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AP Intermediate Chemistry (BIEAP) study materials
AP Intermediate Chemistry (BIEAP) study resources

AP Intermediate Chemistry (BIEAP) practice questions

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Question 1Atomic Structure, Classification of Elements and Periodicity

The ground-state electron configuration of a neutral chromium atom (Z = 24) is best written as:

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

Explanation

Half-filled d subshells gain extra stability from exchange energy and symmetric spin distribution, so one 4s electron moves into 3d. This gives [Ar] 3d5 4s1, not the simple Aufbau order [Ar] 3d4 4s2. The same half-filled preference explains the anomalous configuration of molybdenum.

Question 2Atomic Structure, Classification of Elements and Periodicity

Which set of quantum numbers correctly describes the last electron added to a ground-state sodium atom (Z = 11)?

  1. n = 2, l = 1, m = -1, s = +1/2
  2. n = 3, l = 1, m = 0, s = +1/2
  3. n = 3, l = 0, m = 0, s = +1/2 Correct answer.
  4. n = 3, l = 2, m = 1, s = -1/2

Explanation

Sodium has configuration [Ne] 3s1, so its eleventh electron enters the 3s orbital. A 3s orbital has n = 3, l = 0 and m = 0, and the single electron can be assigned s = +1/2. Any set with l = 1 or l = 2 describes p or d orbitals, which are not occupied last.

Question 3Atomic Structure, Classification of Elements and Periodicity

The number of unpaired electrons in the ground state of a Mn2+ ion (Z = 25) is:

  1. 4
  2. 3
  3. 6
  4. 5 Correct answer.

Explanation

Manganese has configuration [Ar] 3d5 4s2, so Mn2+ loses the two 4s electrons and becomes [Ar] 3d5. By Hund's rule the five d electrons occupy all five d orbitals singly with parallel spins. That gives five unpaired electrons and a half-filled, unusually stable d subshell.

Question 4Atomic Structure, Classification of Elements and Periodicity

In the modern periodic table, the element with atomic number 118 would be placed in which group and block?

  1. Group 2, s-block
  2. Group 17, p-block
  3. Group 18, p-block Correct answer.
  4. Group 1, s-block

Explanation

Elements 2, 10, 18, 36, 54, 86 and 118 complete the p subshell of their period and belong to group 18. Element 118 fills the 7p6 level, making it a p-block noble gas. Group 1 and group 2 elements are s-block metals with one or two outer s electrons.

Question 5Atomic Structure, Classification of Elements and Periodicity

Which property shows a general increase across a period from left to right in the periodic table?

  1. Metallic character
  2. Number of principal shells
  3. Atomic radius
  4. First ionisation enthalpy Correct answer.

Explanation

Across a period the nuclear charge rises while electrons enter the same principal shell, so the effective nuclear charge increases. Outer electrons are therefore held more tightly and first ionisation enthalpy generally increases. Atomic radius and metallic character decrease, while the number of shells stays constant within a period.

Question 6Atomic Structure, Classification of Elements and Periodicity

Which of the following isoelectronic species has the smallest ionic radius?

  1. Cl-
  2. Ca2+ Correct answer.
  3. K+
  4. S2-

Explanation

All four species contain 18 electrons, so the radius depends on nuclear charge. Ca2+ has the highest nuclear charge (Z = 20) and therefore pulls the same 18 electrons in most strongly. Radius decreases in the order S2- > Cl- > K+ > Ca2+, making Ca2+ the smallest.

Question 7Atomic Structure, Classification of Elements and Periodicity

The correct order of increasing first ionisation enthalpy for the elements Be, B, N and O is:

  1. O < N < B < Be
  2. Be < O < B < N
  3. B < Be < O < N Correct answer.
  4. Be < B < N < O

Explanation

The trend across period 2 is broken by electron configurations. B (2s2 2p1) has a lower value than Be (2s2) because removing the shielded 2p electron is easier than breaking a filled 2s subshell. Likewise O (2p4) is lower than N (2p3), which has a stable half-filled p subshell, giving B < Be < O < N.

