Answer:
[tex]24540\frac{mg}{m^3}[/tex]
Explanation:
Hello,
In this case, since the 3% by volume is represented as:
[tex]\frac{3L\ CO}{L\ gas}[/tex]
By using the ideal gas equation we compute the density of CO:
[tex]\rho =\frac{MP}{RT} =\frac{28g/mol*1atm}{0.082\frac{atm*L}{mol*K}*298K}= 0.818g/L[/tex]
Then we apply the conversion factors as follows:
[tex]=\frac{3L\ CO}{100L\ gas}*\frac{0.818g\ CO}{1L\ CO} *\frac{1000mg\ CO}{1g\ CO} *\frac{1000L\ gas}{1m^3\ gas} \\\\=24540\frac{mg}{m^3}[/tex]
Regards.
