Question

How many grams of fluorine gas will exert a pressure of 134 atm in a 3.2-liter container at 40°C?

Answer 1

It will have to be a stout container.

Explanation:

We use the Ideal Gas equation:

`"Mass"=(PV)/(RT)xx38.00*g*mol^-1`

`=(134*atmxx3.2*L)/(0.0821*(L*atm)/(K*mol)xx313*K)=16.7*molxx38.00*g*mol^-1=635*g`

Why did I use `38.00*g*mol^-1` for fluorine gas and not `19.00*g*mol^-1`?