How to calculate the vapor pressure of water? $48~\mathrm{l}$ of dry $\ce{N2}$ passes through $36~\mathrm{g}$ $\ce{H2O}$ at $300~\mathrm{K}$, and this results in a water loss of $1.20~\mathrm{g}$. The vapor pressure of water is?


The solution says, that water loss is observed because of escape of water molecules with $\ce{N2}$ gas. Then it says that water vapours occupy the volume of $\ce{N2}$, i.e  $48~\mathrm{L}$, this part I am not able to understand. Why would water occupy the same volume as that of $\ce{N2}$, it isn't mentioned that these two are in some container of volume of $48~\mathrm{L}$. What is the logic behind this? Then the solution proceeds, by applying $pV=\frac{m\mathcal{R}T}{M}$.
It doesn't matter if there is a container.  I'm picturing bubbling the nitrogen through the water, for example the water is in a flask, and there is a hose going down into the water, maybe a frit at the end of the hose.

The important assumption is that nitrogen reaches equillibrium with the water, which is a big assumption.  

But adding 1.2 grams of water would increase the total volume by about 1 L.  I would use the total volume to answer the question, not 48L, but it will only change the answer by about 2%.