The cell constant of a conductivity cell is given by

A conductivity cell has two platinum electrodes of area $1.2c{m}^{2}1.2\; \backslash \; cm^2$ and 0.92 cm apart. Hence the cell constant is

The conductivity of 0.02 M KI solution is $4.37\times 10^{-4}{\mathrm{\Omega }}^{-1}c{m}^{-1}\mathrm{.}4.37\; \backslash times\; 10\; ^\{-4\}\; \backslash Omega\; ^\{-1\}\; \backslash ,\; cm\; ^\{-1\}.$ Hence its molar conductivity is :

The specific conductance of 0.02 M HCl is $8.2\times 10^{-3}{\mathrm{\Omega }}^{-1}c{m}^{-1}\mathrm{.}8.2\; \backslash times\; 10\; ^\{-3\}\; \backslash ,\; \backslash Omega\; ^\{-1\}\backslash ,\; cm\; ^\{-1\}.$ Hence its molar conductivity is :

Molar conductivity of an electrolyte is given by,

The units of molar conductivity are

If conductivity is expressed in ${\mathrm{\Omega }}^{-1}{m}^{-1}\backslash Omega^\{-1\}\; \backslash \; m^\{-1\}$ and concentration of the electrolytic solution in mol $m^{-3}m^\{-3\}$ then, the molar conductance is given by

Kohlrausch’s law is represented as

The degree of dissociation of a weak electrolyte is given by

${\wedge }_0\backslash wedge\_0$ for $C{H}_{3}COOHCH\_3\; COOH$ is $390.7{\mathrm{\Omega }}^{-1}c{m}^{2}mo{l}^{-1}390.7\; \backslash \; \backslash Omega^\{-1\}\; \backslash \; cm^2\; \backslash \; mol^\{-1\}$ . If ${\wedge }_0\backslash wedge\_0$ for $C{H}_{3}COOKCH\_3\; COOK$, and HBr in ${\mathrm{\Omega }}^{-1}c{m}^{2}mo{l}^{-1}\backslash Omega^\{-1\}\; \backslash \; cm^2\; \backslash \; mol^\{-1\}$ are 115 and 430.4 respectively, then ${\wedge }_0\backslash wedge\_0$ for KBr is