Vapor Pressure of Water

[NeedHelpInChem]

a mix of gases is collected over water at 14C (287.15Kelvin) and has a total pressure of .981 atm and occupies 55mL. How many grams of water escaped into the vapor state? Hint: What is the vapor Pressure of water?

 

This has gotten me completely confused.

n=PV/RT

 

n=(.981*.055)/(.08206*287.15)

 

This makes the molarity .0022897674

 

Since Molarity = Mol/G divide .0022897674 by 16 and you get 1.431104611

 

I know this isn't the answer. Where did i go wrong and why do i need the vapor pressure of water?

 

Which btw is 12 mmHg, which is inevitably .0157894737 atm

 

HELP! I'm going crazy over this question!

[big314mp]

The vapor pressure of a liquid is the partial pressure of that liquid's vapors at a given temperature. So basically what you can do is pretend that the other gasses aren't there (applying the ideas of partial pressure).

 

So you end up with a problem that reads something like this:

 

You have a jar of water vapor at 14C. The jar has a volume of 55ml. The pressure in the jar is 12mmHg (<-here's where you need the vapor pressure). Calculate the grams of water vapor present.