1.

^^_(m) of 0.05 M weak electrolyte is "50 Sm"^(2)mol^(-1),0^(@) of it is "440 Sm"^(2)" mol"^(-1). Calculate alpha (degree of dissociation) of the electrolyte.

Answer»

Solution :Degree of dissociation of an electrolyte is GIVEN by
`ALPHA=(^^_(m))/(^^_(m)^(@))`
`"Given :"^^_(m)="50 sm"^(2)" mol"^(-1)`
`^^_(m)^(@)="440 sm"^(2)" mol"^(-1)`
`"conc "=0.05m`
`alpha=(50)/(440)(cancel(sm^(2)))/(cancel(sm^(2)))(cancel(mol^(-1)))/(cancel(mol^(-1)))`
`alpha=0.


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