The trick here is to realize that the density of the gaseous element is actually five times that of molecular oxygen,
Your strategy here will be to determine the
You can use the ideal gas law equation
#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#
Here you have
to show that the molecular mass of the molecule is five times that of atomic oxygen.
Under similar conditions for pressure and temperature, equal number of moles of gas occupy the same volume. If you take
#n = m/M_M#
to be the number of moles of gas present in a sample, you can say that
#PV = m/M_M * RT#
Rearrange to get
#P * M_M = m/V * RT#
which is equivalent to
#overbrace(m/V)^(color(blue)("density")) = P/(RT) * M_M#
As you know, the ratio between the mass of a substance and the volume it occupies gives you that substance's density. Assuming that you're dealing with ideal gases, the same number of moles of
Now, you have
#m_(O_2)/V = P/(RT) * M_("M O"_2) ->#for oxygen gas
#m_(X_3)/V = P/(RT) * M_("M X"_3) ->#for the triatomic molecule #"X"_3#
You know that
#overbrace(m_ (X_ 3)/color(red)(cancel(color(black)(V))))^(color(purple)("density of X"_ 3)) = 5 xx overbrace(m_ (O_2)/color(red)(cancel(color(black)(V))))^(color(darkgreen)("density of O"_2))#
#color(white)(aaaa)m_(X_3) = 5 xx m_(O_2)#
This means that the difference between these densities lies in the fact that each individual molecule of
Therefore, the molar mass of
#M_ ("M X"_ 3) = 5 xx M_("M O"_2)#
The molar mass of oxygen gas is approximately equal to
#M_ ("M X"_3) = 5 xx "32 g mol"^(-1) = "160 g mol"^(-1)#
The molecule is triatomic, which of course means that it contains
#M_"M X" = M_("M X"_3)/3 = "160 g mol"^(-1)/3 = color(green)(|bar(ul(color(white)(a/a)color(black)("55.3 g mol"^(-1))color(white)(a/a)|)))#
I'll leave the answer rounded to three sig figs.
For the sake of convenient discussion the Symbol of the gaseous element is considered as E .
Since it is tri-atomic one its molecular formula will be written as
If its Relative atomic mass is a then its Relative molecular mass will be 3a .
Now by the given condition
Under the same condition of temperature and pressure
Hence we can write Mass per unit voume in place of density to represent the ratio.
Again by Avogadro's Law It is known that under the same condition of temperature and pressure equal volume of all gases contain same number of molecules. Here we consider that under the same condition of temperature and pressure unit volume of all gases contain n molecules.
So the relation becomes
Dividing both numerator and denominator by
Please Note that Relative Molecular mass,Relative Atomic mass are unit less quantities
As discussed here