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Quiz: Intermolecular Forces and States of Matter

Test your understanding of intermolecular forces, states of matter, and gas laws with these questions.

1. Which type of intermolecular force exists between ALL molecules, regardless of whether they are polar or nonpolar?

  1. London dispersion forces
  2. Dipole-dipole forces
  3. Hydrogen bonding
  4. Ion-dipole forces
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The correct answer is A. London dispersion forces (van der Waals dispersion forces) arise from temporary, instantaneous dipoles in the electron cloud and are universal — they exist in every atom and molecule, polar or nonpolar. Dipole-dipole forces require a permanent dipole, so they are absent in nonpolar molecules. Hydrogen bonding requires H bonded directly to N, O, or F. Ion-dipole forces require an ion to be present.

Concept Tested: London Dispersion Forces

2. A student compares n-pentane (\(\ce{C5H12}\)) to neopentane (\(\ce{C(CH3)4}\)), which have the same molecular formula but different shapes. Which prediction is correct regarding their boiling points?

  1. They have identical boiling points because they have the same molar mass.
  2. n-Pentane has a higher boiling point because its elongated shape gives more surface area for London dispersion forces.
  3. Neopentane has a higher boiling point because its compact spherical shape concentrates electron density.
  4. Neither has measurable intermolecular forces because both are nonpolar.
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The correct answer is B. Although both molecules have the same molar mass (72 g/mol) and the same number of electrons, n-pentane's rod-like shape gives it more surface area for contact with neighboring molecules, increasing London dispersion force strength. Neopentane's compact, nearly spherical shape reduces contact surface area, weakening its dispersion forces. Molar mass alone does not determine boiling point when molecular shape differs.

Concept Tested: Polarizability / London Dispersion Forces

3. Which of the following molecules is capable of hydrogen bonding with other molecules of the same type?

  1. \(\ce{CH3OCH3}\) (dimethyl ether)
  2. \(\ce{CH3F}\) (fluoromethane)
  3. \(\ce{NH3}\) (ammonia)
  4. \(\ce{H2S}\) (hydrogen sulfide)
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The correct answer is C. Hydrogen bonding requires a hydrogen atom covalently bonded directly to N, O, or F. Ammonia (\(\ce{NH3}\)) has three N–H bonds and lone pairs on nitrogen, satisfying both the donor and acceptor requirements. Dimethyl ether has no H attached to oxygen (the H atoms are on carbon). Fluoromethane has a C–F bond, not an H–F bond. Hydrogen sulfide has S–H bonds, but sulfur is not electronegative enough to form true hydrogen bonds.

Concept Tested: Hydrogen Bond Requirements

4. The Kinetic Molecular Theory (KMT) postulate states that average kinetic energy of gas particles is directly proportional to which quantity?

  1. Pressure in atmospheres
  2. Volume of the container
  3. Molar mass of the gas
  4. Absolute temperature in Kelvin
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The correct answer is D. According to KMT, the average kinetic energy of gas particles is directly proportional to absolute temperature: \(KE_{avg} \propto T\) (in Kelvin). This is why temperature must always be expressed in Kelvin when using gas laws. Pressure, volume, and molar mass are related to gas behavior in other ways, but none is directly proportional to average kinetic energy in the KMT postulate.

Concept Tested: Kinetic Molecular Theory

5. A gas at 2.00 atm and 4.00 L is compressed at constant temperature to 1.00 L. What is the new pressure?

  1. 0.500 atm
  2. 4.00 atm
  3. 8.00 atm
  4. 2.00 atm
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The correct answer is C. Boyle's Law states \(P_1V_1 = P_2V_2\) at constant temperature and amount. Substituting: \(P_2 = \frac{P_1 V_1}{V_2} = \frac{(2.00 \text{ atm})(4.00 \text{ L})}{1.00 \text{ L}} = 8.00 \text{ atm}\). When volume decreases by a factor of 4, pressure increases by a factor of 4. Option A inverts the relationship, and option B forgets to account for the volume change.

