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

A

Explanation

`{:(,A_(x)B_(y),rarr,x A^(y+),+,y B^(x-)),("Before diss.",1,,0,,0),("After diss.",1-alpha,,x alpha,,x alpha):}`
`i = l-alpha+x alpha + y alpha` or `(i - l) = alpha (x+y-1)`
`:. alpha = (i-1)/((x+y-1))`