Question

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

Answer 1

`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"`