`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)`

Explanation

Molarity of `N_2H_4` solution
`=0.16/32xx1000/500`=0.01 M
Let the number of `N_2H_4` reacted with water =`alpha`
`{:(,N_2H_4+H_2O hArr , N_2H_5^(+)+, OH^(-)),("Initial con.",0.01, 0,0),("Equi. Conc.",0.01 -alpha, alpha, alpha):}`
`K_b=([N_2H_5^+][OH^-])/[N_2H_4]=(alphaxxalpha)/(0.01-x)`
`4.0xx10^(-6)=alpha^2/(0.01-alpha) =alpha^2/0.01 (0.01 -alpha approx 0.01)`
`therefore alpha^2 =4xx10^(-6) xx0.01=4xx10^(-8)`
`therefore alpha =2xx10^(-4)`
% of `N_2H_4` reacting with water =`(2xx10^(-4)xx100)/0.01` =2.0%