An air sample contains 0.038% #CO_2#. If the total pressure is 758 mm Hg, what is the partial pressure of #CO_2#?

1 Answer
Dec 30, 2017

Answer:

#P_(CO_2)# = #0.288"mmHg"# if #%CO_2# is mole percent. If #%CO_2# is weight percent then one will need the mass values of the other components of the air mixture.

Explanation:

Given #0.038%# (mole/mole) => #X_(CO_2)# = #(0.0038"mole" CO_2)/(100"moles"AIR")# = #0.00038#

#X_(CO_2)# = #P_(CO_2)/P_"Total"# => #P_(CO_2)#= #X_(CO_2)##dot##P_(CO_2)#

= #0.00038(758"mmHg"#)=#0.288"mmHg"#