What mass of carbon dioxide is present in 1.00 m3 of dry air at a temperature of 21 ∘C and a pressure of 649 torr ?

I got #"0.545 g CO"_2(g)#, assuming it is ideal...

If we assume that #"CO"_2# is an ideal gas at this temperature and pressure, then the mols of it are given in the ideal gas law:

#PV = nRT#

• #P# is the pressure in #"atm"#.
• #V# is the volume in #"L"#.
• #n# is the mols of an IDEAL gas.
• #R = "0.082057 L"cdot"atm/mol"cdot"K"# if the units of #P# and #V# are as given above.
• #T# is the temperature in #"K"#.

Convert these to the proper units...

#P = 649 cancel"torr" xx "1 atm"/(760 cancel"torr") = "0.8539 atm"#

#V = 1.00 cancel("m"^3) xx ((100 cancel"cm")/cancel"1 m")^3 xx cancel"1 mL"/cancel("1 cm"^3) xx "1 L"/(1000 cancel"mL")#

#= 1.00 xx 10^3 "L"#

#T = 21 + "273.15 K" = "294.15 K"#

Therefore, the mols of dry air in general, if it is truly ideal, is...

#n = (PV)/(RT)#

#= ("0.8539 atm"cdot1.00 xx 10^3 "L")/("0.082057 L"cdot"atm/mol"cdot"K"cdot"294.15 K")#

#=# #"35.4 mols dry air"#

In dry air, there is a mol fraction of #chi_(CO_2) = 0.0350%#, or #0.000350#. Thus, the mols of #"CO"_2# in dry air is simply

#"35.4 mols dry air" xx 0.000350 = ul("0.00124 mols CO"_2(g))#

and this has a mass of

#color(blue)(m_(CO_2)) = 0.00124 cancel("mols CO"_2) xx ("44.009 g CO"_2)/cancel"1 mol"#

#=# #color(blue)ul("0.545 g CO"_2)#