We can use the **Ideal Gas Law** 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)ρ = m/Vcolor(white)(a/a)|)))" "#

∴ #PM = ρRT# and

#color(blue)(|bar(ul(color(white)(a/a)ρ = (PM)/(RT)color(white)(a/a)|)))" "#

#P = 780.0 color(red)(cancel(color(black)("torr"))) × "1 atm"/(760 color(red)(cancel(color(black)("torr")))) = "1.026 atm"#

#M = "46.01 g/mol"#

#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#

#T = "(37 + 273.15) K" = "310.15 K"#

∴ #ρ = (1.026 color(red)(cancel(color(black)("atm"))) × "46.01 g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.082 06" color(red)(cancel(color(black)("atm")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 310.15 color(red)(cancel(color(black)("K")))) = "1.85 g/L"#