`0.16`g of `N_(2)H_(4)` are dissolved in water and the total volume made upto 500 mL. Calculate the percentage of `N_(2)H_(4)` that has reacted with water in this solution. `(K_(b)` for `N_(2)H_(4)=4.0xx10^(-6))`

Explanation

`{:(,N_(2)H_(4)+,H_(2)OhArr,N_(2)H_(5)^(o+)+,OH^(Theta)),("Before dissociation",1,,0,0),("After dissociation",1-alpha,,alpha,alpha):}`
Also `K_(b) = (Calpha^(2))/((1-alpha))`
Assuming `1-alpha = 1`
`K_(b) = C alpha^(2)`
`[N_(2)H_(4)] = C = (0.16 xx 1000)/(32 xx500) = 0.01`
Given `K_(b) =4 xx 10^(-6)M :. alpha^(2) = (4 xx 10^(-6))/(0.01) = 4 xx 10^(-4)`
`alpha = 2 xx 10^(-2)` i.e., `alpha = 0.2 or 2%`