Question

A sample of argon gas is collected over water at 30.0 degrees C. The level of the water is adjusted until the total pressure in the flask is 771 torr and the volume is 250.0 mL. The vapor pressure of water is 31.8 torr at 30.0 degrees C.?

How many moles of argon are in the flask?
What is the partial pressure of argon in the flask?
What is the total mass of gas in the flask?

Answer 1

`"Partial pressure of Ar"` `=` `(771-31.8)" mm Hg"` `=` `739.2` `"mm Hg"`

Explanation:

Partial pressures are additive. The pressure exerted inside the flask is that due to the argon AND the water vapour.

For `n_(Ar)`, we simply solve the Ideal Gas Equation:

`n=(PV)/(RT)` `=` `((739.2*mm*Hg)/(760*mm*Hg*atm^-1)xx0.2500*L)/(0.0821*L*atm*K^-1*mol^-1xx303*K)`,

the which gives us an answer in `"moles"`, i.e. all the units cancel out except for `n`.

The total mass of GAS in the flask, is that due to this quantity of argon (which has a molar mass of `39.95*g*mol^-1)`, and also the mass of water present as vapour. So you have to do another calculation for the moles of water occupying the given volume at the given temperature with the given pressure of water.