Obtain a relation between conductivity ( k ) and molar conductivi… - Vidyadip

Question

Obtain a relation between conductivity $( k )$ and molar conductivity $\left(\wedge_m\right.$ ).

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Answer

Conductivity or specific conductance $( k )$ is the conductance of $1 cm ^3$ of the solution in C.G.S. units, while molar conductivity is the conductance of a solution containing one mole of an electrolyte. Consider C molar solution, i.e., C moles of an electrolyte present in 1 litre or $1000 cm ^3$ of the solution.
$\therefore$ C moles of an electrolyte are present in $1000 cm ^3$ solution.
$\therefore 1$ mole of an electrolyte is present in $\frac{1000}{ C } cm$
solution.
Now,
$\therefore$ Conductance of $1 cm ^3$ of this solution is $K$,
$\therefore$ Conductance of $\frac{1000}{ C } cm ^3$ of the solution is $\frac{\kappa \times 1000}{C}$
This represents molar conductivity, $\wedge_{ m }$.
$\therefore \wedge_{ m }=\frac{\kappa \times 1000}{C} cm ^2 mol ^{-1}$ (in C.G.S units)
[In case of SI units :
Consider a solution in which $C$ moles of an electrolyte are present in $1 m ^3$ of solution.
Conductivity $k$ is the conductance of $1 m ^3$ of solution.
$\because$ C moles of an electrolyte are present in $1 m ^3$ solution.
$\therefore 1$ mol of an electrolyte is present in $\frac{1}{C}$ solution.
$\because$ Conductance of $1 m ^3$ of this solution is $K$.
$\therefore$ Conductance of $\frac{1}{C} m ^3$ of the solution is $\frac{\kappa}{ C }$
This represents molar conductivity, $\wedge_{ m }$.
$\therefore \Lambda_{ m }=\frac{\kappa}{ C } \Omega^{-1} m ^2 mol ^{-1} \text { (In SI units).] }$

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