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

Question

Jeremiah Adams · Accepted 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.

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"#

Ellie Andrews · Answer

undefined To find the partial pressure of CO2, multiply the total pressure by the percentage of CO2 in the air sample converted to a decimal.

Partial pressure of CO2 = Total pressure * (Percentage of CO2 / 100)
Partial pressure of CO2 = 758 mm Hg * (0.038 / 100) = 0.28764 mm Hg

Therefore, the partial pressure of CO2 in the air sample is approximately 0.28764 mm Hg.