1.

Derive relation between kappa_(a) and Lamda_(m)^(@) for solution of weak electrolyte.

Answer»

SOLUTION :* At CONSTANT temperature, solution of weak electrolyte solution,
If `alpha`=DEGREE of dissociation of solution
c=concentration of solution in MOLARITY
`Lamda_(m)`=MOLAR conductivity of solution
`Lamda_(m)^(@)`=limiting molar conductivity of solution
`therefore alpha=(Lamda_(m))/(Lamda_(m)^(@))=("Molar resistivity of solution")/("Limiting molar resistivity of solution")`
* For weak electrolyte like acetic acid,
`therefore kappa_(a)=(calpha^(2))/((1-alpha))=(cLamda_(m)^(2))/((Lamda_(m)^(@))^(2)(1-(Lamda_(m))/(Lamda_(m)^(@))))`
`kappa_(a)=(cLamda_(m)^(2))/(Lamda_(m)^(@)(Lamda_(m)^(@)-Lamda_(m)))`.


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