A 2.05 L sample of O2(g) was collected over a liquid at a total pressure of 785 torr and 25°C. When the O2(g)

Question

The rest of the question is was dried (vapor of that liquid removed), the gas had a volume of 1.90 L at 25°C and 785 torr. Calculate the vapor pressure of this liquid at 25°C.

Answer 1

Fairly straightforward:

In the wet gas (O2 & water vapor) water vapor is
0.15/2.05 of the mix as a mole fraction. This fraction defines the vapor pressure, which is proportional to mole fraction. So
3/41 * 785 mm Torr = 81 mm appx

Answer 2

The thing you are misssing is the squared rate of a cubical squared trapazoid the 02g of the giant square is 5x66 of pressure is the total of 25% of a gastric acid of h2o would increase in the size of the pressure in the bottle,the vapor will therefore increase as the volume decreases.And thats how you solve the simple matter of L.