Question

An ideal gas has a density of 1.10×10−6 g/cm3 at 1.00×10−3 atm and 80.0 ∘c. identify the gas.

Answer 1

From  ideal  gas  equation  that  is pv =nRt
n=number  of  moles which  can  be  written  as  the  ratio  between the  weight  of  a gas that  is mass and  its  molecular  mass  n=m/Mm
pv=(m/Mm)RT
density  is=mass  per  unit volume
P=m/v by  arranging  the  equation  we  get
R =0.082atm/mol/k
Mm=pRT/P=[(1.10  x10^-6 x1000g/l) xo.082  atm/mol/k x(80+273] /(1.00  x10^-3) =31.84  to the  nearest ten is  32
hence  the  gas  is  oxygen