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