So because the problem asks us to calculate the vapor pressure of water a 25 degrees Celsius, what we can do is we can take the number of moles of total sample and subtract out the portion of the sample of its oxygen to get that 0.27 Moles of that sample was actually water vapor because we have the number of moles of water and the total number of Mel's sample, I can use those two pieces of information along with the total pressure of our sample to figure out the partial pressure of water. After drying the 1.94 leaders again, we can plug in the ideal gas constant in the temperature and get 0.0 Hey 18 moles off oxygen now that the water vapour has been removed. So, after the gases dried, then we can do the same thing, replacing the two leader volume for the reduced volume that is, um, retrieved.
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Um, you know that we have a 785 tour sample that we have to convert toe atmospheres, and we know that that sample before its stride ISS two leaders in volume you can pull you in our ideal gas constant and the temperature, which is 298 Calvin to solve. So over here we can plug in what we know. So to do this, we can use, um, our ideal gas law equation to solve for number of moles, of oxygen and our sample both before and after it's dried. The problem asks you to calculate the vapor pressure of water a 25 degrees Celsius. The gas had a volume of 1.94 leaders at 25 degrees Celsius and 785 tour.
![23.8 mmhg to atm 23.8 mmhg to atm](https://i.ytimg.com/vi/TsMj8UD4Gao/maxresdefault.jpg)
Problem, and 41 says that a two litre sample of oxygen gas was collected over water at a total pressure of 785 tour and 25 degrees Celsius when the oxygen gases dried or the water vapors removed.