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GTLE Subject Specialism Chemistry study materials
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GTLE Subject Specialism Chemistry practice questions

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Question 1Atomic Structure and Periodicity

Which subatomic particle determines the identity of an element?

  1. The nucleon count
  2. The proton Correct answer.
  3. The electron
  4. The neutron

Explanation

The atomic number Z equals the number of protons in the nucleus, and this value uniquely defines the element. Changing the proton count converts the atom into a different element, whereas changing neutrons or electrons does not.

Question 2Atomic Structure and Periodicity

An atom of an element has the ground-state configuration 1s² 2s² 2p⁶ 3s² 3p⁴. In which block of the periodic table does this element appear?

  1. p-block Correct answer.
  2. d-block
  3. f-block
  4. s-block

Explanation

The differentiating electron enters a 3p orbital, and elements whose highest-energy electrons occupy p orbitals are placed in the p-block. The configuration corresponds to sulfur, which sits in group 16 of the p-block.

Question 3Atomic Structure and Periodicity

Which statement correctly describes the trend in first ionisation energy across period 3 from sodium to argon?

  1. It increases smoothly with no irregularities.
  2. It increases overall, with a dip at aluminium and at sulfur. Correct answer.
  3. It decreases steadily because atomic radius increases.
  4. It remains constant because shielding by inner shells is unchanged.

Explanation

Across a period the nuclear charge rises while shielding stays roughly constant, so ionisation energy generally increases. Dips occur at aluminium, where the electron removed comes from a higher-energy 3p orbital, and at sulfur, where removing one 3p electron relieves electron-pair repulsion.

Question 4Atomic Structure and Periodicity

Which ground-state electron configuration is correct for a chromium atom (Z = 24)?

  1. [Ar] 3d⁵ 4s¹ Correct answer.
  2. [Ar] 3d⁶ 4s⁰
  3. [Ar] 3d³ 4s² 4p¹
  4. [Ar] 3d⁴ 4s²

Explanation

Chromium is one of the exceptions to the aufbau order: a half-filled 3d subshell with a single 4s electron is lower in energy than 3d⁴ 4s² because exchange energy and reduced 3d–4s repulsion stabilise it. Experimental evidence confirms [Ar] 3d⁵ 4s¹.

Question 5Atomic Structure and Periodicity

Which property increases from left to right across a period of the periodic table?

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

Explanation

Effective nuclear charge increases across a period while the number of shells stays the same, so valence electrons are held more tightly and first ionisation energy rises. Atomic radius and metallic character both decrease in the same direction.

Question 6Atomic Structure and Periodicity

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

  1. Group 17
  2. Group 2
  3. Group 14 Correct answer.
  4. Group 1

Explanation

A large jump after four electrons removed means the fifth electron must come from a new, inner shell, so the atom has four valence electrons. Four valence electrons place the element in group 14, such as carbon, silicon or germanium.

Question 7Atomic Structure and Periodicity

Which species has the ground-state electron configuration 1s² 2s² 2p⁶?

  1. F⁻
  2. Mg²⁺ Correct answer.
  3. Na⁺
  4. Ne⁺

Explanation

A magnesium atom (Z = 12) loses two electrons to form Mg²⁺, leaving 10 electrons, which gives the configuration 1s² 2s² 2p⁶. This is isoelectronic with neon.

Question 8Atomic Structure and Periodicity

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

  1. Aluminium has more neutrons in its nucleus, which weakens the nuclear attraction for electrons.
  2. Magnesium has a smaller atomic radius, so its outer electron is further from the nucleus.
  3. Aluminium has a larger nuclear charge than magnesium, which reduces the energy needed to remove an electron.
  4. The electron removed from aluminium comes from a 3p orbital, which is higher in energy and better shielded than the 3s orbital used in magnesium. Correct answer.

Explanation

Magnesium loses a 3s electron, whereas aluminium loses a 3p electron. The 3p orbital lies higher in energy and is shielded by the filled 3s subshell, so aluminium's outer electron is easier to remove despite the larger nuclear charge.

