The density of a gaseous mixture of He and `N_2` is found to be `(10)/(22.4) g//1` at STP. The percentage composition of He and `N_2` in this mixture respectively will be

Options

A. `75 % , 25%`
B. `25 % , 75%`
C. `30 % , 70 %`
D. `40 % , 60 %`

Answer

A

Explanation

`because M_("mix") = (10)/(22.4) xx 22.4 = 10`
`therefore` Let the volume of He = `V_(1)` ml
Let the volume of `N_2 = V_2 ` ml
`therefore V_1 + V_2 = 100`
`therefore 10 = ((V_1)/(100)) xx 4 + (V_(2))/(100) xx 28`
`10 + (V_(1))/(100) xx 4 + ((100 - V_(1))/(100)) xx 28`
Solving , `V_(1) = 75%`