Question 8Atomic Structure, Classification of Elements and Periodicity

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

  1. Li and K Correct answer.
  2. F and Ne
  3. Na and Mg
  4. C and N

Explanation

Chemical properties are governed mainly by valence-electron configuration, which repeats down a group. Li and K both belong to group 1 with an ns1 configuration and form similar oxides and hydroxides. Pairs from different groups, or a halogen with a noble gas, behave very differently.

Question 9Atomic Structure, Classification of Elements and Periodicity

The number of radial nodes in a 3p orbital is:

  1. 3
  2. 2
  3. 1 Correct answer.
  4. 0

Explanation

The number of radial nodes equals n - l - 1. For a 3p orbital, n = 3 and l = 1, so the count is 3 - 1 - 1 = 1. The single node arises from the radial part of the wavefunction changing sign once between the nucleus and infinity.

Question 10Atomic Structure, Classification of Elements and Periodicity

Which statement about the shapes and orientations of atomic orbitals is correct?

  1. An s orbital has a spherical shape and its electron density is zero at the nucleus.
  2. A dxy orbital has its lobes lying along the x and y axes.
  3. The three p orbitals differ in energy in a free atom and are filled in order px, py, pz.
  4. The five d orbitals of a given shell are degenerate in a free atom but split in an octahedral ligand field. Correct answer.

Explanation

In a free atom the five d orbitals of a shell have the same energy. When ligands approach octahedrally, the dz2 and dx2-y2 orbitals point at the ligands and are raised, while the others are lowered, so degeneracy is lost. This crystal-field splitting underlies the colour and magnetism of transition-metal complexes.

Question 11Atomic Structure, Classification of Elements and Periodicity

The de Broglie wavelength of an electron accelerated through a potential difference V is proportional to:

  1. 1/sqrt(V) Correct answer.
  2. 1/V
  3. V^2
  4. V

Explanation

The kinetic energy gained is eV, so the momentum is p = sqrt(2meV). Substituting into lambda = h/p gives lambda proportional to 1/sqrt(V). Doubling the accelerating voltage therefore shortens the electron wavelength by a factor of sqrt(2).

Question 12Atomic Structure, Classification of Elements and Periodicity

Which transition in the hydrogen atom emits a photon of the longest wavelength?

  1. n = 2 to n = 1
  2. n = 3 to n = 2 Correct answer.
  3. n = 4 to n = 2
  4. n = 5 to n = 3

Explanation

Photon energy equals the energy gap between the levels, and longer wavelength means lower energy. The n = 3 to n = 2 transition (Balmer series, H-alpha) is the smallest gap among the listed options that ends below n = 3. Transitions ending at n = 1 or falling from higher levels release more energy and shorter wavelengths.

Question 13Atomic Structure, Classification of Elements and Periodicity

According to Hund's rule, the ground-state arrangement of the two 2p electrons of carbon is best described as:

  1. One electron in a 2p orbital and one in the 2s orbital.
  2. Both electrons paired in one 2p orbital with opposite spins.
  3. One electron in each of two different 2p orbitals with parallel spins. Correct answer.
  4. Both electrons in one 2p orbital with parallel spins.

Explanation

Hund's rule states that degenerate orbitals are occupied singly with parallel spins before any pairing occurs. For carbon's 2p2 set, this means one electron in each of two p orbitals with the same spin, minimising electron-electron repulsion. Pairing only begins once each degenerate orbital holds one electron.

Question 14Atomic Structure, Classification of Elements and Periodicity

The correct set of quantum numbers allowed for an electron in a 4d orbital is:

  1. n = 4, l = 3, m = 0
  2. n = 4, l = 2, m = -2 Correct answer.
  3. n = 4, l = 2, m = 3
  4. n = 3, l = 2, m = -1

Explanation

For a 4d orbital n = 4 and l = 2, and the magnetic quantum number may take any integer from -l to +l, that is -2 to +2. The set n = 4, l = 2, m = -2 satisfies all these conditions. Sets with l = 3 or with m outside the -l to +l range are not permitted.