Question 9Atomic Structure and Periodicity

Which pair of elements belongs to the same group of the periodic table?

  1. Carbon and nitrogen
  2. Sodium and magnesium
  3. Potassium and rubidium Correct answer.
  4. Chlorine and argon

Explanation

Group membership is defined by the number of valence electrons. Potassium and rubidium both have one valence electron and occupy group 1, so they share chemical properties.

Question 10Atomic Structure and Periodicity

Which of these is the correct order of increasing first ionisation energy?

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

Explanation

Across period 3 the ionisation energy generally rises, but aluminium sits below magnesium because its outer electron is removed from a 3p orbital. The correct sequence is therefore Na < Al < Mg < Si.

Question 11Atomic Structure and Periodicity

Which statement about the emission spectrum of atomic hydrogen is correct?

  1. It is a continuous spectrum because the electron can have any energy.
  2. It consists of discrete lines whose frequencies correspond to energy differences between quantised levels. Correct answer.
  3. The lines get closer together at high energy because the levels become more widely spaced.
  4. It contains only the Balmer series, which lies in the ultraviolet region.

Explanation

Electron energies in a hydrogen atom are quantised, so photons are emitted only at frequencies that match differences between levels. This produces the characteristic line spectrum rather than a continuum.

Question 12Atomic Structure and Periodicity

The first five ionisation energies of an element are 738, 1451, 7733, 10540 and 13630 kJ mol⁻¹. Which element is most consistent with these data?

  1. Magnesium Correct answer.
  2. Silicon
  3. Aluminium
  4. Sodium

Explanation

The jump between the second and third ionisation energies indicates that only two electrons are removed easily before an inner shell is breached. Two valence electrons identify the element as a group 2 metal, consistent with magnesium.

Question 13Atomic Structure and Periodicity

Which factor is primarily responsible for the decrease in first ionisation energy down group 1?

  1. The nuclear charge decreases down the group.
  2. The outer electron becomes more tightly bound because of the larger nucleus.
  3. The number of valence electrons increases down the group.
  4. The outer electron occupies a shell further from the nucleus, and shielding increases. Correct answer.

Explanation

Down a group each element has an extra filled shell, which both increases the distance between the nucleus and the valence electron and increases shielding. The effective nuclear charge felt by the valence electron therefore falls, so less energy is required to remove it.

Question 14Atomic Structure and Periodicity

Which ion has the largest ionic radius?

  1. S²⁻
  2. Cl⁻
  3. K⁺
  4. P³⁻ Correct answer.

Explanation

P³⁻, S²⁻, Cl⁻ and K⁺ are all isoelectronic with 18 electrons, so their sizes depend on nuclear charge. Phosphorus has the smallest nuclear charge (Z = 15), so it attracts the electrons least and the ion is largest.

Question 15Atomic Structure and Periodicity

Which sublevel is filled immediately after the 4s sublevel in the ground-state configuration of a potassium atom (Z = 19)?

  1. 3p
  2. 4d
  3. 4p
  4. 3d Correct answer.

Explanation

For potassium the 4s orbital is lower in energy than 3d, so the configuration is [Ar] 4s¹. The next electron added goes into the 3d sublevel, which fills from scandium onwards.

Question 16Atomic Structure and Periodicity

Which statement about the relative energies of the 3d and 4s orbitals in a first-row transition metal is correct?

  1. The 4s orbital is lower in energy for the free atom but the 3d orbitals become lower in energy once the atom is ionised. Correct answer.
  2. The 4s orbital is always higher in energy than 3d, which is why 4s fills first.
  3. The 3d orbital is always lower in energy than the 4s orbital, regardless of ionisation.
  4. The 4s and 3d orbitals are exactly degenerate in all transition metals.

