melting requires breaking many strong covalent bonds. This gives it a very high melting point. The other substances are held together by weaker intermolecular or ionic forces.

Question 27Chemical Bonding and Intermolecular Forces

Which of the following correctly describes the hybridisation of the carbon atoms in ethene, C₂H₄, and the geometry around each carbon?

  1. sp² hybridised, trigonal planar Correct answer.
  2. sp² hybridised, tetrahedral
  3. sp³ hybridised, tetrahedral
  4. sp hybridised, linear

Explanation

Each carbon in ethene is bonded to two hydrogen atoms and one other carbon, giving three σ bonds and one π bond. The three σ bonds use sp² hybrid orbitals arranged at 120°, giving a trigonal planar geometry around each carbon.

Question 28Chemical Bonding and Intermolecular Forces

Which of the following lists the intermolecular forces present in liquid hydrogen fluoride, HF, in order of decreasing strength?

  1. London dispersion > hydrogen bonding > dipole–dipole
  2. Hydrogen bonding > dipole–dipole > London dispersion Correct answer.
  3. Dipole–dipole > hydrogen bonding > London dispersion
  4. London dispersion > dipole–dipole > hydrogen bonding

Explanation

Fluorine is the most electronegative element, so HF molecules form strong hydrogen bonds to one another. Dipole–dipole attractions between polar molecules are weaker than hydrogen bonds, and London dispersion forces, arising from temporary fluctuations in electron distribution, are the weakest of the three.

Question 29Chemical Bonding and Intermolecular Forces

The boiling points of the hydrides of Group 16 are: H₂O 100 °C, H₂S −60 °C, H₂Se −41 °C, H₂Te −2 °C. Which statement best explains the trend from H₂S to H₂Te?

  1. Hydrogen bonding becomes stronger as the central atom gets larger
  2. Electronegativity of the central atom increases down the group
  3. The molecules become more polar down the group
  4. London dispersion forces increase as the number of electrons increases Correct answer.

Explanation

From H₂S to H₂Te the molecules become larger and contain more electrons, so the temporary dipoles responsible for London dispersion forces become larger. These stronger dispersion forces require more energy to overcome, so boiling point rises. Water is anomalous because it also forms hydrogen bonds.

Question 30Chemical Bonding and Intermolecular Forces

Which of the following species has a bond angle of approximately 104.5°?

  1. CH₄
  2. H₂O Correct answer.
  3. NH₄⁺
  4. BF₃

Explanation

Water has four electron domains around oxygen: two bonding pairs and two lone pairs. The lone pairs repel more strongly than bonding pairs, compressing the H–O–H angle from the tetrahedral 109.5° to about 104.5°.

Question 31Chemical Bonding and Intermolecular Forces

Which statement explains why graphite conducts electricity while diamond does not?

  1. Diamond contains ionic bonds that trap the electrons
  2. Graphite has a lower melting point, so its electrons move more easily
  3. Graphite contains delocalised electrons within its layers, whereas all diamond electrons are localised in σ bonds Correct answer.
  4. Graphite contains free protons that carry the current

Explanation

In graphite each carbon is sp² hybridised and forms three σ bonds, leaving one electron in a p orbital that is delocalised across the layer. These delocalised electrons can move and carry current. In diamond each carbon is sp³ hybridised with four localised σ bonds, so there are no free electrons.

Question 32Chemical Bonding and Intermolecular Forces

Which of the following molecules has a net dipole moment of zero?

  1. H₂S
  2. BF₃ Correct answer.
  3. SO₂
  4. NH₃

Explanation

BF₃ is trigonal planar with three identical polar B–F bonds arranged at 120°; the bond dipoles cancel by symmetry so the net dipole moment is zero. The other molecules have lone pairs that make their geometries asymmetric, so their bond dipoles do not cancel.

Question 33Chemical Bonding and Intermolecular Forces

Which of the following best explains why the ionic radius of the Mg²⁺ ion is smaller than that of the Na⁺ ion?

  1. Mg²⁺ has a larger number of electrons, so the electron cloud expands
  2. Mg²⁺ has a greater nuclear charge and the same electron configuration, so the electrons are pulled in more strongly Correct answer.
  3. Mg²⁺ has a smaller nuclear charge, so the electrons are less attracted
  4. Mg²⁺ has more electron shells than Na⁺, so the outer electrons are further from the nucleus

Explanation

Mg²⁺ and Na⁺ are isoelectronic, both having the electron configuration of neon. Magnesium has one more proton, so the greater nuclear charge attracts the same number of electrons more strongly, reducing the ionic radius.

Question 34Chemical Bonding and Intermolecular Forces

Which of the following pairs of molecules is held together by hydrogen bonding in the pure liquid state?

  1. H₂S and HCl
  2. CO₂ and SO₂
  3. CH₄ and NH₃
  4. H₂O and CH₃OH Correct answer.

Explanation

Hydrogen bonding requires a hydrogen atom bonded to a highly electronegative atom (N, O or F) and a lone pair on such an atom in a neighbouring molecule. Both water and methanol have O–H groups, so they form hydrogen bonds in the liquid state.

