The degree of dissociation `(alpha)` of a weak electrolyte `A_(x)B_(y)` is related to van't Hoff 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),1,xA^(-y),+,B^(-x)),(1,,0,,0),(1-alpha,,xalpha,,yalpha):}`
`i=(1-alpha+xalpha+yalpha)/(1)`
`i=1+alpha(x+y-1)`
`alpha=(i-1)/(x+y-1)`