For a molecule of diameter 2.5 Å and number density of molecule `2.1 xx 10^(25)` molecule `/m^(3)`, calculate average distance (mean free path) covered by a molecule between two successive collisions.

Explanation

Given, `d=2.5xx10^(-10)m,n=2.1xx10^(25) "molecues"//m^(3),l=?`1
We know that, `l=(1)/(sqrt(2)npid^(2))`
`i.e. l=(0.707)/(2.1xx10^(25)xx3.142xx6.25xx10^(-25))=0.01714xx10^(-5)=1.714xx10^(-8)m=1714`Å