1.

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

Answer»

`alpha=(x+y-1)/(i-1)`
`alpha=(x+y+1)/(i-1)`
`alpha=(i-1)/((x+y-1))`
`alpha=(i-1)/(x+y+1)`

Solution :`{:(,A_(x)B_(y),hArr,XA^(y+),+,yB^(x-)),("MOLES at EQUILIBRIUM",1-alpha,,x alpha,,yalpha):}`
TOTAL no. of moles `=1-alpha+x alpha+y alpha`
`i=(1-alpha+x alpha+y alpha)/(1)`
`:.(i-1)=alpha(x+y-1)`
`:.alpha=(i-1)/((x+y-1))`


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