Your question is a valid analytical chemistry question which has deep roots in history. No, the reason is not that $\ce{P2O5}$ reacts with water to from phosphoric acid and that is why $\ce{P2O5}$ values are quoted as suggested in the comments below the question. This is a very old historical tradition. In older analytical chemistry, chemists had only two main tools to analyse something namely, titration and gravimetry. Oxygen was also used to determine atomic weights, and people were interested in the ratios of element:oxygen ratio. One person even got a Nobel Prize as well. Generally, when the precipitates were heated, what was left was oxides. In really old books (at least 100 year old) you will notice that many analyses were quoted as oxides. For example, this is true in fertilizer industry to quote potassium as $\ce{K2O}$ and so on. Now if you strongly heat a precipitate phosphate, one will not get a $\ce{P2O5}$ but the convention to report elements as oxides continues.
Second part: I will let you solve the rest of the problem: Start from here
1 mol $\ce{P2O5}$ = 2 mol $\ce{H3PO4}$
and do wt percentage conversions to moles. A 70% $\ce{P2O5}$ solution in water means 70 g $\ce{P2O5}$ is present in 100 grams of solution. Do mol conversions. You will also need the density of the solution.