The of ##O_2## under these conditions is is 0.83 g/L.
We can use the to solve this problem.
##color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|))) ##
Since ##moles = mass/molar mass## or ##n = m/M## we can write
##PV = m/MRT##
We can rearrange this to
##PM = m/VRT##
But ##density= mass/volume## or ##color(brown)(|bar(ul(color(white)(a/a) = m/Vcolor(white)(a/a)|))) ##
##PM = RT## and
##color(blue)(|bar(ul(color(white)(a/a) = (PM)/(RT)color(white)(a/a)|))) ##
##P = 1.0 atm; M = 32.00 g/mol; R = 0.082 06 LatmK^-1mol^-1; T = 197.2 C = 470.35 K##
## = (1.0 color(red)(cancel(color(black)(atm))) 32.00 gcolor(red)(cancel(color(black)(mol^-1))))/(0.082 06 color(red)(cancel(color(black)(atm)))Lcolor(red)(cancel(color(black)(K^-1mol^-1))) 470.35 color(red)(cancel(color(black)(K)))) = 0.83 g/L##