The reciprocal of resistivity is known as specific conductance orsimply conductivity. It is denoted by k (kappa). Thus, if k is the specificconductance and G is the conductance of the solution, then $=R= \frac{1}{G}$ and $\rho = \frac{1}{\kappa}$ $\therefore\;\frac{1}{G} = \frac{1}{\kappa} \times \frac{l}{A},$ $\kappa = G \times \frac{l}{A}$ Now, if $l = 1$cm and $A = 1\text{sq.cm}$, then $\kappa$ = G. Hence, conductivity of a solution is defined as the conductance of asolution of 1 cm length and having 1 sq. cm as the area of cross-section.Alternatively,it may be defined as conductance of one centimetre cube ofthe solution of the electrolyte. Molar conductivity of a solution at a dilution V is the conductance of allthe ions produced from 1 mole of the electrolyte dissolved in V cm$^{3}$ofthe solutionwhen the electrodes are one cm apart and the area of theelectrodes is so large that the whole of the solution is contained betweenthem. It is represented by $\Delta_{m}$. $\lambda_{m} = \frac{\kappa A}{l}$ Since $l = 1$ cm and $A = V$ (volume containing 1 mole ofelectrolyte) $\lambda_{m} = \kappa V$Variation of conductivityand molar conductivity withconcentration : Conductivityalways decreases withdecrease in concentration,for both weak and strongelectrolytes. This is becausethe number of ionsper unitvolume that carry the currentin a solution decreases ondilution.Molar conductivity increases with decrease in concentration. This isbecause that total volume, V, of solution containing one mole of electrolytealso increases. It has been foundthat decrease in k on dilution of a solutionis more than compensated by increase in its volume.