Question 35Chemical Bonding and Intermolecular Forces

Which of the following best explains why the melting point of magnesium is higher than that of sodium?

  1. Magnesium has stronger London dispersion forces between discrete molecules
  2. Magnesium contributes two delocalised electrons per atom to the metallic bonding, giving stronger attraction Correct answer.
  3. Magnesium has a giant covalent structure
  4. Magnesium atoms are larger, so they pack more closely

Explanation

Both metals have giant metallic structures in which cations are held by attraction to delocalised electrons. Magnesium contributes two electrons per atom and has a smaller ionic radius, so the electrostatic attraction is stronger and more energy is needed to melt the lattice.

Question 36Chemical Bonding and Intermolecular Forces

Which of the following statements about the bond angle in the NH₂⁻ ion is correct?

  1. The angle is greater than 109.5° because the lone pairs spread out
  2. The angle is close to 104.5° because there are two lone pairs on nitrogen Correct answer.
  3. The angle is 180° because the ion is linear
  4. The angle is exactly 120° because the ion is trigonal planar

Explanation

The NH₂⁻ ion has four electron domains around nitrogen: two N–H bonding pairs and two lone pairs. The two lone pairs repel the bonding pairs more strongly than bonding pairs repel each other, compressing the H–N–H angle to approximately that of water, about 104.5°.

Question 37Chemical Bonding and Intermolecular Forces

A molecule XY₃ has a trigonal pyramidal shape. Which statement is consistent with this observation?

  1. The central atom X has three bonding pairs and one lone pair Correct answer.
  2. The central atom X has two bonding pairs and two lone pairs
  3. The central atom X has three bonding pairs and no lone pairs
  4. The central atom X has four bonding pairs and no lone pairs

Explanation

A trigonal pyramidal shape arises from four electron domains in which one is a lone pair and three are bonding pairs, as in ammonia. The lone pair occupies one vertex of the tetrahedron, leaving the three bonds pointing towards the corners of a triangular base.

Question 38Chemical Bonding and Intermolecular Forces

Which of the following correctly ranks the species in order of increasing bond angle?

  1. H₂O < NH₃ < CH₄ Correct answer.
  2. CH₄ < NH₃ < H₂O
  3. H₂O < CH₄ < NH₃
  4. NH₃ < H₂O < CH₄

Explanation

All three species have four electron domains around the central atom, so the ideal angle is 109.5°. Lone pairs repel more strongly than bonding pairs, so the angle decreases as the number of lone pairs increases: methane (0 lone pairs) 109.5°, ammonia (1 lone pair) 107°, water (2 lone pairs) 104.5°.

Question 39Chemical Bonding and Intermolecular Forces

Which statement best explains why the first ionisation energy of magnesium is higher than that of aluminium?

  1. Aluminium has a smaller atomic radius than magnesium, so its electron is easier to remove
  2. Aluminium has a larger nuclear charge, so its outer electron is held less tightly
  3. Magnesium has a full 3s subshell, making it harder to remove an electron than from aluminium
  4. The electron removed from aluminium is in a 3p orbital, which is higher in energy and better shielded than the 3s electron removed from magnesium Correct answer.

Explanation

Magnesium loses a 3s electron, whereas aluminium loses a 3p electron. The 3p orbital lies further from the nucleus and is shielded by the 3s electrons, so the 3p electron is easier to remove despite aluminium's larger nuclear charge. This accounts for the dip in ionisation energy between groups 2 and 13.

Question 40Chemical Bonding and Intermolecular Forces

Which of the following observations is best explained by the presence of hydrogen bonding?

  1. The dipole moment of HF is greater than that of HCl
  2. The boiling point of HF is higher than that of HCl Correct answer.
  3. The bond enthalpy of H–F is greater than that of H–Cl
  4. The boiling point of HCl is higher than that of HF

Explanation

Hydrogen fluoride molecules form hydrogen bonds because hydrogen is bonded to the highly electronegative fluorine atom. These strong intermolecular attractions raise the boiling point above that of HCl, whose molecules interact only through weaker dipole–dipole and dispersion forces.

Question 41Stoichiometry, Formulae and the Mole Concept

What is the mass of one mole of carbon-12 atoms, and how many atoms does it contain?

  1. 12 g and 6.02 × 10^22 atoms
  2. 24 g and 6.02 × 10^23 atoms
  3. 12 g and 6.02 × 10^23 atoms Correct answer.
  4. 1 g and 6.02 × 10^23 atoms

Explanation

By definition, one mole of carbon-12 has a mass of exactly 12 g and contains the Avogadro number of particles, 6.02 × 10^23. The molar mass in grams per mole is numerically equal to the relative atomic mass.

