At `-50^(@)C` liquid `NH_(3)` has ionic product is `10^(-30)` .How many amide `(NH_(2)^(-))` ions are present per mm`.^(3)` in pure liqudi `NH_(3)`? (Take `N_(A)=6xx10^(23)`)

Explanation

`K=[NH_(4)""^(+)][NH_(2)""^(=)]=10^(-30)`
`[NH_(2)^(-)]=[NH_(4)^(+)]=10^(-15)M " "( :.2NH_(3)hArr NH_(4)^(+)+NH_(2)""^(-))`
No. of `NH_(2)^(-)` ions `NH_(2)^(-)=((10^(-15)mol e)/(L))((1L)/(10^(6)mm^(3)))((6xx10^(23)ions)/(mol e))=600ions//mm^(3)`