A synthetic mixture of nitrogen and Argon has a density of `1.4 g L^(-1)` at `0^@C`. Calculate the average molecular weight. Find out the volume percentage of nitrogen in the mixture.

Explanation

Molecular weight (M) can be obtained from density (d) as, `M=(dRT)/(P )`
Average molecular weight of the mixture =` (1.4 xx 0.0821 xx 273 )/(1) = 31.4`
If the % volume of `N_2 ` is .x.
` 31.4 = ( x xx 28+(100-x)40)/( 100 )implies 12 x = 840 or x=70`
The volume percentage of `N_2` in the mixture = 70