At `- 50^(@) C`, the self - ionization constant (ion product ) of `NH_(3)" is " K_(NH_(3)) = [NH_(4)^(+)] [NH_(2)^(-)] = 10^(-30) M^(2)`. How many amide ions are present per `mm^(3)` of pure liquid ammonia ?

Options

A. `NH_(4)Cl` is completely dissociated in `NH_(3)`
B. `NH_(4)Cl` is partially dissociated in `NH_(3)`
C. `NH_(4)Cl` is not dissociated in `NH_(3)`
D. Boiling piont is not raised

Answer

A

Explanation

`NH_(4)Cl` is a strong acid in liq. `NH_(3)`. Since, it ionizes to give `NH_(4)^(+)` (which is cation of the solvent).
`NH_(4)CliffNH_(4)^(+)+Cl^(-)`.