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

0.16 g of N_(2)H_(4) are dissolved in water and the total volume made

1.	0.16 g of N_(2)H_(4) are dissolved in water and the total volume made upto 500 ml. Calculate the percentageof N_(2)H_(4) that has reacted with water in this solution . The K_(b) for N_(2)H_(4) is 4.0xx10^(-6)M.
Answer» <html><body><p></p>Solution :`N_(2)H_(4)+H_(2)O hArr N_(2)H_(5)^(+) + OH^(-)`<br/> Conc. Of `N_(2)H_(4) = 0.16 g ` in500 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> = 0.<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> g `L^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) = (0.32)/(32) ` mol `L^(-1) = 0.01 M = 10^(-2) M` <br/> Suppose x mol `L^(-1)` is the amount of hydrazine reacted. Then <br/> `{:(,N_(2)H_(4) ,+ ,H_(2)O,hArr,N_(2)H_(5)^(+),+,OH^(-),,),("Initial conc.",10^(-2)M,,,,,,,,),("Conc. after reaction",10^(-2) - x ,,x,,x,,,,):}` <br/> `K_(b) = ([N_(2)H_(5)^(+)][OH^(-)])/([N_(2)H_(4)])` <br/> `4.0xx10^(-6) = (x^(2))/(10^(2)-x) ~= (x^(2))/(10^(2)) or x^(2)=4.0xx10^(-8) or x=2.0xx10^(-4)` mol `L^(-1)` <br/> `:.` % of hydrazine reacted with water `=(2.0xx10^(-4))/(10^(-2))xx100=2%`</body></html>