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How is Lambda(m)^(oo) of weak electroyte...

How is `Lambda_(m)^(oo)` of weak electroytes calculated according to Kohlrausch's law ?

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`Lambda_(m)^(oo)` of weak electrolytes is calculated in terms of `Lambda_(m)^(oo)` of strong electrolytes. For example,
`Lambda_(m)^(oo)CH_(3)COOH=Lambda_(m)^(oo)CH_(3)COONa+Lambda_(m)^(oo)HCl-Lambda_(m)^(oo)NaCl`
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Why is not possible to determine Lambda_(m)^(oo) for weak electrolytes by extrapolation ?

Conductors allow the passage of electric current through them. Metallic and electrolytic are the two type of conductors. Current carriers in metallic and electrolytic conductors are free electrons and free ions respectively. Specific conductance or consuctivity of the electrolyte solution is given by the following relation : kappa=c xx l/A where, c=1//R is the conductance and l//A is the cell constant. Molar conductance (Lambda_(m)) and equivalence conductance (Lambda_(e)) of an electrolyte solution are calculated using the following similar relations : Lambda_(m)= kappa xx 1000/M Lambda_(e)= kappa xx 1000/N Where, M and N are the molarity and normality of the solution respectively. Molar conductance of strong electrolyte depends on concentration : Lambda_(m)=Lambda_(m)^(@)-b sqrt(c) where, Lambda_(m)^(@)= molar conductance at infinite dilution c= concentration of the solution b= constant The degrees of dissociation of weak electrolytes are calculated as : alpha=Lambda_(m)/Lambda_(m)^(@)=Lambda_(e)/Lambda_(e)^(@) For which of the following electrolytic solution Lambda_(m) and Lambda_(e) are equal ?

Why is it not possible to determine Lambda_(m)^(oo) for weak electrolytes graphically ? Explain.

Conductors allow the passage of electric current through them. Metallic and electrolytic are the two type of conductors. Current carriers in metallic and electrolytic conductors are free electrons and free ions respectively. Specific conductance or consuctivity of the electrolyte solution is given by the following relation : kappa=c xx l/A where, c=1//R is the conductance and l//A is the cell constant. Molar conductance (Lambda_(m)) and equivalence conductance (Lambda_(e)) of an electrolyte solution are calculated using the following similar relations : Lambda_(m)= kappa xx 1000/M Lambda_(e)= kappa xx 1000/N Where, M and N are the molarity and normality of the solution respectively. Molar conductance of strong electrolyte depends on concentration : Lambda_(m)=Lambda_(m)^(@)-b sqrt(c) where, Lambda_(m)^(@)= molar conductance at infinite dilution c= concentration of the solution b= constant The degrees of dissociation of weak electrolytes are calculated as : alpha=Lambda_(m)/Lambda_(m)^(@)=Lambda_(e)/Lambda_(e)^(@) The conductance of a solution of an electrolyte is equal to that of its specific conductance. The cell constant of the conductivity cell is equal to :

Conductors allow the passage of electric current through them. Metallic and electrolytic are the two type of conductors. Current carriers in metallic and electrolytic conductors are free electrons and free ions respectively. Specific conductance or consuctivity of the electrolyte solution is given by the following relation : kappa=c xx l/A where, c=1//R is the conductance and l//A is the cell constant. Molar conductance (Lambda_(m)) and equivalence conductance (Lambda_(e)) of an electrolyte solution are calculated using the following similar relations : Lambda_(m)= kappa xx 1000/M Lambda_(e)= kappa xx 1000/N Where, M and N are the molarity and normality of the solution respectively. Molar conductance of strong electrolyte depends on concentration : Lambda_(m)=Lambda_(m)^(@)-b sqrt(c) where, Lambda_(m)^(@)= molar conductance at infinite dilution c= concentration of the solution b= constant The degrees of dissociation of weak electrolytes are calculated as : alpha=Lambda_(m)/Lambda_(m)^(@)=Lambda_(e)/Lambda_(e)^(@) Which of the following equality holds good for the strong electrolytes ?

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