Explanation

For isolated first-row transition-metal atoms the 4s orbital is occupied first because it is slightly lower in energy. On ionisation, electrons are removed from 4s before 3d because the 3d orbitals drop below 4s once the atom loses electrons.

Question 17Atomic Structure and Periodicity

Which of the following gives the correct number of protons, neutrons and electrons in the ion ⁵⁶Fe³⁺?

  1. 30 protons, 26 neutrons, 23 electrons
  2. 26 protons, 30 neutrons, 23 electrons Correct answer.
  3. 26 protons, 30 neutrons, 29 electrons
  4. 23 protons, 30 neutrons, 26 electrons

Explanation

Iron has atomic number 26, so the ion contains 26 protons; the mass number 56 gives 56 − 26 = 30 neutrons. The 3+ charge means three electrons have been lost, leaving 26 − 3 = 23 electrons.

Question 18Atomic Structure and Periodicity

Which statement correctly describes the trend in electronegativity across period 2 from lithium to fluorine?

  1. It increases because effective nuclear charge increases while the number of shells stays constant. Correct answer.
  2. It remains constant because all period 2 elements have the same valence shell.
  3. It increases because the number of occupied shells increases.
  4. It decreases because atomic radius increases.

Explanation

Electronegativity measures the ability of an atom to attract a shared pair of electrons. Across period 2 the effective nuclear charge rises while the valence shell remains the same, so the attraction for bonding electrons increases from lithium to fluorine.

Question 19Atomic Structure and Periodicity

A hydrogen atom in the n = 4 level emits a photon and drops to the n = 2 level. Which series does this emission line belong to?

  1. The Brackett series
  2. The Balmer series Correct answer.
  3. The Paschen series
  4. The Lyman series

Explanation

Transitions ending at n = 2 produce the Balmer series, which lies mainly in the visible region. A drop from n = 4 to n = 2 is one of the Balmer lines.

Question 20Atomic Structure and Periodicity

Which statement best explains why the second ionisation energy of an element is always greater than its first ionisation energy?

  1. The second ionisation energy includes the energy released when the first electron is removed.
  2. The second electron has more kinetic energy because the atom has been heated.
  3. The second electron comes from a higher-energy shell in every case.
  4. The second electron is removed from a positively charged ion, so the electrostatic attraction it experiences is stronger. Correct answer.

Explanation

Once one electron has been removed, the remaining species is a cation and the electrostatic attraction on the next electron is greater. Removing a second electron therefore requires more energy, regardless of which shell it occupies.

Question 21Chemical Bonding and Intermolecular Forces

Which statement best describes the bonding in solid sodium chloride?

  1. Non-polar molecules held by London dispersion forces
  2. A metallic lattice with delocalised electrons
  3. Discrete NaCl molecules held by covalent bonds
  4. A lattice of Na+ and Cl- ions held together by electrostatic attraction Correct answer.

Explanation

Sodium chloride forms an ionic lattice in which each sodium atom has transferred an electron to chlorine, giving Na+ and Cl- ions. The lattice is stabilised by the electrostatic attraction between oppositely charged ions acting in all directions. There are no discrete NaCl molecules in the solid.

Question 22Chemical Bonding and Intermolecular Forces

Which molecule has a net dipole moment of zero?

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

Explanation

CO2 is linear, so its two polar C=O bond dipoles are equal and point in exactly opposite directions, cancelling to zero. NH3 and H2O are bent and HCl is linear but has only one polar bond, so each has a non-zero resultant dipole. Molecular geometry, not just bond polarity, determines the net dipole.

Question 23Chemical Bonding and Intermolecular Forces

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

  1. Ion-dipole interactions
  2. Hydrogen bonding Correct answer.
  3. Metallic bonding
  4. London dispersion forces

Explanation

Oxygen is much more electronegative than sulfur and is small, so water forms extensive hydrogen bonds between molecules. Hydrogen sulfide lacks these strong directional interactions and boils far lower. Hydrogen bonding requires H attached to N, O or F.