Question 15Atomic Structure, Classification of Elements and Periodicity

Which factor is chiefly responsible for the very high first ionisation enthalpy of helium compared with hydrogen?

  1. Helium has a lower effective nuclear charge per electron than hydrogen.
  2. Helium has two electrons in the same shell, each screening the other almost completely.
  3. Helium has a filled 1s subshell and a doubled nuclear charge acting on electrons in the same shell. Correct answer.
  4. Helium has a larger atomic radius than hydrogen.

Explanation

Helium's nuclear charge is twice that of hydrogen while its two electrons occupy the same n = 1 shell with little mutual screening. Each electron therefore experiences a much stronger effective nuclear charge, and the filled 1s subshell adds extra stability. Both effects make helium's first ionisation enthalpy far higher than hydrogen's.

Question 16Atomic Structure, Classification of Elements and Periodicity

An element has the outer electronic configuration 4s2 4p5. The element and its position in the periodic table are:

  1. Krypton, group 18, period 4
  2. Selenium, group 16, period 4
  3. Bromine, group 17, period 4 Correct answer.
  4. Iodine, group 17, period 5

Explanation

The highest principal quantum number is 4, so the element lies in period 4. The total of seven valence electrons (4s2 4p5) places it in group 17, the halogens, and counting to Z = 35 identifies it as bromine. An np5 configuration is the hallmark of the halogen group.

Question 17Atomic Structure, Classification of Elements and Periodicity

Which comparison of atomic or ionic size is correct?

  1. Mg2+ > Na+
  2. Na+ > Na
  3. Cl- > Cl Correct answer.
  4. Na > K

Explanation

Adding an electron to a chlorine atom increases electron-electron repulsion while the nuclear charge stays the same, so the anion is larger than the parent atom. Hence Cl- is larger than Cl. Atomic radius increases down a group, and cations are smaller than their parent atoms because electrons are lost without a change in nuclear charge.

Question 18Atomic Structure, Classification of Elements and Periodicity

The successive ionisation enthalpies of an element show a very large jump between the third and fourth values. The element most likely belongs to:

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

Explanation

A large jump between the third and fourth ionisation enthalpies means the fourth electron must be removed from a stable inner shell after three valence electrons are gone. An element with three valence electrons sits in group 13. Removing electrons from a filled inner shell requires far more energy than removing valence electrons.

Question 19Atomic Structure, Classification of Elements and Periodicity

The correct order of increasing electronegativity for the elements Na, Si, P and Cl is:

  1. Cl < P < Si < Na
  2. Si < Na < P < Cl
  3. Na < Si < P < Cl Correct answer.
  4. Na < P < Si < Cl

Explanation

Electronegativity increases across a period as effective nuclear charge grows and atomic radius shrinks. For period 3 the order is therefore Na < Si < P < Cl, with chlorine the most electronegative and sodium the least. The same left-to-right trend underlies the decrease in metallic character.

Question 20Atomic Structure, Classification of Elements and Periodicity

The Heisenberg uncertainty principle is best expressed as:

  1. The product of the uncertainties in position and momentum is at least h/4pi. Correct answer.
  2. Energy and time can both be measured exactly for a bound electron.
  3. The uncertainty in position can be made zero by using light of sufficiently short wavelength.
  4. An electron in an atom must always move in a well-defined circular orbit.

Explanation

Heisenberg showed that position and momentum cannot both be known precisely, with the product of their uncertainties at least h/4pi. The limit is not a failure of instruments but a fundamental property of matter described by wavefunctions. It is why Bohr's fixed circular orbits were replaced by probability distributions.

Question 21Chemical Bonding and Molecular Structure

According to VSEPR theory, what is the shape of the molecule SF4?

