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

Solution :`{:(A_(x)B_(y),RARR,XA^(y+),+,yB^(x-)),(1-alpha,,x alpha,,y alpha):}`
`i=1-alpha+x alpha+ y alpha`
`i=1+alpha(x+y-1)`
`alpha=(i-1)/((x+y-1))`


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