Question

unit 3 college board questions
18.

the molecular formula and molar mass of two straight - chain hydrocarbons are listed in the table above. based on the information in the table, which compound has the higher boiling point, and why is that compounds boiling point higher?
(a) ( c_{4}h_{10} ), because it has more hydrogen atoms, resulting in more hydrogen bonding
(b) ( c_{4}h_{10} ), because it has more electrons, resulting in greater polarizability and stronger dispersion forces
(c) ( c_{2}h_{6} ), because its molecules are smaller and they can get closer to one another, resulting in stronger dispersion forces
(d) ( c_{2}h_{6} ), because its molecules are more polar, resulting in stronger dipole - dipole attractions
19.

the table above shows the structural formulas and molar masses for three different compounds. which of the following is a list of the compounds in order of increasing boiling points?
(a) butane < 1 - propanol < acetone
(b) butane < acetone < 1 - propanol
(c) 1 - propanol < acetone < butane
(d) acetone = butane < 1 - propanol

Explanation:

• For question 18:
• Both \(C_{2}H_{6}\) and \(C_{4}H_{10}\) are non - polar hydrocarbons. The main intermolecular force is London dispersion force.
• London dispersion force is related to the number of electrons (or molar mass). \(C_{4}H_{10}\) has a higher molar mass (\(58\ g/mol\) vs \(30\ g/mol\) for \(C_{2}H_{6}\)), so it has more electrons.
• More electrons mean greater polarizability and stronger London dispersion forces. Stronger intermolecular forces lead to a higher boiling point.
• Option A is wrong because neither \(C_{2}H_{6}\) nor \(C_{4}H_{10}\) can form hydrogen bonds (no \(N - H\), \(O - H\), or \(F - H\) bonds). Option C is wrong because smaller molecules (\(C_{2}H_{6}\)) have weaker London dispersion forces. Option D is wrong because both are non - polar.
• For question 19:
• Butane: It is a non - polar hydrocarbon. The only intermolecular force is London dispersion force.
• Acetone: It is a polar molecule (\(C = O\) bond creates a dipole). It has dipole - dipole forces in addition to London dispersion forces. Dipole - dipole forces are stronger than London dispersion forces (for molecules of similar molar mass).
• 1 - propanol: It can form hydrogen bonds (\(O - H\) bond). Hydrogen bonds are stronger than dipole - dipole forces.
• So, the order of increasing boiling points (based on increasing strength of intermolecular forces) is Butane (weakest London dispersion) \(<\) Acetone (dipole - dipole + London dispersion) \(<\) 1 - propanol (hydrogen bonds + dipole - dipole+ London dispersion)

Answer:

1. B. \(C_{4}H_{10}\), because it has more electrons, resulting in greater polarizability and stronger dispersion forces
2. B. Butane \(<\) acetone \(<\) 1 - propanol