  1. Square planar
  2. See-saw Correct answer.
  3. Trigonal bipyramidal
  4. Tetrahedral

Explanation

Sulfur in SF4 has four bond pairs and one lone pair, giving five electron domains that adopt a trigonal bipyramidal arrangement. The lone pair occupies an equatorial position, so the molecular shape is described as see-saw.

Question 22Chemical Bonding and Molecular Structure

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

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

Explanation

In CCl4 the four C–Cl bond dipoles are arranged tetrahedrally and cancel exactly, so the molecular dipole moment is zero. Polarity of a molecule depends on both bond polarity and the geometry of the molecule.

Question 23Chemical Bonding and Molecular Structure

Which type of hybridisation is present at the carbon atoms of ethyne (C2H2)?

  1. sp3
  2. sp Correct answer.
  3. sp3d
  4. sp2

Explanation

Each carbon in ethyne forms one single bond to hydrogen and a triple bond to the other carbon, giving two sigma bonds and hence two hybrid orbitals. Two hybrid orbitals correspond to sp hybridisation, leaving two unhybridised p orbitals for the pi bonds.

Question 24Chemical Bonding and Molecular Structure

Which of the following species has a bond order of 2.5 according to molecular orbital theory?

  1. N2
  2. O2+ Correct answer.
  3. O2
  4. O2−

Explanation

The O2+ ion has nine valence electrons, giving the configuration with one electron removed from an antibonding pi orbital. Bond order equals (bonding − antibonding)/2, which gives (8 − 3)/2 = 2.5 for O2+.

Question 25Chemical Bonding and Molecular Structure

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

  1. Hydrogen bonding Correct answer.
  2. Dipole–dipole interactions
  3. Ion–dipole interactions
  4. London dispersion forces

Explanation

Water molecules form extensive hydrogen bonds because hydrogen is bonded to a highly electronegative oxygen atom with lone pairs. These strong directional interactions require more energy to overcome, raising the boiling point well above that of H2S.

Question 26Chemical Bonding and Molecular Structure

Which of the following bonds is the most polar?

  1. H–Br
  2. H–Cl
  3. H–I
  4. H–F Correct answer.

Explanation

Bond polarity increases with the electronegativity difference between the two bonded atoms. Fluorine is the most electronegative halogen, so the H–F bond has the largest electronegativity difference and the greatest polarity.

Question 27Chemical Bonding and Molecular Structure

The bond angle in a water molecule is approximately 104.5°, which is smaller than the ideal tetrahedral angle. Which statement best explains this observation?

  1. The O–H bonds are ionic, which reduces the angle.
  2. Hydrogen atoms repel each other more strongly than lone pairs do.
  3. The oxygen atom is sp2 hybridised, which naturally gives a smaller angle.
  4. The two lone pairs on oxygen repel the bonding pairs more strongly than bonding pairs repel each other. Correct answer.

Explanation

VSEPR theory ranks repulsions as lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. The two lone pairs on oxygen push the O–H bonding pairs closer together, compressing the angle below the tetrahedral 109.5°.

Question 28Chemical Bonding and Molecular Structure

Which of the following molecules has a linear shape?

  1. CO2 Correct answer.
  2. SO2
  3. NO2
  4. H2O

Explanation

The central carbon in CO2 has two bonding domains and no lone pairs, so the two double bonds are arranged 180° apart to give a linear molecule. The other species have lone pairs on the central atom that bend the geometry.

Question 29Chemical Bonding and Molecular Structure

Which of the following correctly lists the species in order of increasing carbon–oxygen bond length?

  1. CO2 < CO < CO3^2−
  2. CO < CO3^2− < CO2
  3. CO < CO2 < CO3^2− Correct answer.
  4. CO3^2− < CO2 < CO

Explanation

Bond length decreases as bond order increases. Carbon monoxide has a bond order of 3, carbon dioxide has two double bonds with bond order 2, and the carbonate ion has bond order about 1.33 due to resonance delocalisation, so the order of increasing length is CO < CO2 < CO3^2−.

