Unit 3 Practice: Medium
Intermolecular Forces and Properties · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
Rank CH₄, HCl, and H₂O from weakest to strongest dominant intermolecular attraction.
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Answer:
CH₄ < HCl < H₂O
Explanation: CH₄ is nonpolar, so its main attraction is London dispersion.
HCl is polar, so it has dipole-dipole attractions.
H₂O can hydrogen bond, which is especially strong for molecules of similar size.
Question 2
Why does H₂O have a much higher boiling point than H₂S?
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Answer: H₂O forms strong hydrogen bonds.
Explanation: Oxygen is highly electronegative, and hydrogen is directly bonded to O in water. H₂S does not form comparably strong hydrogen bonding in the usual AP Chemistry model.
Question 3
Why do larger hydrocarbons generally have higher boiling points than smaller hydrocarbons?
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Answer: Larger hydrocarbons have stronger London dispersion forces.
Explanation: As molecular size and electron count increase, polarizability increases. Stronger attractions require more energy to overcome during boiling.
Question 4
Why can a straight-chain molecule have a higher boiling point than a highly branched isomer with the same molecular formula?
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Answer: The straight-chain molecule can have greater surface contact.
Explanation: Greater surface contact allows stronger London dispersion interactions between neighboring molecules. Branching can make a molecule more compact and reduce this contact.
Question 5
Which evaporates faster: a liquid with weak IMFs or one with strong IMFs?
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Answer: The liquid with weaker IMFs
Explanation: Its molecules require less energy to escape the liquid surface and enter the gas phase.
Question 6
Which has the higher vapor pressure: a liquid with strong IMFs or weak IMFs?
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Answer: The liquid with weak IMFs
Explanation: More molecules can escape into the vapor phase, so the gas above the liquid produces greater pressure.
Question 7
Why does water boil below 100°C at high altitude?
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Answer: External atmospheric pressure is lower.
Explanation: Boiling occurs when vapor pressure equals external pressure.
At high altitude, the external pressure is lower, so this condition is reached at a lower temperature.
Question 8
A gas has P = 1.00 atm, V = 5.00 L, and T = 300 K. Calculate the number of moles.
Use R = 0.08206 L·atm·mol⁻¹·K⁻¹.
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Answer: 0.203 mol
Work:
PV = nRT
Solve for n:
n = PV/RT
n = (1.00)(5.00) / [(0.08206)(300)]
n ≈ 0.203 mol
Question 9
A gas occupies 2.00 L at 300 K. What volume will it occupy at 450 K if pressure and moles remain constant?
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Answer: 3.00 L
Work:
Use Charles's law:
V₁/T₁ = V₂/T₂
2.00/300 = V₂/450
V₂ = 3.00 L
Explanation: At constant pressure, volume is directly proportional to Kelvin temperature.
Question 10
A mixture contains 2.00 mol N₂ and 3.00 mol O₂ at a total pressure of 5.00 atm. Find the partial pressure of N₂.
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Answer: 2.00 atm
Work:
Total moles:
2 + 3 = 5 mol
Mole fraction N₂:
XN₂ = 2/5 = 0.400
Then:
PN₂ = XN₂Ptotal
= (0.400)(5.00)
= 2.00 atm
Question 11
Why do lighter gas particles move faster than heavier gas particles at the same temperature?
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Answer: They have the same average kinetic energy but smaller mass.
Explanation: Because:
KE = 1/2 mv²
if m decreases while average KE stays the same, v must increase.
Question 12
Under what conditions do gases behave most ideally?
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Answer: High temperature and low pressure
Explanation: At high temperature, particle kinetic energy is large compared with intermolecular attractions.
At low pressure, particles are farther apart, so attractions and particle volume matter less.
Question 13
Why do real gases deviate from ideal behavior at high pressure?
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Answer: Gas particles are forced close together, so particle volume and intermolecular interactions become significant.
Explanation: The ideal model assumes particles have negligible volume and do not attract each other. These assumptions become less accurate at high pressure.
Question 14
Why do real gases often deviate at low temperature?
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Answer: Intermolecular attractions become more important.
Explanation: At lower temperature, particles move more slowly. Attractions between them can then significantly influence their motion and pressure.
Question 15
Calculate the molarity of a solution containing 0.500 mol solute in 2.00 L solution.
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Answer: 0.250 M
Work:
M = n/V
M = 0.500/2.00
= 0.250 M
Question 16
How many moles of solute are in 250. mL of 0.400 M solution?
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Answer: 0.100 mol
Work:
Convert:
250. mL = 0.250 L
Then:
n = MV
n = (0.400)(0.250)
= 0.100 mol
Question 17
A 2.00 M solution is diluted from 50.0 mL to 200. mL. Find the final concentration.
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Answer: 0.500 M
Work:
M₁V₁=M₂V₂
(2.00)(50.0)=M₂(200.)
M₂ = 0.500 M
Question 18
A particle diagram represents dissolved CaCl₂. If it contains 8 Ca²⁺ ions, how many Cl⁻ ions should it contain?
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Answer: 16 Cl⁻ ions
Explanation: CaCl₂ dissociates in a ratio:
1 Ca²⁺ : 2 Cl⁻
So:
8 × 2 = 16
Question 19
In TLC, a compound travels 3.0 cm while the solvent front travels 6.0 cm. Calculate Rf.
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Answer: 0.50
Work:
Rf = distance traveled by solute / distance traveled by solvent front
Rf = 3.0/6.0
= 0.50
Question 20
A solution's absorbance doubles while path length and molar absorptivity stay constant. What happens to concentration?
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Answer: The concentration doubles.
Explanation: From:
A = εbc
if ε and b are constant, absorbance is directly proportional to concentration.