The resistance of a 0.5 M solution of an electrolyte enclosed between two platinum electrodes 1.5 cm apart and having an area of `2*0cm^(3)` was found to be 30 ohm. Calculate the molar conductivity of the electrolyte.

Explanation

Given M= 0.5M, l= 1.5cm, `a=2 cm^(2), R= 30` Ohm, so `C= (1)/(R )= (1)/(30)` Ohm
`K= C xx (l)/(a) = (1)/(30) xx (1.5)/(2)`
Now Molar conductivity `Lamda_(m)= (K xx 1000)/(M)= (1)/(30) xx (1.5)/(2) xx (1000)/(2.5)= 50 "Ohm"^(-1) cm^(2) mol^(-1)`