Give the expression for the radius of a molecule.

Explanation

Consider 1 mole of water, molecules pertaining to the number `6.023 xx 10^(23)`.
(2xx"atomic mass of" H_(2) + "atomic mass of" O_(2))`
`=(2xx1+1xx16)` gramme = 18g.
0.018 kg.
Hence mass of 1 molecule `=(0.018)/(6.023xx10^(23))kg`
`m = 3xx10^(26)kg.`
Volume of 1 molecule of water `=(m)/(p)=(3xx10^(-26))/(1000)=30xx10^(-30)m^(3)=((4)/(3))piR^(3).`
`R^(3)=(3)/(4pi)xx30xx10^(-30)`.
`therefore " " R=2xx10^(-10)m=2overset(**)(A).`