Van Der Waals constant of water and benzene ($a$) are $5.537$ and $18.82$ respectively. I have read that the value of Van Der Waals constant indicates the extent of attractive forces. By this, shouldn't it be greater for water instead because water has hydrogen bonding? Also, what are the factors(with priority) on which the Van Der Waals constant depends on ?
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The parameter $a$ from the van der Waals equation
$$\left(p+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$$
is not to be taken only as the measure of strength of intermolecular bonding.
Rather, it is overall measure of significance and frequency of intermolecular interaction.
It may be based on purely elastic electrostatic interactions, where only the weakest van der Waals force may ( or even may not ) be involved.
The probability 2 benzene molecules interact is bigger than for 2 water molecules.
Notes also that for n-hexane is $\pu{a = 24.71 L^2\cdot bar mol^{2-}}$. It has bigger value than benzene, in spite of the weaker interaction due the lack of $\pi$ bonding. But its linear molecule is stratched with all the hydrogen atoms in many directions, so collisions are more probable.
