Difference between water and ethanol density for an object I'm just slightly confused. 
Say that I had an object that floated 19.4m in water from the bottom of the object to the surface. 
Now I was going to change the fluid to ethanol, which has a density of 789kg $m^{-3}$
Would this mean that the object sinks deeper so the calculation to find the new sinking depth be $19.4 \times (\frac{1030}{789}) = 25.3 \ metres$ or the opposite?
 A: Well, that sort of answer requires the object to present a roughly-constant cross-section, which you have not explicitly stated.
When an object is "floating out of the water" for any fluid of density $\rho,$ you can draw an imaginary plane which coincides with the water's surface everywhere else but "cuts through" the object. The space between this imaginary plane and the water has some "displaced volume" $V,$ and the object will sink into the water until $V = M/\rho.$ Another way to write this is $V \rho = M.$
For a constant-cross-section $A$ we can say that $V = A h$ and this indeed leads to $A h_1 \rho_1 = A h_2 \rho_2$  hence $h_2 = h_1 \rho_1 / \rho_2,$ just as you've done.
Suppose instead the cross-section is that of a triangular prism which points straight down: then we would have $V = \alpha h^2$ and this would then cause $h_2 = h_1 \sqrt{\rho_1 / \rho_2},$ for example. A more realistic model for most ships, for example, would be a parabolic hull sitting in the water, so that $h = \lambda w^2$ and then $V = \int dh~\ell~w,$ for constant $\ell$ this is $\alpha h^{3/2}$ and the result is then $h_2 = h_1 (\rho_1/\rho_2)^{2/3}.$
The bottom line is that all we can say is that $h_2 = h_1~f(\rho_1/\rho_2)$ for some $f$ (and we can say that by purely dimensional analysis). The selection of the exact function $f$ will depend sharply on geometry. Physics precluding negative-volumes from being made underwater just restricts it to be a monotonically increasing function but does not say, e.g., that $f(x) = x$.
A: Archimedes principle says:

the upward buoyant force that is exerted on a body immersed in a fluid is equal to the weight of the fluid that the body displaces

Or as an equation:
$$ \rho V = mg $$
where $V$ is the volume displaced and $\rho$ is the density of the liquid. A quick rearrangement of this gives:
$$ V = \frac{mg}{\rho} $$
So if you reduce the density $\rho$ the volume displaced $V$ goes up i.e. your object sinks lower.
Relating $V$ to the depth the object sinks depends on the shape of the object.
