The resistance of 0.5N solution of an electrolyte ina conductivity cell was found to be `25Omega`. Calculate the equivalent conductivity of the solution if the electrodes in the cell are 1.6cm apart & having an area of cross section `3.2cm^(2)`

Explanation

`R = rho (l)/(a)`
`rho= (R_(a))/(l)= (75 xx 3.6)/(1.8)= 150Omega cm`
`k= (1)/(rho) = (1)/(150)= 0.0067 Omega^(-1) cm^(-1)`
`k_(m)= (k xx 1000)/(C )= (0.0067 xx 1000)/(0.25)`
`=26.8 cm^(2) mol^(-1)`
`R= rho (l)/(a)`
`rho = (R_(rho))/(l)=(25 xx 3.2)/(1.6)= 50`
`k= (1)/(50)= 0.02 r^(-1) cm^(-1)`
`Lamda_(eq)= (k xx 1000)/(C_(eq))`
`=(0.02 xx 1000)/(0.5) = 40S cm^(2) ("gap")^(-1)`