If \[k\]is the conductivity, \[G\]is the conductance and \[l\]is the distance between 2 electrodes of a conductivity cell then, the molar conductivity is defined as:
A.\[\;Gl\]
B.\[kl\]
C.\[G{l^2}\]
D.\[k{l^2}\]

Hint: Molar conductivity is conductance of solution containing 1 mole of electrolyte and numerically it becomes equal to specific conductivity divided by concentration of the solution. After this, replacing the parameters by different relations it will lead directly to the answer.

Complete Step by step Solution:
In the given question, molar conductivity is asked and its relation numerically with given parameters in the question. Now, the molar conductivity is defined as conductance of the solution containing 1 mole of the given electrolyte. In general, the molar conductivity of a solution becomes the conductance of the volume of solution containing unit mole of electrolyte that is placed between 2 electrodes a centi-meter apart and with unit cross section area.
Molar conductivity is represented as ${ \wedge _m}$
Molar conductivity $({ \wedge _m}) = \dfrac{k}{C}$ where k = specific conductivity and C = concentration of the solution.
Now we know the relation i.e. \[R{\text{ }} = {\text{ }}\rho {\text{ }}\dfrac{l}{A}\]
Inverting this relation, we get \[\dfrac{1}{R} = \dfrac{1}{\rho } \times \dfrac{A}{l}\]
Or also we can write \[G{\text{ }} = {\text{ }}k{\text{ }}\dfrac{A}{l}\] ,Because inverse of Resistance is G
From this we can write \[k{\text{ }} = {\text{ }}G{\text{ }}\dfrac{l}{A}\]
Also, we know that Concentration = number of moles present in volume of solution
\[C{\text{ }} = {\text{ }}\dfrac{n}{V}\]
Volume can be written as product of area and length i.e. \[V{\text{ }} = {\text{ }}A \times l\]
So, putting all these values in molar conductivity equation (when 1 Molar solution is there i.e. n = 1) we get
${ \wedge _m} = \dfrac{{G \times (\dfrac{l}{A})}}{C} = \dfrac{{G \times l}}{A} \times (\dfrac{V}{n}) = \dfrac{{G \times l \times A \times l}}{{A \times n}} = G{l^2}$
So, we see that Molar conductivity is related to conductance as this relation i.e. ${ \wedge _m} = G{l^2}$

So, the correct answer is C i.e. \[G{l^2}\]

Note: There is also another conductivity called specific equivalent conductivity and which is as similar as molar and the only difference in equivalent conductivity, the solution is taken as 1 gram equivalent and also often students get confused between the units of conductivity, conductance, specific conductance, resistivity etc. So, all these should be well cleared to answer similar types of questions easily.