Is the density of the solution significantly different from the density of the solvent?
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.