Question

If 1.93 * 10^-3 mol of argon occupies a 65.8-mL container at 20 degrees Celsius, what is the pressure (in torr)?

Answer 1

`P~=0.706*atm-=??"Torr"`

Explanation:

From the Ideal Gas equation, which you must be getting sick of.......

`P=(nRT)/V=(1.93xx10^-3*molxx0.0821*(L*atm)/(K*mol)xx293*K)/(65.8*mLxx10^-3*L*mL^-1)`

`-=0.706*atm`

And here the trick is to know that `1000*mL-=1*L`, i.e. `1*mL-=10^-3*L`.

Anyway, you know the relationship between `"atmospheres"` and `mm*Hg` so have at it.............