Question

`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 1

Correct Answer - A::B
`{:(,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%`