Of the floowing compound, which would you exoect to have the lowest vapor pressure at room temperature:
CHCl3
CH2Cl2
CH3Cl
CCl4
From my result I think it is CH3cl beacuse
of the dispersion force it has the lowest force
To determine which compound would have the lowest vapor pressure at room temperature, we need to consider the intermolecular forces acting between the molecules.
Vapor pressure is a measure of the tendency of a substance to evaporate. The stronger the intermolecular forces, the lower the vapor pressure.
In this case, we need to compare the strength of intermolecular forces among the given compounds - CHCl3, CH2Cl2, CH3Cl, and CCl4.
CHCl3, CH2Cl2, and CH3Cl are all halogenated hydrocarbons and have dipole-dipole forces between their molecules, which arise due to the difference in electronegativity between the carbon and the halogen atoms. These dipole-dipole forces increase with increased molecular polarity.
Now, let's compare the molecular polarities of these compounds. The chlorine atom is more electronegative than carbon, causing it to attract electrons more strongly and creating a partial negative charge. As a result, the carbon atom becomes partially positive.
CHCl3 has three chlorine atoms, resulting in the highest molecular polarity among the given compounds. CH2Cl2 has two chlorine atoms, CH3Cl has one chlorine atom, and CCl4 has four chlorine atoms.
Hence, according to the strength of intermolecular forces, CHCl3 would have the highest boiling point among the given compounds. As a result, it would also have the lowest vapor pressure at room temperature.