Question 30Chemical Bonding and Molecular Structure

Which of the following compounds contains both ionic and covalent bonds?

  1. CH4
  2. NaCl
  3. H2
  4. NaOH Correct answer.

Explanation

Sodium hydroxide contains an ionic bond between the Na+ cation and the OH− anion, while the O–H bond within the hydroxide ion is covalent. A compound with both types of bonding must contain a polyatomic ion.

Question 31Chemical Bonding and Molecular Structure

The bond angle in BF3 is 120°. Which hybridisation and geometry are consistent with this observation?

  1. sp3d hybridisation and trigonal bipyramidal geometry
  2. sp hybridisation and linear geometry
  3. sp2 hybridisation and trigonal planar geometry Correct answer.
  4. sp3 hybridisation and trigonal pyramidal geometry

Explanation

Boron in BF3 has three bonding domains and no lone pairs, so it uses three sp2 hybrid orbitals pointing to the corners of an equilateral triangle. This trigonal planar arrangement gives bond angles of 120°.

Question 32Chemical Bonding and Molecular Structure

Which of the following explains why the first ionisation energy of nitrogen is greater than that of oxygen?

  1. Oxygen has a higher nuclear charge than nitrogen.
  2. Oxygen has fewer valence electrons than nitrogen.
  3. Nitrogen has a larger atomic radius than oxygen.
  4. Nitrogen has a half-filled 2p subshell, which confers extra stability. Correct answer.

Explanation

Nitrogen has the configuration 2p3 with a half-filled p subshell, which is unusually stable because each orbital is singly occupied with parallel spins. Removing an electron from this stable arrangement requires more energy than removing the paired electron in oxygen's 2p4 configuration.

Question 33Chemical Bonding and Molecular Structure

Which of the following molecules exhibits only London dispersion forces as intermolecular forces?

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

Explanation

Methane is a non-polar tetrahedral molecule with no net dipole moment, so the only intermolecular attractions are London dispersion forces arising from instantaneous dipoles. The other molecules are polar and experience dipole–dipole attractions.

Question 34Chemical Bonding and Molecular Structure

Which of the following species is paramagnetic according to molecular orbital theory?

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

Explanation

O2 has two unpaired electrons in its degenerate antibonding pi* orbitals, which makes it paramagnetic and explains its magnetic behaviour. The other molecules have all their electrons paired in the molecular orbital diagram.

Question 35Chemical Bonding and Molecular Structure

Which of the following sets of bond angles is expected for PCl5?

  1. 90° and 120° Correct answer.
  2. 109.5° only
  3. 90° and 180°
  4. 107° and 120°

Explanation

PCl5 has five bonding domains and no lone pairs, adopting a trigonal bipyramidal geometry. The equatorial bonds are at 120° to one another, while the axial bonds are at 90° to the equatorial bonds.

Question 36Chemical Bonding and Molecular Structure

Which of the following statements about the resonance structures of the nitrate ion, NO3−, is correct?

  1. One N–O bond is a double bond and the other two are single bonds, giving different lengths.
  2. The nitrogen atom is sp3 hybridised in nitrate.
  3. The three N–O bonds are identical in length and intermediate between single and double bonds. Correct answer.
  4. All three N–O bonds are pure double bonds.

Explanation

Resonance delocalises the double-bond character over all three N–O positions, so each bond has a bond order of about 1.33. Consequently all three bonds are identical in length and lie between typical single and double bond lengths.

Question 37Chemical Bonding and Molecular Structure

Which of the following molecules has the smallest bond angle at the central atom?

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

Explanation

The number of lone pairs on the central atom determines how strongly the bonding pairs are pushed together. Water has two lone pairs, giving an angle of about 104.5°, which is smaller than ammonia's 107° and methane's 109.5°.

