0.96 g of HI were, heated to attain equilibrium 2HI hArr H_(2)+I_(2). The reaction mixture on titration requires 15.7 mL of N//10 hypo solution. Calculate the degree of dissociation of HI.

0.96 g of HI were, heated to attain equilibrium 2HI hArr H_(2)+I_(2).

1.	0.96 g of HI were, heated to attain equilibrium 2HI hArr H_(2)+I_(2). The reaction mixture on titration requires 15.7 mL of N//10 hypo solution. Calculate the degree of dissociation of HI.
Answer» Solution :`{:(,2HI,hArr,H_(2),+,I_(2)),("moles at",0.96/128,,0,,0),(t=0,,,,,),(,=7.5xx10^(-3),,0,,0),("moles",(7.5xx10^(-3)-x)x/2,,,,x/2),("at EQUILIBRIUM",,,,,):}` mEq of `I_(2)` at equilibrium =mEq of hypo used for reaction mixture. `w_(I_(2))/Exx1000=15.7xx1/10` `:. (w/E)` of `I_(2)=1.57xx10^(-3)` `:.` moles of `I_(2)` formed `=(1.57xx10^(-3))/2=0.785xx10^(-3)` `x/2=0.785xx10^(-3) rArr x=1.57xx10^(-3)` `:.` Degree of dissociation of HI `("or " alpha_(HI))` `=("moles dissociated")/("moles TAKEN")=(1.57xx10^(-3))/(7.5xx10^(-3))` `alpha_(HI)=0.209` or `20.9%`