Question 24Chemical Bonding and Intermolecular Forces

Which species has a trigonal planar shape according to VSEPR theory?

  1. BF3 Correct answer.
  2. PCl3
  3. H2O
  4. NH3

Explanation

BF3 has three bonding pairs and no lone pairs on boron, giving three electron domains that adopt a 120° trigonal planar arrangement. The other species all contain a lone pair on the central atom, which distorts them from trigonal planar geometry.

Question 25Chemical Bonding and Intermolecular Forces

Which property is characteristic of substances held together by London dispersion forces only?

  1. Strong directional bonding between neighbouring molecules
  2. High electrical conductivity in the solid state
  3. Low melting points that increase with molecular size Correct answer.
  4. Very high melting points arising from electrostatic attraction

Explanation

London dispersion forces are weak, instantaneous dipole-induced dipole attractions, so such substances melt at low temperatures. Because polarisability rises with the number of electrons, dispersion forces and melting points increase with molecular size. There is no directional or electrostatic lattice stabilisation in these molecular solids.

Question 26Chemical Bonding and Intermolecular Forces

Which of the following is an example of a network covalent solid?

  1. Diamond Correct answer.
  2. Sodium chloride
  3. Copper
  4. Iodine

Explanation

Diamond consists of carbon atoms each covalently bonded to four neighbours in a continuous three-dimensional lattice. This giant covalent network gives diamond extreme hardness and a very high melting point. The other substances are ionic, molecular or metallic respectively.

Question 27Chemical Bonding and Intermolecular Forces

According to molecular orbital theory, why is the bond order of O2 equal to 2?

  1. Eight valence electrons completely fill the bonding orbitals
  2. Two electrons occupy the sigma 2p orbital and four occupy the pi 2p orbitals
  3. Six bonding electrons and two antibonding electrons give a net of four bonding electrons Correct answer.
  4. Two unpaired electrons occupy antibonding orbitals

Explanation

For O2 the molecular orbital configuration has six electrons in bonding orbitals and two in antibonding pi* orbitals. Bond order equals half the difference between bonding and antibonding electrons, giving (6 - 2)/2 = 2. This also explains the two unpaired electrons that make O2 paramagnetic.

Question 28Chemical Bonding and Intermolecular Forces

Which molecule is expected to be paramagnetic in the gas phase?

  1. CO
  2. N2
  3. O2 Correct answer.
  4. F2

Explanation

O2 has two degenerate antibonding pi* orbitals each containing one unpaired electron, making the molecule paramagnetic. N2, F2 and CO all have their valence molecular orbitals filled in pairs, so they are diamagnetic. Paramagnetism is a direct consequence of unpaired electrons in the molecular orbital scheme.

Question 29Chemical Bonding and Intermolecular Forces

Why does the boiling point of the noble gases increase from helium to radon?

  1. Hydrogen bonding becomes possible for the heavier gases
  2. Electronegativity increases, strengthening permanent dipoles
  3. The number of polarisable electrons increases, strengthening London dispersion forces Correct answer.
  4. Covalent bond strength within each atom increases

Explanation

Noble gas atoms are non-polar, so only London dispersion forces act between them. As the electron count and atomic size increase down the group, polarisability rises and the instantaneous dipoles become stronger. Hence more energy is needed to separate the atoms, raising the boiling point.

Question 30Chemical Bonding and Intermolecular Forces

Which pair of molecules can form hydrogen bonds with each other?

  1. Methane and water
  2. Chlorine and bromine
  3. Ethanol and water Correct answer.
  4. Carbon dioxide and methane

Explanation

Hydrogen bonding requires a hydrogen atom covalently bonded to N, O or F acting as donor, and a lone pair on N, O or F acting as acceptor. Ethanol has an O-H group and water has both O-H donors and oxygen lone pairs, so hydrogen bonds form between them. The other pairs lack a suitable donor.

