1.

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))

Answer»

Solution :`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)`


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