Question

What is the formula for equivalent conductivity and how is it different from the molar conductivity formula?

Answer 1

Hint :Equivalent conductivity describes the conductance shown by that volume of solution which contains one gram equivalent of electrolyte. On the other hand, molar conductivity describes the conductance shown by that volume of solution which contains one mole of electrolyte. Equivalent factor of the electrolyte generally describes the total charge existing either on anion or cation in one formula unit.

Complete Step By Step Answer:
Equivalent conductivity describes the conductance shown by that volume of solution $\left(Vc{m}^3\right)$ which contains one equivalent of electrolyte and is given by the following formula.
 $\mathrm{\Lambda }=\kappa \times V----\left(1\right)$
Where $\mathrm{\Lambda }$ and $\kappa$ denote equivalent conductivity and specific conductance, respectively.
Since normality defines the number of gram equivalents of the electrolyte present per liter of solution and is represented by the following formula:
 $N={\displaystyle \frac{Eq}{V}}\times 1000------\left(2\right)$
Where $Eq$ represents the number of gram equivalents of electrolyte.
Normality for one gram equivalent of the electrolyte is given as
 $N={\displaystyle \frac{1000}{V}}$
 $\Rightarrow V={\displaystyle \frac{1000}{N}}$
Put this value in equation (1), then we get
 $\mathrm{\Lambda }=\kappa \times {\displaystyle \frac{1000}{N}}-----\left(3\right)$
The above equation represents the formula for equivalent conductivity.
On the similar lines, molar conductivity is represented by the following formula:
 $\mathrm{\Lambda }m=\kappa \times {\displaystyle \frac{1000}{M}}-----\left(4\right)$
Where $\mathrm{\Lambda }m$ and $M$ denote molar conductivity and molarity of the electrolytic solution, respectively.
By considering the definition of normality and molarity into account, the relationship between them for the electrolytic solution is given as
 $M={\displaystyle \frac{N}{equivalentfactoroftheelectrolyte}}$
Put this value in equation (4), then we get
  $\mathrm{\Lambda }m=\kappa \times {\displaystyle \frac{1000}{N}}\times equivalentfactoroftheelectrolyte$
Put the value of equivalent conductivity in the above equation, then we get
 $\mathrm{\Lambda }m=\mathrm{\Lambda }\times equivalentfactoroftheelectrolyte$
The above equation describes the difference or relation between molar conductivity and equivalent conductivity. Therefore, we conclude that multiplication of equivalent conductivity with an equivalent factor of the electrolyte gives the molar conductivity.

Note :
It is important to note that the multiplication of equivalent conductivity with an equivalent factor of the electrolyte gives the molar conductivity. Similarly, multiplication of normality with an equivalent factor of the electrolyte gives the molarity of the electrolytic solution. Equivalent factor of the electrolyte is determined by evaluating the total charge existing either on anion or cation in one formula unit.