Assuming a nitrogen molecule spherical in shape and occupying the radius `200` pm calculate
(a) The volume of single molecule of gas
(b) the percentage of empty space in one mole of `N_(2)` gas at `NTP` ? .

Explanation

Step I. Volume of single `N_(2)` molecule
Radius of the molecule `(r) = 2 xx 10^(-10) m = 2 xx 10^(-8) m`
Volume of the molecular `(V) = (4)/(3)pir^(3) = 4/3 xx 22/7 xx (2 xx10^(-8) cm)^(3) = 3.35 xx 10^(-23) cm^(3)`.
Step II. Volume of the empty space
Volume of single molecule `= 3.35 xx 10^(-23) cm^(3)`
No. of molecules in one mole of the gs `= 6.022 xx 10^(23)`
Volume occupied by `6.022 xx 10^(23)` molecules `= (3.35 xx 10^(-23) cm^(3)) xx (6.022 xx 10^(23))`
`= 20.2 cm^(3)`
But the total volume of one mole of gas at N.T.P. `= 22400 cm^(3)`
The percentage of empty space `= ("Volume of the empty space")/("Total volume") xx 100`
`= (22379.8)/(22400) xx 100 = 99.9 %`