Question 42Stoichiometry, Formulae and the Mole Concept

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula? (Ar: C = 12, H = 1, O = 16)

  1. C2H6O
  2. CHO
  3. C2H4O2
  4. CH2O Correct answer.

Explanation

Dividing each percentage by the atomic mass gives C : H : O = 3.33 : 6.7 : 3.33, which simplifies to 1 : 2 : 1. The empirical formula is therefore CH2O.

Question 43Stoichiometry, Formulae and the Mole Concept

How many moles of ions are produced when 0.50 mol of aluminium sulfate, Al2(SO4)3, is completely dissolved in water?

  1. 2.50 mol Correct answer.
  2. 1.00 mol
  3. 0.50 mol
  4. 5.00 mol

Explanation

Each formula unit gives 2 Al3+ ions and 3 SO4^2- ions, a total of 5 ions. Therefore 0.50 mol of Al2(SO4)3 yields 0.50 × 5 = 2.50 mol of ions.

Question 44Stoichiometry, Formulae and the Mole Concept

Which statement correctly describes the molar volume of an ideal gas at s.t.p. (273 K and 1 atm)?

  1. It is 22.4 dm3 mol-1 but only for diatomic gases.
  2. It is 22.4 dm3 mol-1 only at 298 K.
  3. It is 24.0 dm3 mol-1 and depends on the gas.
  4. It is 22.4 dm3 mol-1 and is the same for all gases. Correct answer.

Explanation

Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of moles. At s.t.p. this volume is 22.4 dm3 per mole, independent of the identity of the gas.

Question 45Stoichiometry, Formulae and the Mole Concept

What is the concentration, in mol dm-3, of a solution made by dissolving 4.0 g of sodium hydroxide in water and making the volume up to 250 cm3? (Mr of NaOH = 40)

  1. 0.10 mol dm-3
  2. 1.0 mol dm-3
  3. 0.40 mol dm-3 Correct answer.
  4. 4.0 mol dm-3

Explanation

The amount of NaOH is 4.0 / 40 = 0.10 mol. The volume is 250 cm3 = 0.250 dm3, so the concentration is 0.10 / 0.250 = 0.40 mol dm-3.

Question 46Stoichiometry, Formulae and the Mole Concept

Hydrated copper(II) sulfate, CuSO4·5H2O, has Mr = 250. What is the percentage by mass of water of crystallisation in the compound? (Mr of H2O = 18)

  1. 64.0%
  2. 25.0%
  3. 18.0%
  4. 36.0% Correct answer.

Explanation

Five water molecules contribute 5 × 18 = 90 to the molar mass of 250. The percentage is (90 / 250) × 100 = 36.0%.

Question 47Stoichiometry, Formulae and the Mole Concept

In the reaction 2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O, what volume of chlorine at s.t.p. is produced from 0.020 mol of KMnO4?

  1. 2.24 dm3
  2. 5.60 dm3
  3. 1.12 dm3 Correct answer.
  4. 0.56 dm3

Explanation

The equation shows 2 mol KMnO4 give 5 mol Cl2, so 0.020 mol gives 0.050 mol Cl2. At s.t.p. this is 0.050 × 22.4 = 1.12 dm3.

Question 48Stoichiometry, Formulae and the Mole Concept

25.0 cm3 of a 0.100 mol dm-3 solution of a diprotic acid H2A is exactly neutralised by 20.0 cm3 of sodium hydroxide solution. What is the concentration of the sodium hydroxide?

  1. 0.125 mol dm-3
  2. 0.0625 mol dm-3
  3. 0.250 mol dm-3 Correct answer.
  4. 0.500 mol dm-3

Explanation

Moles of H2A = 0.0250 × 0.100 = 2.50 × 10^-3 mol, which can supply 5.00 × 10^-3 mol of H+. This needs 5.00 × 10^-3 mol NaOH in 0.0200 dm3, giving 0.250 mol dm-3.

Question 49Stoichiometry, Formulae and the Mole Concept

A gaseous hydrocarbon contains 85.7% carbon by mass and has a molar mass of 42 g mol-1. What is its molecular formula? (Ar: C = 12, H = 1)

  1. C3H6 Correct answer.
  2. CH2
  3. C2H4
  4. C3H8

Explanation

The empirical ratio is C : H = 85.7/12 : 14.3/1 = 7.14 : 14.3 = 1 : 2, so the empirical formula is CH2 with mass 14. The molar mass 42 gives n = 3, so the molecular formula is C3H6.

Question 50Stoichiometry, Formulae and the Mole Concept

When 5.00 g of a carbonate of a divalent metal, MCO3, is heated it loses 2.20 g of carbon dioxide. What is the relative atomic mass of M? (Mr of CO2 = 44)

  1. 40 Correct answer.
  2. 24
  3. 65
  4. 56

Explanation

Moles of CO2 lost = 2.20 / 44 = 0.0500 mol, equal to the moles of MCO3 decomposed. The molar mass of MCO3 is 5.00 / 0.0500 = 100 g mol-1, so Ar(M) = 100 − 60 = 40.