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First, we need to determine the molecular mass for each of the substances. Since volatility is affected by molecular mass, this will help us determine the substances' volatility order. - \(\mathrm{CH}_{4}\): \(12.01 + (4 \times 1.01) = 16.05 \, g/mol\) - \(\mathrm{CBr}_{4}\): \(12.01+(4 \times 79.904) = 331.73 \, g/mol\) - \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\): \(12.01+ (2 \times 1.01) + (2 \times 35.453) = 84.93 \, g/mol\) - \(\mathrm{CH}_{3} \mathrm{Cl}\): (\(12.01+(3 \times 1.01)+35.453 = 50.48 \, g/mol\) - \(\mathrm{CHBr}_{3}\): \(12.01+(3 \times 79.904) + 1.01= 252.74 \, g/mol\) - \(\mathrm{CH}_{2} \mathrm{Br}_{2}\): \(12.01+(2 \times 1.01)+(2 \times 79.904) = 173.84 \, g/mol\)

For each substance, we will determine the type of intermolecular force present. There are mainly three types of intermolecular forces: London dispersion forces (LDF), dipole-dipole forces, and hydrogen bonding. All substances have LDF, but the strength might vary. We will also consider other types of forces if applicable and use them to estimate boiling points and volatility. - \(\mathrm{CH}_{4}\): London dispersion forces - \(\mathrm{CBr}_{4}\): London dispersion forces - \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\): London dispersion forces and dipole-dipole forces - \(\mathrm{CH}_{3} \mathrm{Cl}\): London dispersion forces and dipole-dipole forces - \(\mathrm{CHBr}_{3}\): London dispersion forces and dipole-dipole forces - \(\mathrm{CH}_{2} \mathrm{Br}_{2}\): London dispersion forces and dipole-dipole forces

Since larger molecules with higher molecular mass have stronger LDFs, substances with greater molecular mass will have lower volatility. Similarly, dipole-dipole forces will also lead to lower volatility as they are stronger than LDFs. Considering these factors, we can place the given substances in the following order of increasing volatility: \(\mathrm{CBr}_{4} < \mathrm{CHBr}_{3} < \mathrm{CH}_{2} \mathrm{Br}_{2} < \mathrm{CH}_{2} \mathrm{Cl}_{2} < \mathrm{CH}_{3} \mathrm{Cl} < \mathrm{CH}_{4}\)

As the volatility of the substances increases, their boiling points are expected to decrease. The reason being, more volatile substances escape to the gas phase more easily, requiring less energy (i.e., lower boiling point) to boil. With this in mind, the boiling points would decrease in the following order: \(\mathrm{CBr}_{4} > \mathrm{CHBr}_{3} > \mathrm{CH}_{2} \mathrm{Br}_{2} > \mathrm{CH}_{2} \mathrm{Cl}_{2} > \mathrm{CH}_{3} \mathrm{Cl} > \mathrm{CH}_{4}\)

The varying boiling points in this series can be explained by the presence and strength of their intermolecular forces. London dispersion forces are present in all of the substances, but their strength increases with increasing molecular mass. Additionally, the substances that also possess dipole-dipole forces result in stronger intermolecular forces leading to higher boiling points. In this series, the strongest intermolecular forces are found in \(\mathrm{CBr}_{4}\) due to its high molecular mass, causing the highest boiling point. On the other hand, \(\mathrm{CH}_{4}\) has the weakest intermolecular forces (only LDF present) and the lowest molecular mass, hence it has the lowest boiling point. The boiling points of other substances can be similarly explained based on the molecular mass and the type of intermolecular forces present.