Unit 3 practice test: Properties of Substances and Mixtures

Doubling T from 200 K to 400 K doubles P from 0.821 to 1.641 atm, and the line points back to zero pressure at 0 K: P = (nR/V) × T.

• Correct: Pressure is directly proportional to kelvin temperature.: Right: P/T is constant, as PV = nRT predicts at fixed n and V.
• Pressure is inversely proportional to kelvin temperature.: Inverse proportion would make P fall as T rises.
• Pressure rises with temperature, but slower at high T.: The ratio P/T is the same at every point, so the rise is steady, not slowing.
• Pressure is directly proportional to Celsius temperature.: In °C the line would not pass through zero; only kelvin gives proportionality.

Shifting every temperature by 273.15 slides the line sideways without bending it, so it still reaches zero pressure at 0 K, which is −273.15 °C.

• Correct: A straight line that reaches zero pressure at −273 °C, not at 0 °C.: Right: same straight line, shifted.
• A straight line passing through the origin, the point 0 °C and 0 atm.: At 0 °C (273 K) the gas still has a large pressure, so the line cannot pass through the origin.
• A curve that bends upward, because Celsius degrees are smaller than kelvins.: A Celsius degree is the same size as a kelvin, so the slope does not change and no bend appears.
• A horizontal line, because Celsius temperatures do not affect pressure.: Pressure depends on temperature in any scale; only the zero point differs.

At fixed V, P = (nR/V)T. Doubling n doubles the slope nR/V, so every pressure doubles.

• Correct: Twice as steep: at each temperature the pressure doubles.: Right: P is proportional to n at fixed V and T.
• Half as steep: at each temperature the pressure halves.: More gas particles hit the walls more often, so pressure rises, not falls.
• The same line, because the container and temperatures are the same.: The amount of gas changed, so the pressure changes.
• Shifted 200 K to the right, but with the same steepness.: Adding gas changes the slope; every line still starts at 0 K.

Read the group 14 series: CH₄ −161.5 °C, SiH₄ −111.9 °C, GeH₄ −88.5 °C, SnH₄ −51.8 °C, a steady rise.

• Correct: It rises steadily, from −162 °C to −52 °C.: Right: each heavier hydride boils higher.
• It falls steadily, from −52 °C to −162 °C.: This reverses the direction of the trend shown on the graph.
• It falls from period 2 to period 3, then rises.: That pattern belongs to the group 15 and 16 series, whose period 2 members are unusual.
• It stays within a few degrees of −110 °C.: The values span more than 100 °C, so the series is far from flat.

Hydrogen bonds need H bonded to N, O or F. NH₃ and H₂O have them, so their boiling points jump above the trend; CH₄ is nonpolar and follows the dispersion trend.

• Correct: CH₄ has no H bonded to N, O or F, so it cannot form hydrogen bonds; NH₃ and H₂O can.: Right: the high points come from hydrogen bonding, which methane cannot do.
• CH₄ is the heaviest molecule in its group, which keeps its boiling point near the others.: CH₄ is the lightest in its group, which is why it boils lowest.
• CH₄ has a strong polar C–H bond that cancels the hydrogen bonds between its molecules.: C–H bonds are nearly nonpolar, and methane is a symmetric, nonpolar molecule.
• Carbon has more electrons than nitrogen or oxygen, so its hydride has the strongest dispersion forces.: Carbon has fewer electrons (6) than nitrogen (7) or oxygen (8).

From H₂S (18 electrons) to H₂Te (54 electrons) the molecules become much more polarizable, so dispersion forces increase. Their dipoles actually get smaller, because S, Se and Te are less electronegative than O, so the dispersion increase dominates.

• Correct: Down the group the molecules have more electrons, so dispersion forces grow and outweigh small changes in polarity.: Right: rising dispersion forces dominate in this series.
• Down the group the molecules form more hydrogen bonds, because S, Se and Te each have two lone pairs that can accept them.: Hydrogen bonds need H bonded to N, O or F; S, Se and Te do not qualify.
• Down the group the molecules become more polar, because Te is more electronegative than S.: Electronegativity falls down the group, so the molecules become less polar, not more.
• Down the group the H–X covalent bonds get stronger and longer, so more energy is needed to separate the molecules from one another.: Bonds inside the molecules do not break on boiling, and these bonds get weaker down the group.

Electron count still sets dispersion strength. Water wins overall because each molecule also forms strong hydrogen bonds, which H₂Te cannot. The comparison must name every force in both substances.

• Correct: H₂Te does have stronger dispersion forces, but water's hydrogen bonds add an attraction large enough to outweigh that difference.: Right: the extra force, hydrogen bonding, explains water's high boiling point.
• The student is right: water has fewer electrons but stronger dispersion forces because its molecules are smaller.: Smaller molecules with fewer electrons are less polarizable, so their dispersion forces are weaker.
• H₂Te boils lower because its covalent bonds are weaker, so its molecules fall apart sooner when heated.: Boiling does not break the H–Te bonds; the molecules leave the liquid whole.
• Water boils higher because it is heavier per molecule than H₂Te, which offsets its smaller electron count.: H₂O (18.02 g/mol) is much lighter than H₂Te (129.6 g/mol), so mass cannot explain it.