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» <html><body><p><br/></p>Solution :`{:(,2HI,hArr,H_(2),+,I_(2)),("moles at",<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.96/128,,0,,0),(t=0,,,,,),(,=7.5xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>),,0,,0),("moles",(7.5xx10^(-3)-x)x/2,,,,x/2),("at <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a>",,,,,):}` <br/> mEq of `I_(2)` at equilibrium =mEq of hypo used for reaction mixture. <br/> `w_(I_(2))/Exx1000=15.7xx1/10` <br/> `:. (w/E)` of `I_(2)=1.57xx10^(-3)` <br/> `:.` moles of `I_(2)` formed `=(1.57xx10^(-3))/2=0.785xx10^(-3)` <br/> `x/2=0.785xx10^(-3) rArr x=1.57xx10^(-3)` <br/> `:.` Degree of dissociation of HI `("or " alpha_(HI))` <br/> `=("moles dissociated")/("moles <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a>")=(1.57xx10^(-3))/(7.5xx10^(-3))` <br/> `alpha_(HI)=0.209` or `20.9%`</body></html>