Question 38Chemical Bonding and Molecular Structure

Which of the following correctly describes the hybridisation of the carbon atoms and the number of sigma and pi bonds in the molecule CH2=CH–C≡N?

  1. Both alkene carbons are sp, the nitrile carbon is sp, and the molecule has 4 sigma and 3 pi bonds.
  2. Both alkene carbons are sp2, the nitrile carbon is sp, and the molecule has 5 sigma and 3 pi bonds. Correct answer.
  3. Both alkene carbons are sp3, the nitrile carbon is sp2, and the molecule has 6 sigma and 2 pi bonds.
  4. Both alkene carbons are sp2, the nitrile carbon is sp2, and the molecule has 5 sigma and 2 pi bonds.

Explanation

Each double-bonded carbon has three sigma bonds and is sp2 hybridised, while the triple-bonded nitrile carbon has two sigma bonds and is sp hybridised. Counting sigma bonds gives C–H, C–H, C=C sigma, C–CN sigma and C–N sigma for five, and the double bond contributes one pi while the triple bond contributes two, giving three pi bonds.

Question 39Chemical Bonding and Molecular Structure

The bond dissociation enthalpy of F2 is unexpectedly lower than that of Cl2. Which explanation is most consistent with this observation?

  1. Fluorine has a higher bond order than chlorine in the diatomic molecule.
  2. F2 is a gas while Cl2 is a liquid at room temperature, which directly determines bond enthalpy.
  3. Chlorine has a lower electronegativity, so its bond is stronger.
  4. Fluorine has a smaller atomic radius, so its lone pairs in adjacent atoms repel strongly, weakening the F–F bond. Correct answer.

Explanation

The very small size of fluorine atoms brings the lone pairs on the two bonded atoms close together, and the resulting repulsion offsets the expected increase in bond strength from orbital overlap. Chlorine atoms are larger, so lone-pair repulsion is less significant and the Cl–Cl bond is stronger.

Question 40Chemical Bonding and Molecular Structure

A molecule of the type AB4 is found to have a net dipole moment of zero, while AB3 has a measurable dipole moment. Which combination of geometries is most consistent with these observations?

  1. AB4 is tetrahedral and AB3 is trigonal planar with a lone pair on A. Correct answer.
  2. AB4 is see-saw and AB3 is trigonal pyramidal.
  3. AB4 is tetrahedral with one lone pair and AB3 is linear.
  4. AB4 is square planar and AB3 is trigonal planar with no lone pair.

Explanation

A tetrahedral AB4 molecule has four identical bond dipoles that cancel exactly, giving zero net dipole moment. A trigonal planar AB3 species with a lone pair on the central atom is actually bent or pyramidal, so its bond dipoles do not cancel and a net dipole moment results.

Question 41States of Matter: Gases and Liquids

A sealed rigid vessel contains an ideal gas at constant temperature. If the amount of gas is doubled while the volume is held constant, what happens to the pressure?

  1. It stays the same
  2. It becomes four times larger
  3. It halves
  4. It doubles Correct answer.

Explanation

At constant temperature and volume, pressure is directly proportional to the number of moles (Avogadro's law combined with the ideal gas equation pV = nRT). Doubling n therefore doubles p, since V, T and R are unchanged.

Question 42States of Matter: Gases and Liquids

Which statement correctly describes the behaviour of gas molecules according to the kinetic molecular theory of gases?

  1. Molecules attract each other strongly except during collisions
  2. Molecules occupy a large fraction of the container volume
  3. Molecular speeds are identical at a given temperature
  4. Collisions between molecules are perfectly elastic Correct answer.

Explanation

The kinetic model assumes that intermolecular forces are negligible and that collisions with each other and with the walls are perfectly elastic, so kinetic energy is conserved. The volume of the molecules themselves is also treated as negligible compared with the container volume.

Question 43States of Matter: Gases and Liquids

A sample of gas occupies 300 mL at 1.0 atm. At constant temperature, what volume does it occupy when the pressure is raised to 2.0 atm?

