A gas mixture composed of `N_(2) and O_(2)` gases has a density of `1.17g*L^(-1)` at `27^(@)C` and 1 atm pressure. Calculate the mass percents of `N_(2) and O_(2)` in the mixture. Assume that the gas mixture behaves like an ideal gas.

Explanation

Given P=1 atm and T=(273+27)K=300K
`therefore M=(dRT)/(P)=(1.17xx0.0821xx300)/(1)=28*8g*mol^(-1)`
`therefore`Average molar mass of the gas mixture`=28.8g*mol^(-1)`
In the mixture, if the mole-fraction of `N_(2)` be x, then the mole-fraction of `O_(2)` is (1-x).
`therefore M=28.8g*mol^(-1)=[28xx x+32(1-x)]g*mol^(-1)`
Mole-fraction of `N_(2)=0.8` and that of `O_(2)=1-0.8=0.2`
Mass percent of
`N_(2)=(0.8xx28)/(0.8xx28+0.2xx32)xx100=77.77%` and mass
persent of `O_(2)=(100-77.77)%=22.23%`