The resistance of 0.5 N solution of an electrolyte in a conductivity cell was found to be 25 ohm. Calculate the equivalent conductivity of the solution if the electrodes in the cell are 1.6 cm apart and have an area of `3.2 cm^(2)`

Answer

`40 ohm^(-1) cm^(2) "equiv"^(-1)`

Explanation

Cell constant = `(l)/(a) = (1.6)/(3.2) = 0.5 cm^(-1)`
`kappa = (1)/(25) xx 0.5 = 2.0 xx 10^(-2) S cm^(-1)`
`Lambda = (2.0 xx 10^(-2) xx 1000)/(0.5) = 40 S cm^(2) "equiv"^(-1)`