Limiting molar conductivity of an electr...

Limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of anions and cations. This law was given by

Faraday

Ostwald

Davy

Kohlrausch

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___________ Law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.

Limiting molar conductivity of NaBr is

STATEMENT 1 Using koihlrausch's law of independent migration of ions it is possible to calculate wedge^(@) for anyelectrolyte form the lambda^(@) of individual ions STATEMENT 2 : Limiting molar molar conductivity of na electrolyte can be repersented as the sum of hte individual contribution of the anion and cation of hte electrolyte STATEMENT 3: When concentration appraches zero molar conductivity reaches the lowest limit

The molar conductivity of strong electrolyte

At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductance of electrolyte, irrespective of the nature of the other ion with which it is associated. the molar conductance of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions. A_(x)B_(y) rarr xA^(y+)+yB^(x-) Lambda_(m)^(@)(A_(x)B_(y))=xlambda_(A^(y+))^(@)+ylambda_(B^(x-))^(@) where, x and y are the number of cations and anions respectively. The degree of ionisation 'alpha' of weak electrolyte can be calculated as : alpha=Lambda_(m)/Lambda_(m)^(@) The unit of molar conductance of an electrolyte solution will be :

At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductance of electrolyte, irrespective of the nature of the other ion with which it is associated. the molar conductance of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions. A_(x)B_(y) rarr xA^(y+)+yB^(x-) Lambda_(m)^(@)(A_(x)B_(y))=xlambda_(A^(y+))^(@)+ylambda_(B^(x-))^(@) where, x and y are the number of cations and anions respectively. The degree of ionisation 'alpha' of weak electrolyte can be calculated as : alpha=Lambda_(m)/Lambda_(m)^(@) The molar conductances at infinite dilution for electrolytes BA and CA are 140 and 120 ohm^(-1) cm^(2) mol^(-1) . If the molar conductance at infinite dilute dilution of BX is 198 ohm^(-1) cm^(2) mol^(-1) , then at infinite dilution, the molar conductance of CX is :

At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductance of electrolyte, irrespective of the nature of the other ion with which it is associated. the molar conductance of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions. A_(x)B_(y) rarr xA^(y+)+yB^(x-) Lambda_(m)^(@)(A_(x)B_(y))=xlambda_(A^(y+))^(@)+ylambda_(B^(x-))^(@) where, x and y are the number of cations and anions respectively. The degree of ionisation 'alpha' of weak electrolyte can be calculated as : alpha=Lambda_(m)/Lambda_(m)^(@) The ionic conductances of Al^(3+) and SO_(4)^(2-) ions at infinite dilution are x and y ohm^(-1) cm^(2) mol^(-1) respectively. If Kohlrausch's law is valid then molar conductance of aluminium sulphate at infinite dilution will be :

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