In each of the following groups of substances, pick the one that has the given property. Justify your answer. a. highest boiling point: HBr, Kr, or \(\mathrm{Cl}_{2}\) b. highest freezing point: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{NaCl}\) , or HF c. lowest vapor pressure at $25^{\circ} \mathrm{C} : \mathrm{Cl}_{2}, \mathrm{Br}_{2},\( or \)\mathrm{I}_{2}$ d. lowest freezing point: \(\mathrm{N}_{2}, \mathrm{CO},\) or \(\mathrm{CO}_{2}\) e. lowest boiling point: \(\mathrm{CH}_{4}, \mathrm{CH}_{3} \mathrm{CH}_{3},\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) f. highest boiling point: \(\mathrm{HF}, \mathrm{HCl},\) or \(\mathrm{HBr}\) g. lowest vapor pressure at $25^{\circ} \mathrm{C} : \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CCH}_{3}$ or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

Step by step solution

01

a. Highest Boiling Point

To determine the highest boiling point, we consider the strength of the intermolecular forces. HBr has dipole-dipole forces, Kr has London Dispersion forces, and Cl2 also have London Dispersion Forces. However, HBr is a polar molecule with stronger dipole-dipole forces so it will have the highest boiling point among the three substances. Answer: HBr

02

b. Highest Freezing Point

When looking at the highest freezing point, we need to consider the type of bonding and the intermolecular forces. H2O has hydrogen bonding, NaCl has ionic bonding, and HF has hydrogen bonding. NaCl has the strongest ionic bond which leads to a much higher freezing point compared to H2O and HF. Answer: NaCl

03

c. Lowest Vapor Pressure at 25°C

The substance with the lowest vapor pressure at 25°C is expected to have stronger intermolecular forces. Cl2, Br2, and I2 are all nonpolar molecules having London Dispersion Forces. As molecular weight increases, the London Dispersion Forces also increase. Hence, I2 has the highest molecular weight and thus the strongest London Dispersion forces, leading to the lowest vapor pressure. Answer: I2

04

d. Lowest Freezing Point

To determine the lowest freezing point, we'll consider intermolecular forces. N2, CO, and CO2 are all nonpolar molecules with London Dispersion forces, but CO and CO2 also have dipole-dipole forces. As molecular weight increases, London Dispersion forces increase. N2 has the lowest molecular weight, so it will have the weakest intermolecular forces and the lowest freezing point. Answer: N2

05

e. Lowest Boiling Point

For the lowest boiling point, we'll again consider intermolecular forces. CH4, CH3CH3, and CH3CH2CH3 are all nonpolar molecules with London Dispersion forces. The molecule with the smallest molecular weight will have the weakest intermolecular forces and the lowest boiling point. CH4 has the smallest molecular weight among the three. Answer: CH4

06

f. Highest Boiling Point

HF, HCl, and HBr are polar molecules with dipole-dipole forces, and since HF also forms hydrogen bonds, it should have a higher boiling point than the other two. Answer: HF

07

g. Lowest Vapor Pressure at 25°C

CH3CH2CH3, CH3CCH3, and CH3CH2CH2OH can be compared considering their intermolecular forces. CH3CH2CH3 and CH3CCH3 have London Dispersion forces, while CH3CH2CH2OH has hydrogen bonding. Since hydrogen bonding is stronger than London Dispersion forces, CH3CH2CH2OH will have the lowest vapor pressure at 25°C. Answer: CH3CH2CH2OH

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