Is the density of the solution significantly different from the density of the solvent? I am working on a mathematical model for the dissolution of coffee in water. I need to know some precise details about the effects of dissolution on the density of the mixture of the substances (that is, the density of the solution). The system is isothermal and the pressure is constant. I just want to focus on mass, volume, density, and concentration as the relevant variables.

In the drawing (below), there is a graphical representation of the system in question (in the bottom image, I should have mentioned concentration of water and coffee instead of density). We have a container of water with a single coffee solid and I've drawn in 20 molecules for both the water and the coffee. The volumes are obviously not to scale (it looks like approximately 50% of the coffee has dissolved in the second pic, not 20%!).

Start

So at the beginning, $t = 0$, no coffee has dissolved so far and the solution liquid is simply the water. We have the following information for the density and the volume of the coffee, water, and the solution, respectively

$\rho_c = 1500$ kg m$^{-3}$

$\rho_w = 1000$ kg m$^{-3}$

$\rho_s = \rho_w = 1000$ kg m$^{-3}$

$V_c = 0.1$ m$^{-3}$

$V_w = 0.5$ m$^{-3}$

$V_s = V_w = 0.5$ m$^{-3}$

The solution is just the water at the start.

Finish

At, say, $t = 100$, 20% of the coffee has dissolved and has diffused out evenly into the solution and the system is in equilibrium.

Now the solution contains both water and coffee. I realise that the chemical reaction dissolution and interactions at a molecular level between the coffee and the water have an impact on the volume and density of the solution. I need information on the density of the solution to decide whether the compressible Navier-Stokes or the incompressible Navier-Stokes govern the flow in the solution.

What is the density  and volume of the solution? If experimental measurement is required for this can you instead give me an approximate answer. 

I need to know how much of a change in solution density occurs from $t = 0$ to $t = 100$.
This depends a great deal both on the solute and the concentration.

At the concentrations that we normally drink coffee at, you can almost certainly ignore the difference in density of the solution. But ccording to your question you are trying to dissolve 0.1m3 coffee (I assume the minus sign is an error) in 0.5m3 water. That's 100L of coffee in 500L of water, a 100L/600L=16.667% volume/volume solution! A lot of very strong coffee! 

As a general rule, the density rho is given as:

rho = (mass component 1 + mass component 2) / (volume component 1 + volume component 2 )

This assumes (correctly) that masses can be added. Strictly speaking, the volume of a solution is not equal to the sum of the volumes of its components, as there is a slight volume change on dissolution, but this is normally insignificant.

The hard part is to know the solid density of coffee, because it cannot be readily measured, as coffee comes in powder form. If you are not using instant coffee there is the problem that the husks of the grains will not dissolve. But in your case, you are told the density of the coffee is 1500kg/m3

So you can easily work out the masses of water and coffee, and then the density of your enormous vat of extra strong coffee.

Hint: it's going to be more than water, but not different enough to make any difference regarding the correct navier stokes model to follow.

EDIT: on re-reading, it seems you are assuming that only 20% of your 0.1m3 of coffee will dissolve: that is 0.02m3 in 0.5m3 of water, an approximate concentration of 4%. That's not quite as absurdly strong as before. I was a little confused because 0.1m3 is 20% of 0.5m3. However I'm not sure how you decided that you would reach equilibrium with 20% of the available coffee dissolved.