We can use the Ideal Gas Law to solve this problem:.
#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#
Since #n = m/M#, we can rearrange this equation to get
#pV = (m/M)RT#
And we can solve this equation to get
#M = (mRT)/(pV)#
In your problem,
#m = "283.3 g" #
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "(27 + 273.15) K" = "300.15 K"#
#p = "3.2 atm"#
#V = "30 L"#
∴ #M = ("283.3 g" × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 300.15 color(red)(cancel(color(black)("K"))))/(3.2 color(red)(cancel(color(black)("atm"))) ×30 color(red)(cancel(color(black)("L")))) = "72.7 g/mol"#
The molecular mass of #"X"_2# is 72.7u.
The atomic mass of #"X"# is 36.3 u.
The atomic mass of #"Cl"# is 35.45 u, and the atomic mass of #"Ar"# is 39.95 u.
The atomic mass of #"X"# is closer to that of #"Cl"#, and #"Cl"# also forms diatomic #"Cl"_2# molecules.
Element #"X"# is chlorine.
