Answer to Question #66735 in Inorganic Chemistry for Divyash Chhetri

Question

A flask contains 3.00 moles of nitrogen and 3.00 moles of neon. How many grams of argon must be pumped into the flask in order to make the partial pressure of argon twice that of neon?

Answer 1

The partial pressure of individual gas in the gas mixture can be expressed the following way:
Pi = Ptotal *xi,
Where
Pi – is the partial pressure of individual gas in the mixture;
Ptotal – is the total pressure of the gas mixture;
xi – is the mole fraction of individual gas in the mixture.

Mole fraction is number of moles of individual component divided by total number of moles in the mixture:
xi = ni / ntotal.

Therefore
Pi = Ptotal * ni / ntotal.

We see that partial pressure of the gas component is directly proportional to the number of moles of that gas in the mixture.
So if we want the partial pressure of argon to be twice that of neon, than number of moles of argon should also be twice number of moles of neon.
It is known that mixture contains 3.00 moles of neon. So we need to add 6.00 moles of argon.
Argon has monoatomic molecule, so the mass of 1 mole of argon in grams equals to its atomic weight which is 39.95.
Then 6.00 moles of argon have a mass (6.00 * 39.95) = 239.70 g.

Answer:
239.70 grams of argon must be pumped into the flask.