At `0^(@)C` the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar What is the molecular mass of the oxide ? .
Using the expression, `d=(MP)/(RT)`, at the same temperature and for same density,
`underset((“Gaseous oxide”))M_(1)P_(1) ” “= ” ” underset((N_(2))M_(2)P_(2)` (as R is constant)
`:. ” ” M_(1)xx2=28xx5 (“Molecular mass of “N_(2)=28 u)`
or `” ” M_(1)=70 u`
The density `(rho)` of a gas `= PM//RT`
Here R and T are the constants for the gases
For nitrogen, `P = 5 “bar”, M = 28 g mol^(-1)`
`:. rho_(N_(2)) = (PM)/(RT) = ((5 “bar”) xx (28 g mol^(-1)))/(R xx T)`
For gaseous oxide, `P = 2 “bar” , M = ?`
`:. rho_(“oxide”) = (PM)/(RT) = ((2 “bar”) xx M)/(R xx T)`
According to available data , `rho_(N_(2)) = rho _(“oxide”)`
or `(5 “bar”) xx (28 g mol^(-1)) = (2 “bar”) xx M`
`:. M = ((5 “bar”) xx (28 g mol^(-1)))/((2 “bar”)) = 70 g mol^(-1)`.