Write a note on molar conductivity `(Lamda_(m))` of solution.

Explanation

* The conductivity of solutions of different electrolytes in the same solvent and ata given temperature differs due to charge and size of the ions in which they dissociate, the concentration of ions or ease with which the ions move under a potential gradient. it, therefore becomes necessary to define a physically more meaningful quantity called molar conductivity.
* It is denoted by the symbol `Lamda_(m)` (Greek, lambda).
* It is related to the conductivity of the solution by the equation:
`(Lamda_(m)=(k)/(c))" S "m^(2)mol^(-1) or Lamda_(m)=(1000kappa)/(c)`
Where, `kappa`=conductivity of solution (its unit S `m^(-1)`)
c=concentration of solution (its unit is mol `m^(-3)`)
`Lamda_(m)`=molar conductivity of solution
`therefore Lamda_(m)=(k(S" "m^(-1)))/(c("mol "m^(-3)))=(k)/(c)S" "m^(2)mol^(-1)`
So unit of `Lamda_(m)` is S `m^(2)mol^(-1)`
But 1 mol `m^(-)=1000(L//m^(3))=` molarity (mol/L)
`therefore Lamda_(m)(S" "cm^(2)mol^(-1))=(k(S" "m^(-1)))/((1000Lm^(-3))xx"molarity "(mol//L^(-1)))`
* If unit of `kappa` is S `cm^(-1)` and concentration is mol `m^(-3)` then, unit of `Lamda_(m)` will be S `cm^(2)mol^(-1)`.
So calculation will be:
`therefore Lamda_(m)=(kappa(S" "cm^(-1))xx1000(cm^(3)//L))/("molarity"(mol//L))`
`therefore Lamda_(m)=(k)/(c)"S "cm^(2)mol^(-1) or Omega^(-1)cm^(2)mol^(-1)`
Thus, 1 S `m^(2)mol^(-1)=10^(4)S" "cm^(2)mol^(-1)`
or 1 S `cm^(2)mol^(-1)=10^(-4)S" "m^(-2)mol^(-1)`