The molecular mass of air, at standard pressure and temperature, is approximately 28.97 g/mol. Calculate the mass of 3.33 moles of air.
First, complete the unit conversion using dimensional analysis:
A • B/C
A: 3.33 mil air
B: 28.87 g air
C: 1 mol air

We are given:

Molar mass of air = 28.97 grams/mole

Amount of Air in question = 3.33 moles

Mass of 3.33 moles:

We know that:

mass = molar mass * number of moles

replacing the variables

mass = 28.97 * 3.33 moles

[the 'mole' in the numerator and the denominator will cancel out]

mass = 28.97 * 33.3 grams

mass = 96.47 grams

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

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