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0.16g of N(2)H(4) are dissolved in water...

`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)lt)`

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`{:(,N_(2)H_(4)+H_(2)OhArr,N_(2)H_(5)^(+),+OH^(-)),("Before dissociation", 1,0,0),("After dissociation",1-alpha,alpha,alpha):}`
Also `K_(b)=(Calpha^(2))/((1-alpha))`
Assuming `1-alpha=~1`
`K_(b)=Calpha^(2)`
`[NH_(2)H_(4)]=C=(0.16xx1000)/(32xx500)=0.01`
Given `K_(b)=4xx10^(-6)M`
`:. alpha^(2)=(4xx10^(-6))/(0.01)=4xx10^(-4)`
`:. alpha=2xx10^(-2)`
`i.e., alpha= 0.02` or `2%`
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