Question 31Chemical Bonding and Intermolecular Forces

The bond angle in water is approximately 104.5°, while the ideal tetrahedral angle is 109.5°. What is the main reason for this reduction?

  1. Oxygen has a higher electronegativity than hydrogen
  2. The O-H bonds are longer than typical tetrahedral bonds
  3. Repulsion from two lone pairs on oxygen is greater than bonding-pair repulsion Correct answer.
  4. The molecule is linear and the angle is set by geometry

Explanation

In VSEPR theory, lone pairs occupy more angular space than bonding pairs because they are held by only one nucleus. In water the two lone pairs on oxygen repel the bonding pairs more strongly, compressing the H-O-H angle below the tetrahedral value. The electronegativity difference explains polarity, not the angle compression.

Question 32Chemical Bonding and Intermolecular Forces

Which statement correctly describes the bonding in a metal such as magnesium?

  1. Cations are surrounded by a sea of delocalised valence electrons Correct answer.
  2. Electrons are localised in covalent bonds between adjacent atoms
  3. Discrete Mg2 molecules are held by dispersion forces
  4. Anions and cations alternate in a rigid ionic lattice

Explanation

Metallic bonding is modelled as a lattice of positive ions immersed in a delocalised sea of valence electrons. This delocalisation explains electrical and thermal conductivity and malleability. The bonding is non-directional and not ionic or molecular.

Question 33Chemical Bonding and Intermolecular Forces

Which explanation best accounts for the very high melting point of magnesium oxide compared with sodium chloride?

  1. MgO has stronger London dispersion forces between formula units
  2. MgO contains hydrogen bonds between oxide ions
  3. MgO has a higher lattice enthalpy because its ions are doubly charged and smaller Correct answer.
  4. MgO contains covalent bonds between Mg and O atoms

Explanation

Lattice enthalpy increases with the product of the ionic charges and decreases with ion size. Mg2+ and O2- carry double charges and are smaller than Na+ and Cl-, so the electrostatic attraction in the MgO lattice is much stronger. More energy is therefore needed to melt MgO.

Question 34Chemical Bonding and Intermolecular Forces

Which molecule has a bond angle closest to 120°?

  1. NH3
  2. CH4
  3. H2O
  4. BF3 Correct answer.

Explanation

BF3 has three bonding domains and no lone pairs, so VSEPR predicts a trigonal planar arrangement with 120° angles. CH4 is tetrahedral at about 109.5°, while NH3 and H2O are compressed below that by lone-pair repulsion.

Question 35Chemical Bonding and Intermolecular Forces

Which factor best explains why the ionic radius of Na+ is smaller than that of the Na atom?

  1. Loss of the valence electron reduces electron-electron repulsion while the nuclear charge stays the same Correct answer.
  2. The nucleus loses a proton when the electron is removed
  3. The ion gains an extra electron, increasing the effective nuclear charge
  4. The remaining electrons occupy higher-energy orbitals

Explanation

A sodium atom loses its single 3s electron to form Na+, leaving the filled 2p shell as the outermost level. The nuclear charge is unchanged but one whole shell and one electron are removed, so electron-electron repulsion falls and each remaining electron experiences a stronger effective nuclear charge. The ion is therefore smaller than the atom.

Question 36Chemical Bonding and Intermolecular Forces

Which observation provides the strongest evidence that a compound is ionic rather than molecular?

  1. It dissolves in a non-polar solvent
  2. It melts at a low temperature
  3. Its aqueous solution conducts electricity Correct answer.
  4. It exists as a gas at room temperature

Explanation

Ionic compounds dissociate into mobile ions in water, so their aqueous solutions conduct electricity. Molecular substances generally do not produce mobile ions on dissolving. Low melting points, solubility in non-polar solvents and volatility are all more typical of molecular substances.

Question 37Chemical Bonding and Intermolecular Forces

Which molecule has the largest permanent dipole moment?

