Auto-ionisation of liquid `NH_(3)` is
`2NH_(3) hArr NH_(4)^(o+) +NH_(2)^(Theta)`
with `K_(NH_(3)) = [NH_(4)^(o+)] [NH_(2)^(Theta)] = 10^(-30) at -50^(@)C` Number fo amide ions `(NH_(2)^(Theta))`, present per `mm^(3)` of pure liquied `NH_(3)` is

Options

A. 600
B. 300
C. 200
D. 100

Answer

A

Explanation

`[NH_(2)^(-)]=[NH_(4)^(+)]`
`:. [NH_(2)^(-)]=(10^(-30))^((1)/(2))=10^(-15)M`
`:. `No. of `NH_(2)^(-1)` in 1 mL `=10^(-15)xx 10^(-3)`
`=10^(-18) `mol
`=6.02 xx 10^(23) xx 10^(-18) =6.02 xx 10^(5)`
`:. `No. of `NH_(2)^(-)` per `m m^(3)`
`=("NO. of "NH_(2)^(-)"ion in 1 " cm^(2))/(10^(3))=(6.02 xx 10^(5))/(10^(3))~~600`