Liquid `NH_(3)` ionises to a slight extent. At `-50^(@)C`, its ionic product `K_(NH_(3)) = [overset(Θ)Nh_(4)] [overset(Θ)NH_(2)]` is `10^(-30)`. How many amide ions, `overset(o+)NH_(2)` are present per `mm^(3)` of pure liquid `NH_(3)`?

Explanation

`2NH_(3) hArr overset(oplus)(N)H_(4)+ overset(oplus)(N)H_(2)`
`K_(NH_(3)) = [overset(o+)NH_(4)] [overset(Θ)NH_(2)] = 10^(-30)`
Since each ion ins produced in equal number of moles
`:. [overset(Θ)NH_(2)] = [overset(o+)NH_(4)] = 10^(-15)M`
Number of `"ions/ mm"^(3)`
`= ((10^(-15) "mol")/(L)) ((1L)/(10^(6)mm^(3))) ((6xx10^(23) ions)/("mol"))`
`= (10^(-15)xx1xx6 xx 10^(23))/(10^(6)) = 600 "ions mm"^(-3)`