This is intended as an addendum to the other answers, not a full answer in itself.$\newcommand{\b}[2]{\ce{#1\bond{...}#2}}$
Let's say we have a solution of molecule A, and a solution of molecule B. I'm denoting the strength of intramolecular forces by $\b AA,\b BB$ etc.
Whenever we create a real (i.e. non-ideal) solution, we classify the solution on the basis of deviation from Raoult's law:
Positive deviation
$\Delta V_\text{mix}>0$
$\Delta H_\text{mix}>0$
$VP>VP_\text{expected}$
$\b AB<\b AA, \b BB$. (cohesive forces are stronger than adhesive forces)
Example: water and benzene.
Here, the adhesive forces are weaker. This leads to a net "expansion" of the mixture, and, it is endothermic due to the fact that we are partially replacing strong cohesive "bonds" with weaker, adhesive ones.
Negative deviation
$\Delta V_\text{mix}<0$
$\Delta H_\text{mix}<0$
$VP<VP_\text{expected}$
$\b AB>\b AA, \b BB$. (cohesive forces are weaker than adhesive forces)
Example: ethanol and chloroform.
Here, the adhesive forces are stronger. This leads to a net "contraction" of the mixture, and, it is exothermic due to the fact that we are partially replacing weak cohesive "bonds" with stronger, adhesive ones.
So yes, the enthalpy is not additive for non-ideal solutions.