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

Volume of 1 molecule ` = (4)/(3) pi r^(3)` (spherical shape)
`( :' 200 "pm" = 2xx10^(-8) cm) = (4)/(3) xx (22)/(7) xx [2xx10^(-8)]^(3)`
` = 3.35 xx 10^(-23) cm^(3)` per molecule
(b) Now volume of `N` molecules of `N_(2)`
`= 3.35 xx 10^(-23) xx 6.02 xx 10^(23)`
`= 20.177 = 20.20 cm^(3)` per mol
At `NTP` volume of 1 mole ` = 22400 cm^(3)`
Thus empty space ` = 22400 - 20.20 = 22379.8 cm^(3)`
Thus % empty space `= ("Empty volume")/("Available volume") xx 100`
` = (22379.8)/(22400) xx 100 = 99. 9%` .