Does a dehumidifier make air heavier or lighter? The above question arose to me when considering to buy a dehumidifier to use it in a two story house.
Think first of a heater: I would place it downstairs, because warmer air, lighter, would climb upstairs on its own (and relatively colder air, heavier, would go downstairs), making it clear that it should better work downstairs. On the contrary, putting it upstairs, heat would not circulate properly.
But, being it a dehumidifier, where should I better make it work?
It depends if dehumidified air is heavier or lighter, than the original air.
But I am undecided:
If I think that dry air is (approximately) a mixture of 80% N2 (molecular weight 28) and O2 (32), and water vapour H2O, which only weighs 18, being the lightest fraction, I've always thought that at the same temperature, humid air would be lighter, and conversely dehumidified air would be heavier.
But normally a dehumidifier would render all heat Q1 subtracted from the humid air (to make the water condense) plus the work W used to make the implied refrigerator work, as heat (Q2=Q1+W) in the output air; making it much warmer than incoming air. In fact after consulting a Mollier diagram I am persuaded that output air, even after subtracting water vapour is, due to the higher temperature, lighter. 
Following this reasoning it seems to me that any mixture of this air with the original one, shall still be lighter.
Therefore I should make my dehumidifier work downstairs, much like a heater.
But this field (humid air thermodynamics) is not one where my knowledge is so sound, so I'm not completely sure of which answer is the right one.
 A: Dehumidifiers make air heavier when temperature's held constant.  However, dehumidifiers that aren't air-conditioners tend to dump their heat pump's waste heat back into the processed air, raising its temperature.  The below partial answer considers the constant-temperature case, but doesn't yet address the issue of increased air temperature; more in the comments below.

Wikipedia provides the same argument that you did based on molecular mass, i.e. that density's$$
{\rho}_{\text{humid air}}=\frac{p_{\text{dry air}}M_{\text{dry air}}+p_{\text{water}}M_{\text{water}}}{RT},
$$where the molar mass of dry air,$$
M_{\text{dry air}}{\approx}28.964\frac{\mathrm{g}}{\mathrm{mol}},
$$is greater than the molar mass of water vapor,$$
M_{\text{water}}{\approx}18.016\frac{\mathrm{g}}{\mathrm{mol}}.
$$
There's some non-ideality in the situation, e.g. water molecules are highly polar and so they don't behave perfectly like the ideal gas model assumes, but overall it's close enough at typical temperatures and pressures.
Other symbols:


*

*$p_{\text{dry air}}$ is the partial pressure of dry air;

*$p_{\text{water}}$ is the partial pressure of water vapor;

*$R$ is the universal gas constant;

*$T$ is temperature.