  1. 600 mL
  2. 300 mL
  3. 150 mL Correct answer.
  4. 75 mL

Explanation

Boyle's law states that at constant temperature pV is constant, so p1V1 = p2V2. Substituting gives 1.0 × 300 = 2.0 × V2, hence V2 = 150 mL. Doubling the pressure halves the volume.

Question 44States of Matter: Gases and Liquids

Which set of conditions corresponds to the critical point of a substance?

  1. The temperature at which a solid melts under its own vapour pressure
  2. The temperature and pressure at which three phases coexist in equilibrium
  3. The temperature above which a liquid cannot exist, at the pressure where liquid and gas densities become equal Correct answer.
  4. The pressure at which a gas liquefies at the normal boiling point

Explanation

At the critical point the meniscus between liquid and vapour disappears because the densities of the two phases become identical. Above the critical temperature a substance cannot be liquefied by pressure alone, however great.

Question 45States of Matter: Gases and Liquids

Equal volumes of hydrogen and oxygen are kept in separate vessels at the same temperature and pressure. Which statement is correct?

  1. The hydrogen molecules have a lower average kinetic energy
  2. The oxygen vessel contains more molecules
  3. Both vessels contain the same number of molecules Correct answer.
  4. The hydrogen vessel contains more molecules

Explanation

Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. The identity of the gas does not affect this count, although the masses differ because the molar masses differ.

Question 46States of Matter: Gases and Liquids

A gas diffuses through a porous plug. If the molar mass of gas A is four times that of gas B, what is the ratio of the rate of diffusion of A to that of B?

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

Explanation

Graham's law states that the rate of diffusion is inversely proportional to the square root of molar mass. Since the molar mass ratio is 4, the rate ratio is 1/√4 = 1/2, so A diffuses half as fast as B.

Question 47States of Matter: Gases and Liquids

In the van der Waals equation for a real gas, what does the constant 'a' account for?

  1. The attractive forces between gas molecules Correct answer.
  2. The number of moles of gas present
  3. The absolute temperature of the gas
  4. The finite volume occupied by the gas molecules

Explanation

The term a(n/V)² is added to the measured pressure to correct for intermolecular attractions that reduce the pressure exerted on the walls. The constant b, by contrast, corrects for the finite volume occupied by the molecules.

Question 48States of Matter: Gases and Liquids

At 27 °C a fixed mass of gas at constant pressure occupies 500 mL. What volume does it occupy at 127 °C at the same pressure?

  1. Approximately 375 mL
  2. Approximately 500 mL
  3. Approximately 667 mL Correct answer.
  4. Approximately 2360 mL

Explanation

Charles's law gives V1/T1 = V2/T2 with temperatures in kelvin: 300 K and 400 K. Thus V2 = 500 × 400/300 ≈ 667 mL, an increase consistent with heating at constant pressure.

Question 49States of Matter: Gases and Liquids

Which property of a liquid is directly responsible for capillary rise in a narrow glass tube?

  1. The balance between cohesive forces within the liquid and adhesive forces with the glass Correct answer.
  2. The absence of any intermolecular forces in the liquid
  3. Its low density compared with the surrounding air
  4. Its high compressibility

Explanation

Capillary rise occurs when adhesive forces between the liquid and the tube wall exceed cohesive forces among the liquid molecules, so the liquid wets the glass and climbs. Surface tension then pulls the column upward until the weight of the column balances the upward force.

Question 50States of Matter: Gases and Liquids

A mixture contains 2 mol of gas X and 3 mol of gas Y in a total pressure of 5 atm. What is the partial pressure of gas X?

  1. 3 atm
  2. 5 atm
  3. 2 atm Correct answer.
  4. 1 atm

Explanation

By Dalton's law, the partial pressure of a component equals its mole fraction times the total pressure. The mole fraction of X is 2/5 = 0.4, so its partial pressure is 0.4 × 5 = 2 atm.