`0.16 g N_(2)H_(4)` is dissoolved in `H_(2)O` and total volume is made upto `500 mL`. Calculate the percentage of `N_(2)H_(4)` that has reacted with `H_(2)O` in this solution. `K_(b)` for `N_(2)H_(4) = 4.0 xx 10^(-6)M`.

Explanation

Molarity of `N_(2)H_(4)` solution `=("Mass of "N_(2)H_(4))/("Molar mass of "N_(2)H_(4)) xx (100)/(500)`
`=((0.16 g))/(32 g " mol"^(-1)) xx (100)/(500) =0.01M`
Let x mole on `N_(2)H_(4)` be dissolved in water .
`{:(,N_(2)H_(4)+H_(2)O,hArr,N_(2)H_(5)^(+) ,+, OH^(-)),("Initial conc", 0.01 "mol" ,,0,,0),("Eqm. conc",(0.01 -x)"mol",, x"mol",,x"mol"):}`
`K_(b) =[[N_(2)H_(5)^(+)][OH^(-)]]/[[N_(2)H_(4)]] =(x xx x)/((0.01 -x))`
`4.0 xx 10^(-6) =(x^(2))/((0.01 -x)) =(x^(2))/(0.01)`
`x^(2) =4.9 xx 10^(-6) xx 10^(-2) " or " x= (4.0 xx 10^(-8))^(1//2)" "[:. 0.01 -x~~ 0.01]`
`= 2.0 xx 10^(-4) mol`
`:.`Precentage of `N_(2)H_(4)` reacted with water `=((2.0 xx 10^(-4) "mol"))/((0.01 "mol")) xx 100 =2.0%`