Concept Tested: Boyle's Law

6. A gas sample occupies 3.00 L at 300 K. At constant pressure, what volume will it occupy at 450 K?

  1. 2.00 L
  2. 3.00 L
  3. 4.50 L
  4. 6.75 L
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The correct answer is C. Charles's Law states that \(V/T\) is constant at fixed pressure and moles: \(V_2 = \frac{V_1 T_2}{T_1} = \frac{(3.00 \text{ L})(450 \text{ K})}{300 \text{ K}} = 4.50 \text{ L}\). Temperature increased by a factor of 1.5, so volume also increases by 1.5. A common error is using Celsius instead of Kelvin — always convert to Kelvin before applying Charles's Law.

Concept Tested: Charles's Law

7. A sealed container holds a mixture of gases: 0.50 mol \(\ce{N2}\), 0.30 mol \(\ce{O2}\), and 0.20 mol \(\ce{Ar}\), with a total pressure of 2.00 atm. What is the partial pressure of \(\ce{O2}\)?

  1. 0.60 atm
  2. 0.30 atm
  3. 1.00 atm
  4. 2.00 atm
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The correct answer is A. By Dalton's Law, the partial pressure of each gas equals its mole fraction times total pressure. The mole fraction of \(\ce{O2}\): \(\chi_{O_2} = \frac{0.30}{0.50 + 0.30 + 0.20} = \frac{0.30}{1.00} = 0.30\). Therefore \(P_{O_2} = 0.30 \times 2.00 \text{ atm} = 0.60 \text{ atm}\). Option B confuses moles with partial pressure, forgetting to multiply by total pressure.

Concept Tested: Dalton's Law / Partial Pressures

8. Which of the following properties of a liquid INCREASES as the strength of intermolecular forces increases?

  1. Vapor pressure
  2. Evaporation rate
  3. Viscosity
  4. Compressibility
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The correct answer is C. Viscosity — a liquid's resistance to flow — increases with stronger IMFs because molecules are more strongly attracted to each other, making it harder for them to slide past one another. Vapor pressure and evaporation rate both decrease with stronger IMFs, because molecules are held more tightly in the liquid phase and escape less readily. Gases, not liquids, are described by compressibility; liquids are already nearly incompressible.

Concept Tested: Viscosity / Liquid Properties

9. Under which conditions does a real gas deviate MOST significantly from ideal gas behavior?

  1. High temperature and low pressure
  2. Low temperature and high pressure
  3. High temperature and high pressure
  4. Low temperature and low pressure
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The correct answer is B. Real gases deviate most from ideal behavior at low temperature (where slow-moving molecules experience significant IMFs and may condense) and high pressure (where molecules are forced so close together that their finite volume becomes a significant fraction of the total container volume). High temperature and low pressure are conditions where the ideal gas law works best, because molecules move fast and are widely separated.

Concept Tested: Deviations from Ideality / Real Gases

10. The Van der Waals constant \(a\) for \(\ce{NH3}\) is 4.17 L²·atm/mol², while for \(\ce{H2}\) it is 0.244 L²·atm/mol². What does this comparison tell us?

  1. \(\ce{NH3}\) molecules are physically larger than \(\ce{H2}\) molecules.
  2. \(\ce{H2}\) exerts greater pressure on container walls than \(\ce{NH3}\) at the same conditions.
  3. \(\ce{NH3}\) has a lower boiling point than \(\ce{H2}\) because it deviates more from ideality.
  4. \(\ce{NH3}\) has stronger intermolecular attractions than \(\ce{H2}\).
Show Answer

The correct answer is D. In the Van der Waals equation, the constant \(a\) corrects for the pressure reduction caused by intermolecular attractions — a larger \(a\) means stronger attractive forces between molecules. \(\ce{NH3}\) undergoes hydrogen bonding, giving it far stronger IMFs than \(\ce{H2}\), which has only weak London dispersion forces. The constant \(b\) (not \(a\)) reflects molecular size. \(\ce{NH3}\) actually has a much higher boiling point than \(\ce{H2}\), consistent with stronger IMFs.

Concept Tested: Van der Waals Equation / IMF Strength