Account for the volatilities of #HF#, and #"carbon dioxide"#. Why should #CO_2# be MORE volatile than the lighter #HF# molecule?

A chemist interprets data. The relevant data here are the fact that #HF# has a normal boiling point of #19.5# #""^@C#. On the other hand, #CO_2# sublimes at #-78# #""^@C# at #1*atm#. Why the discrepancy, especially as carbon dioxide is a bigger molecule, and thus has greater opportunity for dispersion forces?

When hydrogen is bound to a strongly electronegative element such as oxygen, nitrogen, or fluorine, the heteroatom strongly polarizes electron density towards itself to give a dipole that we could represent as #""^(delta+)H-stackrel(delta-)O-H^(delta+)# or #""^(delta+)H-F^(delta-)# or #""^(delta+)H-stackrel(delta-)NH_2#. The dipoles align to effect the phenomenon of hydrogen bonding, which acts a potent molecular force and is responsible for the elevated boiling points of these hydrides in comparison with say, #PH_3#, or #CH_4#, which are very volatile.

As a chemist, as a physical scientist, you should look up the normal boiling points of the hydrides mentioned here, and see if the data are consistent with the argument proposed.

In general, the stronger the intermolecular forces between the molecules of a compound, the higher the boiling point of that compound.

Why is this? Recall that different intermolecular forces have different relative strengths. The order, from weakest to strongest, of the most common intermolecular forces are

Dispersion forces < Dipole-induced dipole interaction < Dipole- dipole interaction < Hydrogen bonding

Hydrogen bonding is therefore the strongest intermolecular force, and will therefore tend to have the highest boiling point (excluding ion-dipole and ion-ion (ionic bonding) forces), and it occurs between the #"H"# atom in a polar bond (most often #"H"-"F"#, #"H"-"O"#, and #"H"-"N"#) and a nonbonding electron pair on usually #"F"#, #"O"#, or #"N"# in another molecule.

For #"HF"#, hydrogen bonding exists between the #"H"# atom of one molecule and the #"F"# atom of another molecule.

The intermolecular forces present between #"CO"_2# molecules are dispersion forces (because the structure of the #"CO"_2# molecule shows that it is nonpolar), the weakest of the intermolecular forces, and therefore will have the lowest boiling point.