The degree of dissociation `(alpha)` of a weak electrolyte, `A_(x)B_(y)` is related to van't Hoff's factor `(i)` by the expression:

Options

A. `alpha=(i-1)/((x+y-1))`
B. `alpha=(i-1)/(x+y+1)`
C. `alpha=(x+y-1)/(i-1)`
D. `alpha=(x+y+1)/(i-1)`

Answer

1

Explanation

`{:(,A_(x)B_(y),hArr,xA^(y+),+,yB^(x-)),("Total moles before ionization",1mol,,0mol,,0mol),("Total moles after ionization",(1-alpha)mol,,(xalpha)mol,,(yalpha)mol):}`
`i=("Total moles after ionization")/("Total moles before ionization")`
`=((1-alpha)+(xalpha)+(yalpha))/1`
`i=1+(x+y-1)alpha`
or `alpha=(i-1)/((x+y-1))`