Question

How do you find the number of grams of `CO_2` that exert a pressure of 785 mmHg at a volume of 325 L and a temperature of 32°C?

Answer 1

Approx. `600*g` of carbon dioxide gas.

Explanation:

We know (or should know) that `n=(PV)/(RT)`, and further that `1*atm-=760*mm*Hg`.

And thus `n=((785*mm*Hg)/(760*mm*Hg*atm^-1)xx325*L)/(0.0821*L*atm*K^-1*mol^-1xx305*K)~=13*mol`.

And `13*molxx44.0*g*mol^-1~=600*g`.

From where did I get the figure `44.0*g*mol^-1`?

My own feeling is that you should never use a column of mercury to measure pressure ABOVE atmospheric pressure. You will end up getting mercury EVERYWHERE. A column of mercury is suitable for measuring LOW pressures. These days, mercury is rarely present in a laboratory due to safety concerns.