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The equivalent conductance (Lambda(eq.))...

The equivalent conductance `(Lambda_(eq.))` is given by the relation

A

`Lambda_(eq.) = 1000(kappa)/(N)`

B

`Lambda_(eq.) = 1000(N)/(kappa)`

C

`Lambda_(eq.) = 1000kappa//N`

D

`Lambda_(eq.) = 1000(kappa)/(N)^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
The magnitude of the conductance, (i.e. the reciprocal of resistance) of an electrolyte dependds mainly on three facors: the number of ions, mafnitude of chanfe on each ion and the ionic mobility (or absolute ionic velocity, the distance travelled by an ion per second under a potential graident of one volt per `cm`. The conductance of an electrolyte may be measured either in terms of molar conductance or equivalent conductance. But to compare the conductance of two solutions, equivalent conductance (now called equivalent of different electrolutes involves the same number of electrons (i.e. Avogadro constant of electrons) in accordance with Faraday's second law of electrolysis while one mole of different electrolytes may or may not involve the same number of electrons.
In other words, the solutions, each containing one equivalent of different electrolytes, are equivalent in terms of moles of electrons being carried.
`Lambda_(eq.) = kappa(Scm^(-1))/(N(eq. L^(-1))) xx 1000(cm^(3))/(L)`
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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)^(@) For which of the following electrolytic solution Lambda_(m) and Lambda_(e) are equal ?

Knowledge Check

  • The unit of equivalent conductivity (Lambda_(eq.)) are

    A
    `ohm^(-1)cm^(2)eq^(-1)`
    B
    `ohm^(-2)cm^(2)`
    C
    `ohm^(-1)cm^(-1)`
    D
    `ohm^(-1)cm^(-2)`

  • The specific conductance (kappa) of an electrolyte of 0.1 N concentration is related to equivalent conductance (Lambda_(e)) by the following formula :

    A
    `Lambda_(e)= kappa`
    B
    `Lambda_(e)= 10kappa`
    C
    `Lambda_(e)= 100kappa`
    D
    `Lambda_(e)= 10000kappa`

  • 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 correct order of equivalent conductances at infinite dilution of LiCl, NaCl and KCl is :

    A
    `LiCl gt NaCl gt KCl`
    B
    `KCl gt NaCl gt LiCl`
    C
    `NaCl gt KCl gt LiCl`
    D
    `LiCl gt KCl gt NaCl`

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