  1. CO2
  2. CHCl3 Correct answer.
  3. CCl4
  4. CH4

Explanation

In CHCl3 the C-H and C-Cl bond dipoles are not symmetrically arranged, so they add to give a substantial net dipole. CH4, CCl4 and CO2 are highly symmetric, so their individual bond dipoles cancel almost completely. The result is that chloroform has by far the largest dipole moment of the four.

Question 38Chemical Bonding and Intermolecular Forces

Which statement about the bonding in the carbonate ion, CO3 2-, is correct?

  1. It is held together only by London dispersion forces
  2. It contains three C-O single bonds and a lone pair on carbon
  3. It contains one short C=O double bond and two long C-O single bonds
  4. It contains three identical C-O bonds with bond order about 1.33 Correct answer.

Explanation

In carbonate the carbon is bonded to three oxygen atoms with no lone pair, giving a trigonal planar shape. The pi electrons are delocalised over all three C-O positions, so the bonds are identical with a bond order between one and two, about 1.33. Resonance averaging explains the equal bond lengths observed.

Question 39Chemical Bonding and Intermolecular Forces

Which comparison of the bond order of the species N2, N2+ and N2- is correct?

  1. N2- has the highest bond order and N2+ the lowest
  2. All three species have the same bond order
  3. N2+ has the highest bond order and N2 the lowest
  4. N2 has the highest bond order and N2- the lowest Correct answer.

Explanation

N2 has eight valence electrons in bonding orbitals and two in the sigma* antibonding orbital, giving a bond order of 3. Removing an electron from the bonding sigma orbital in N2+ gives a bond order of 2.5, while adding an electron to a pi* antibonding orbital in N2- gives a bond order of 2.5 as well, but its higher antibonding population lowers the net bond order below that of N2. Hence N2 has the highest bond order and N2- the lowest of the three.

Question 40Chemical Bonding and Intermolecular Forces

Which factor best explains why the melting point of MgO is higher than that of CaO, even though both are ionic oxides?

  1. MgO has stronger London dispersion forces between ions
  2. CaO contains covalent bonds that reduce its stability
  3. Mg2+ is smaller than Ca2+, so the lattice enthalpy of MgO is greater Correct answer.
  4. CaO has a higher charge product than MgO

Explanation

Both oxides contain O2- ions, but Mg2+ has a smaller ionic radius than Ca2+. Lattice enthalpy is proportional to the product of ionic charges divided by the sum of ionic radii, so the smaller interionic distance in MgO produces a stronger lattice and a higher melting point. The charge product is the same for both compounds.

Question 41Stoichiometry and Chemical Calculations

What is the mass of one mole of calcium carbonate, CaCO3? (Ar: Ca = 40, C = 12, O = 16)

  1. 84 g
  2. 56 g
  3. 68 g
  4. 100 g Correct answer.

Explanation

The molar mass is the sum of the relative atomic masses of all atoms in the formula: 40 + 12 + (3 × 16) = 100, so one mole of CaCO3 has a mass of 100 g.

Question 42Stoichiometry and Chemical Calculations

How many moles of particles are present in 1.204 × 10^24 formula units of sodium chloride? (Avogadro constant L = 6.02 × 10^23 mol^-1)

  1. 6.0 mol
  2. 1.0 mol
  3. 0.5 mol
  4. 2.0 mol Correct answer.

Explanation

The amount in moles equals the number of particles divided by the Avogadro constant: 1.204 × 10^24 ÷ 6.02 × 10^23 = 2.0 mol.

Question 43Stoichiometry and Chemical Calculations

In the balanced equation 2H2 + O2 → 2H2O, what is the mole ratio of hydrogen to oxygen consumed?

  1. 2:1 Correct answer.
  2. 4:1
  3. 1:2
  4. 1:1

Explanation

The coefficients in a balanced equation give the reacting mole ratio directly. Hydrogen has coefficient 2 and oxygen has coefficient 1, so the ratio H2:O2 is 2:1.

