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

To calculate `kappa`
`R=(1)/(kappa)(l)/(a)`
`kappa=(1)/(R)xx(l)/(a)`
Here `R=30Omega,I=1.5cm,and a=2*0cm^(2)`
`therefore kappa=(1)/(30Omega)xx(1*5cm)/(2cm^(2))=(1)/(40)Omega^(-1)cm^(-1)`
`=0*025Omega^(-1)cm^(-1)`
Step II. To calculate `lamda_(m)`
`lamda_(m)=(kappaxx1000cm^(3)L^(-1))/(C ("Molarity"))`
Here molarity
`C=0*5M=0*5mol" "L^(-1)`
`therefore lamda_(m)=((0*025Omega^(-1)cm^(-1))xx(1000cm^(3)L^(-1)))/(0*5mol" "L^(-1))`
`=50Omega^(-1)cm^(2)mol^(-1)`