Nitrogen molecule `(N_(2))` has radius of about 0.2 nm. Assuming that nitrogen molecule is spherical in shape, calcualte
(a) volume of a single molecule of `N_(2)`.
(b) the percentage of empty space in one mole of `N_(2)` gas at S.T.P.

Explanation

(a) The volume of a sphere `=(4)/(3)pir^(3)`
where r is the radius of the sphere.
For `N_(2)` molecule, `r=0.2nm=0.2xx10^(-9)m=2xx10^(-8)cm`
Volume of a molecule of `N_(2)=(4)/(3)xx(22)/(7)xx(2xx10^(-8))^(3)cm^(3)`
`=3.35xx10^(-23)cm^(3)`
(b) To calculate the empty space, let us first find the total volume of 1 mole `(6.02xx10^(23)" molecules")` of `N_(2)`
`"Volume of "6.02xx10^(23)" molecules of "N_(2)=3.35xx10^(-23)xx6.02xx10^(23)=20.17cm^(3)`
`"Now, volume occupied by 1 mole of gas at S.T.P. "="22.4 litre "=22400 cm^(3)`
`"Empty volume"="Total volume of gas "-"Volume occupied by molecules"`
`=(22400-20.17)cm^(3)`
`=22379.83cm^(3)`
`therefore" Percentage empty space"=("Empty space")/("Total volume")xx100`
`=(22379.83)/(22400)xx100`
`=99.9%`
Thus, `99.9%` of space of 1 mole of `N_(2)` at S.T.P. is empty.