Question 44Stoichiometry and Chemical Calculations

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. C2H4O
  2. CHO
  3. CH2O Correct answer.
  4. C2H6O

Explanation

Dividing each percentage by the atomic mass gives mole ratios C 3.33, H 6.7, O 3.33; dividing by the smallest (3.33) gives 1:2:1, so the empirical formula is CH2O.

Question 45Stoichiometry and Chemical Calculations

What volume of 0.200 mol dm^-3 sodium hydroxide solution is required to neutralise exactly 25.0 cm^3 of 0.100 mol dm^-3 hydrochloric acid? (HCl + NaOH → NaCl + H2O)

  1. 50.0 cm^3
  2. 5.00 cm^3
  3. 25.0 cm^3
  4. 12.5 cm^3 Correct answer.

Explanation

Moles of HCl = 0.100 × 0.0250 = 2.50 × 10^-3 mol, and the 1:1 equation means the same moles of NaOH are needed. Volume = moles ÷ concentration = 2.50 × 10^-3 ÷ 0.200 = 0.0125 dm^3 = 12.5 cm^3.

Question 46Stoichiometry and Chemical Calculations

A gas occupies 240 cm^3 at room temperature and pressure. What amount in moles does this represent? (Molar gas volume = 24.0 dm^3 mol^-1 at r.t.p.)

  1. 0.100 mol
  2. 0.0100 mol Correct answer.
  3. 1.00 mol
  4. 10.0 mol

Explanation

Convert the volume to dm^3 (240 cm^3 = 0.240 dm^3) and divide by the molar gas volume: 0.240 ÷ 24.0 = 0.0100 mol.

Question 47Stoichiometry and Chemical Calculations

What is the percentage by mass of nitrogen in ammonium sulfate, (NH4)2SO4? (Ar: N = 14, H = 1, S = 32, O = 16)

  1. 21.2% Correct answer.
  2. 28.0%
  3. 14.0%
  4. 10.6%

Explanation

The molar mass is (2 × 18) + 32 + 64 = 132 g mol^-1 and the mass of nitrogen is 2 × 14 = 28 g. Percentage = (28 ÷ 132) × 100 = 21.2%.

Question 48Stoichiometry and Chemical Calculations

In the reaction N2 + 3H2 → 2NH3, 6.0 mol of hydrogen reacts with excess nitrogen. What amount of ammonia is formed?

  1. 4.0 mol Correct answer.
  2. 2.0 mol
  3. 6.0 mol
  4. 18.0 mol

Explanation

The equation shows 3 mol H2 produce 2 mol NH3, so the ratio is 2/3. Thus 6.0 mol H2 gives 6.0 × (2/3) = 4.0 mol NH3.

Question 49Stoichiometry and Chemical Calculations

In a titration, 25.0 cm^3 of a hydrochloric acid solution required 20.0 cm^3 of 0.150 mol dm^-3 sodium carbonate solution. The reaction is Na2CO3 + 2HCl → 2NaCl + H2O + CO2. What is the concentration of the acid?

  1. 0.120 mol dm^-3
  2. 0.0600 mol dm^-3
  3. 0.480 mol dm^-3
  4. 0.240 mol dm^-3 Correct answer.

Explanation

Moles of Na2CO3 = 0.150 × 0.0200 = 3.00 × 10^-3 mol; the equation shows twice as many moles of HCl are needed, giving 6.00 × 10^-3 mol. Dividing by 0.0250 dm^3 gives 0.240 mol dm^-3.

Question 50Stoichiometry and Chemical Calculations

A 2.00 g sample of an impure carbonate is found to contain 1.60 g of calcium carbonate. What is the percentage purity of the sample?

  1. 1.25%
  2. 20.0%
  3. 125%
  4. 80.0% Correct answer.

Explanation

Percentage purity is the mass of the pure substance divided by the mass of the sample, multiplied by 100: (1.60 ÷ 2.00) × 100 = 80.0%.